Skip to main content
$$\require{cancel}$$

# 01. Concepts and Principles

• • Contributed by Paul D'Alessandris
• Professor (Engineering Science and Physics) at Monroe Community College

This project was supported, in part, by the National Science Foundation. Opinions expressed are those of the author and not necessarily those of the Foundation. Model Specifics We are now ready to consider how the actual size and shape of an object affects its motion. The most immediate ramification of this Model is that the objects we study can now rotate. However, there are still three important restrictions to the types of motions will we investigate. The object is rigid. The size and shape of the object under investigation do not change during the motion. The rotation is planar. This restriction may take some explaining. Imagine an object spinning in place. The axis (real or imaginary) about which the object spins is referred to as the rotation axis. Note that all points that lie on the rotation axis do not move. All other points on the object, however, exhibit circular motion around the rotation axis. If the object, in addition to spinning, is also moving, this model will restrict us to motions where the center-of-mass moves exclusively in the plane perpendicular to the rotation axis. I will refer to this as planar rotation. For example, a yo-yo is an object that typically moves in a plane perpendicular to its rotation axis. Thus, we can study the motion of most yo-yos. The motion of a wheel is typically in a plane perpendicular to its rotation axis. Thus, we can study the motion of most wheels. The motion of the earth around the sun, however, is beyond this model’s capabilities since the rotation axis of the earth is tilted relative to the plane of the earth’s motion. A wobbling top is also beyond this model’s capabilities (although a steadily spinning top is not). The object is classical. Kinematics Concepts and Principles Rotation about a Fixed Axis (Spinning) Imagine a rigid body constrained to rotate about a fixed axis. Non-physicists would say that the object is spinning. Place the origin of a coordinate system at the location of the rotation axis. Examine an arbitrary point on the object, denoted by the position and Notice that as the object spins, this point undergoes circular motion (denoted by the dashed line). Although the actual object may be of irregular shape, as it spins every point on the object undergoes circular motion. Therefore, to understand the kinematics of spinning objects we first must understand the kinematics of circular motion, so let’s examine circular motion in detail. Defining the Angular Kinematic Variables The position of an object moving along a circular path of radius R (whether part of a spinning object or a separate point particle) can always be described by the following relations, and assuming the origin of the coordinate system is placed at the center of the circle.  specifies the angular position of the object, and is typically measured in radians. It specifies where along the circle the object is located. For example, if  is a constant, the object’s position along the circle is constant (the object doesn’t move). If  varies, the object’s position along the circle varies. The rate at which  is changing, , is termed the angular velocity of the object and denoted by , the Greek letter omega. Thus, Since  is the rate at which the angular position is changing, it will have units of radians/second, or rad/s. The rate at which  is changing, , is termed the angular acceleration of the object and denoted by  the Greek letter alpha. Thus, Since  is the rate at which the angular velocity is changing, it will have units of rad/s2. Notice that since angular position, angular velocity, and angular acceleration are defined with exactly the same inter-relationships as “regular” position, velocity, and acceleration, the kinematic equations involving  and  must follow exactly the same form as the kinematic equations involving r, v, and a. In other words, and must be valid relationships between the angular variables. Deriving the Relationships between Angular and Linear Kinematics Now that we have the inter-relationships between the angular quantities out of the way, let’s determine the relationships between the angular variables and the linear variables. At left is illustrated the position of the object (r1 and r2) at two instants separated by t, as well as the change in the position (r) between these times. The length of r is approximately equal to the arc length between r1 and r2, especially when  is very small. From geometry, the arc length is equal to the product of the radius of the circle and the enclosed angle. Thus, Since the change in position takes place over a time interval t, dividing both sides by t yields Thus, the magnitude of the velocity of an object undergoing circular motion is the product of the radius and the angular velocity. Also note that the direction of the velocity is tangent to the circular path of the object. We will refer to this direction as the ‘’ direction, since it is in the direction of increasing angular position, . In short, Now let’s determine a relationship for the acceleration of the particle. At left is illustrated the velocity of the object (v1 and v2) at two instants separated by t. Examining the change in the velocity allows us to construct the acceleration. Notice that the acceleration can be thought of as having two components: one directed radially inward toward the center of the circle (aR) and one directed tangent to the circular path (a). The ‘R’, or radial, direction will refer to the direction of increasing radius, i.e., away from the center of the circle. Thus aR will turn out to be negative since it points toward the center of the circle. Let’s look at the radial component of acceleration first. The radial component of the acceleration is due to the changing direction of the velocity. As the velocity changes direction, the arrow representing the velocity rotates. If the speed of the object is constant, the tip of this arrow traces a circular path. The magnitude of vR is approximately equal to the arc length between v1 and v2 (ignoring the change in length of v2), especially when  is very small. From geometry, the arc length is equal to the product of the “radius” of the circle, v, and the enclosed angle. Thus, Since the change in velocity takes place over a time interval t, dividing both sides by t yields Substituting, results in: As noted above, this acceleration points toward the center of the circular path, in the direction of decreasing radius, so our final result for the radial acceleration is: Now let’s examine the tangential component of the acceleration. The tangential component of the acceleration is due to the changing magnitude of the velocity. Ignoring the change in direction of the velocity (since we’ve taken care of that in computing aR), Since the change in velocity takes place over a time interval t, dividing both sides by t yields Thus, the acceleration of an object undergoing circular motion has two components. If the object is speeding up or slowing down,  does not equal zero and there is an acceleration component in the tangential direction. However, even if the object is moving at constant speed there is an acceleration component in the radial direction, pointing toward the center of the circle. By virtue of traveling in a circle, the velocity of an object continually changes its orientation and this change in orientation results in an acceleration toward the center of the circle. Everything described above is true for both a point on a spinning rigid body as well as for a particle undergoing circular motion. However, for a rigid body, since each and every point on the object has to complete an entire cycle around the rotation axis in the same amount of time, every point must undergo circular motion with the same angular speed, , and the same angular acceleration, . Since every point on a rigid body must have the same angular speed and the same angular acceleration, we will speak of the angular speed and angular acceleration of the object, rather than of each point on the object. Rotation and Translation To describe the motion of an object undergoing pure rotation (spinning), we have to describe the circular motion each point on the object undergoes. What must we do if the object is simultaneously rotating and translating (moving in a plane perpendicular to the rotation axis), like a wheel rolling down an incline? The answer lies in the independence of these two types of motion. In much the same manner as we attacked kinematics in two dimensions by independently analyzing the horizontal and vertical motions, we will attack rigid-body kinematics by independently analyzing the rotational and translational motions. In short, we will model any motion of a rigid-body as a superposition of a translation of the object’s center-of-mass (CM) (which we will analyze by particle kinematics) and a rotation about an axis passing through the CM (which we will analyze by the kinematics of spinning, detailed above). For example, examine the motion of the thin rod between t1 and t2. Although the rod may have been spinning crazily through space between these two times, we can model its motion as a superposition of a simple translation of its CM without rotation (denoted by the vector r), which leaves the rod in the orientation denoted by the dashed lines, and a simple rotation about an axis through its CM without translation (denoted by ), which leaves the rod in its proper, final orientation. If we imagine the time difference (t2 - t1) shrinking toward zero, hopefully it becomes plausible that we can model any motion through this method. In summary, to describe the motion of an arbitrary rigid body we will break the motion down into a pure translation of the CM and a pure rotation about the CM. We will use particle kinematics to describe the translational portion of the motion and the kinematics of circular motion to describe the rotational portion. The velocity (or acceleration) of any point on the object is then determined by the sum of the velocity (or acceleration) due to the translation and the velocity (or acceleration) due to the rotation. Kinematics Analysis Tools Pure Rotation After the off button is pressed, a ceiling fan takes 22 s to come to rest. During this time, it completes 18 complete revolutions. To analyze this situation, we should first carefully determine and define the sequence of events that take place. At each of these instants, let’s tabulate what we know about the motion. Since we are dealing with pure rotation, the relevant kinematic variables are the angular position, velocity, and acceleration of the fan. Also, let’s take the direction that the fan is initially rotating to be the positive direction. Event 1: The ‘off’ button is pressed t1 = 0 s 1 = 0 rad 1 = 12 = Event 2: The ceiling fan stops t2 = 22 s 2 = 18 (2) = 113 rad 2 = 0 rad/s Using the angular versions of the kinematic equations yields: Now substitute this expression into the other equation: Substitute this result back into the original equation: Notice that the sign of the angular acceleration is negative. This indicates that the angular acceleration is in the opposite direction of the angular velocity, as it should be since the fan is slowing down. Also note that every point on the ceiling fan has the angular velocity and angular acceleration calculated above. Circular Motion An automobile enters a U-turn of constant radius of curvature 95 m. The car enters the U-turn traveling at 33 m/s north and exits at 22 m/s south. Since the car moves along a circular path, we will not break the position, velocity and acceleration vectors into x- and y-components. Rather, we will use polar coordinates, where we view these vectors as having either radial (R) or tangential () components. The position is solely in the radial direction (which is written as rR, read as “position in the radial direction”, or simply the radius of the circular path), the velocity is solely in the tangential direction (v), and the acceleration has components in both directions (aR and a). Additionally, we can tabulate the angular position, angular velocity, and angular acceleration of the car.

Event 1: The auto enters the turn. t1 = 0 s r1R = 95 m 1 = 0 rad v1 = 33 m/s 1 = a1R = a12 = 12 = Event 2: The auto exits the turn. t2 = r2R = 95 m 2 =  rad v2 = 22 m/s 2 = a2R = Notice that since the car completely reverses its direction of travel, it must have traveled halfway around a circular path. Thus, 2 =  rad.

The unknown variables listed above are not all independent. In fact, we know we can go back and forth between position, velocity, and acceleration in polar coordinates and angular position, angular velocity, and angular acceleration using the relationships we derived above. Using these relationships, we can quickly determine the angular velocity of the car at the two events: and the radial acceleration of the car: All that’s left to determine is the time it takes for the car to complete the u-turn and the tangential acceleration and angular acceleration it has as it performs this turn. Since there are only two variables in the set of angular quantities, we can use the two kinematic equations, written in terms of angular quantities, to finish our analysis.

Now substitute this expression into the other equation: Substitute this result back into the original equation: The tangential acceleration can be determined from: Thus, the car has acceleration sin both the radial and tangential directions can it goes around the u-turn. The tangential acceleration corresponds to the rate at which the speed of the car decreases, and is due to the driver stepping on the brakes as she goes around the turn. The radial acceleration is due to the change in direction of the car as it goes around the turn, and is ultimately due to the frictional forces acting on the tires of the car which prevent it from sliding off the road. When the car initially enters the turn, this acceleration is quite large and the required frictional force to stay on the road is large. As the car slows, this acceleration decreases as the required frictional force to stay on the road decreases. Connecting Pure Rotation to Pure Translation The device at right is used to lift a heavy load. The free rope is attached to a truck which accelerates from rest at a rate of 1.5 m/s2. The inner radius of the pulley is 20 cm and the outer radius is 40 cm. The load must be raised 15 m. The coordinate system chosen indicates that the block moving upward, the pulley rotating clockwise, and the truck moving to the right are all positive. There are three different objects that we should be able to describe kinematically; the truck, the pulley, and the block. Let’s tabulate everything we know about each object: Truck Event 1: Truck begins to move t1 = 0 s r1 = 0 m v1 = 0 m/s a12 = 1.5 m/s2 Event 2: Block raised 15 m t2 = r2 = v2 = Pulley Event 1: Truck begins to move t1 = 0 s 1 = 0 rad 1 = 0 rad/s 12 = Event 2: Block raised 15 m t2 = 2 = 2 = Block Event 1: Truck begins to move t1 = 0 s r1 = 0 m v1 = 0 m/s a12 = Event 2: Block raised 15 m t2 = r2 = 15 m v2 = The truck and block exhibit translational motion, so we only have to tabulate linear variables. The pulley spins, so the relevant variables are the angular variables. Notice that just because the block moves 15 m doesn’t mean the truck moves 15 m. Also, the acceleration of the block is not equal to the acceleration of the truck. However, these variables are related to each other since both objects are attached to the same pulley. Also notice that we can’t currently solve this problem. Each of the objects has three unknown quantities. However, since the kinematics of the three objects are related, we will be able to solve the problem once we’ve worked out the exact relationship between each object’s kinematics. Let’s start with the acceleration. Assuming the rope from the truck does not slip on the pulley, the point on the pulley in contact with the rope must be accelerating at the same rate as the truck. Notice that this acceleration is tangent to the pulley, and this rope is located 0.4 m from the center of the pulley. Therefore, from The pulley must have an angular acceleration of 3.75 rad/s2 since it is attached to the truck. In addition, assuming the rope from the block does not slip on the pulley, the point on the pulley in contact with this rope must be accelerating at: Since the point on the pulley attached to the block is accelerating at 0.75 m/s2, the block itself must be accelerating at 0.75 m/s2. In a nutshell, what we’ve done is used the acceleration of the truck to find the angular acceleration of the pulley, and then used the angular acceleration of the pulley to find the acceleration of the block. This chain of reasoning is very common. Now that we know the block’s acceleration, we can solve the problem. Applying the two kinematic equations to the block yields: Now that we know the final speed of the block, we can find the final angular speed of the pulley and the final speed of the truck: Again assuming the rope from the block does not slip on the pulley, the point on the pulley in contact with the rope must be moving at the same rate as the block. Notice that this velocity is tangent to the pulley, and this rope is located 0.2 m from the center of the pulley. Therefore, from The pulley must have an angular velocity of 23.7 rad/s since it is attached to the block. In addition, assuming the rope from the truck does not slip on the pulley, the point on the pulley in contact with this rope must be moving at: The truck is moving at 9.48 m/s when the block reaches 15 m. We can relate the displacement of the block to the angular displacement of the pulley and the displacement of the truck using where r can be thought of as either the displacement of the object attached to the rope or the arc length over which the rope is wrapped around the pulley (since the amount of rope wrapped around a pulley is exactly equal to the displacement of the object attached to the rope), and  is the angular displacement of the pulley. Therefore, relating the block to the pulley yields The pulley must have turned through 75 rad since it is attached to the block. Relating the pulley to the truck results in: The truck moved 30 m while the block moved 15 m. Notice that in all cases, the values of the kinematic variables for the truck are exactly twice the values for the block. This is not a coincidence. Since the truck is attached to the pulley at twice the distance from the axle that the block is attached, all of its kinematic variables will have twice the value. Once you understand why this is the case, you can use this insight to simplify your analysis. Rotating and Translating Accelerating from rest, a Cadillac Sedan de Ville can reach a speed of 25 m/s in a time of 6.2 s. During this acceleration, the Cadillac’s tires do not slip in their contact with the road. The diameter of the Cadillac’s tires is 0.80 m. Let’s examine the motion of one of the Cadillac’s tires. Obviously, the tire both translates and rotates. We will imagine the motion to be a superposition of a pure translation of the CM of the tire and a pure rotation about the CM of the tire. To analyze the situation, we should define the sequence of events that take place and tabulate what we know about the motion at each event. Since we are dealing with translation as well as rotation, we will need to keep track of both the linear kinematic variables and the angular kinematic variables. Let’s take the positive x direction to be the direction that the Cadillac translates and the positive  direction to be the direction in which the tires rotate. Event 1: The Cadillac begins to accelerate t1 = 0 s r1 = 0 m 1 = 0 rad v1 = 0 m/s 1 = 0 rad/s a12 = 12 = Event 2: The Cadillac reaches 25 m/s t2 = 6.2 s r2 = 2 = v2 = 25 m/s 2 = First, let’s examine the translational portion of the tire’s motion. What about the rotational kinematics of the tire? The motion of the tire is the superposition of the pure translational motion of the CM and the pure rotational motion about the CM . Thus, the velocity of any point on the tire due to rotation is given by where R is the distance from the rotation axis, i.e., the distance from the CM. The key insight into studying the rotation of the tire is to realize that at any instant the velocity of the point on the tire in contact with the road is zero because the tire never slips in its contact with the road. Thus, if the CM of the tire is moving forward at 25 m/s, the point on the bottom of the tire must be moving backward relative to the CM at the exact same speed in order for its velocity relative to the ground to be zero! Thus, only a very particular value for  will allow the tire to roll without slipping. From above, Therefore, at the end of the acceleration the tire has an angular speed of 62.5 rad/s. We can find the angular acceleration and angular position of the tire by using the linear variables determined above and “translating” those values into angular variables, or by using the angular versions of the kinematic equations. Thus, the wheel rotates through 194 rad, or 30.8 revolutions, while accelerating. Kinematics Activities Construct motion diagrams for the motions described below. a. A satellite has been programmed to circle a stationary space station at a radius of 10 km and a constant angular speed of 0.02 rad/s. b. A rider on a merry-go-round, 3 m from the axis, is traveling at 4 m/s. The merry-go-round slows, and the rider reaches a speed of 0.5 m/s in 11 seconds. c. In a device built to acclimate astronauts to large accelerations, astronauts are strapped into a pod that is swung in a 6 m radius circle at high speed. The linear speed of the pod is increased from rest to a speed of 17 m/s in a time interval of 25 seconds. A car drives around a semi-circular U-turn at constant speed. Set the origin of the coordinate system at the center-of-curvature of the U-turn. a. Using polar coordinates, sketch position, velocity, and acceleration vs. time graphs for this motion. b. Sketch angular position, angular velocity, and angular acceleration vs. time graphs for this motion. A car drives around a semi-circular U-turn at decreasing speed. Set the origin of the coordinate system at the center-of-curvature of the U-turn. a. Using polar coordinates, sketch position, velocity, and acceleration vs. time graphs for this motion. b. Sketch angular position, angular velocity, and angular acceleration vs. time graphs for this motion. A girl pedals her bicycle at constant speed along a level path. Examine the motion of a point on the rim of her tire and a point midway along a spoke for one complete cycle. Set the origin of the coordinate system at the center of the tire. a. Using polar coordinates, sketch position, velocity, and acceleration vs. time graphs for each point. b. Sketch angular position, angular velocity, and angular acceleration vs. time graphs for each point. A girl pedals her bicycle, from rest, at a constantly increasing speed along a level path. Examine the motion of a point on the rim of her tire and a point midway along a spoke for one complete cycle, starting from rest. Set the origin of the coordinate system at the center of the tire. a. Using polar coordinates, sketch position, velocity, and acceleration vs. time graphs for each point. b. Sketch angular position, angular velocity, and angular acceleration vs. time graphs for each point. A block is attached to a rope that is wound around a pulley. Sketch velocity vs. time and acceleration vs. time graphs for a point on the rim of the pulley and for the block. Set the origin of the coordinate system at the center of the pulley. a. Sketch the polar components of the point on the rim’s motion and the Cartesian components of the block’s motion when the block is lowered at constant speed. b. Sketch the polar components of the point on the rim’s motion and the Cartesian components of the block’s motion when the block is released from rest. The two wheels at right are coupled together by an elastic conveyor belt that does not slip on either wheel. Examine the motion of a point on the rim of the large wheel and a point on the rim of the small wheel when the conveyor belt’s speed increases from rest. Set the origin of a coordinate system at the center of each pulley. a. Using polar components, sketch velocity and acceleration vs. time graphs for each point. b. Sketch angular velocity and angular acceleration vs. time graphs for each point. The two gears at right have equal teeth spacing and are well-meshed. Examine the motion of a point on the rim of the large gear and a point on the rim of the small gear when the small gear’s speed increases from rest in the counterclockwise direction. Set the origin of a coordinate system at the center of each gear and let counterclockwise be the positive direction. a. Using polar components, sketch velocity and acceleration vs. time graphs for each point. b. Sketch angular velocity and angular acceleration vs. time graphs for each point. Below are angular position vs. time graphs for six different objects. a. Rank these graphs on the basis of the angular velocity of the object. Rank positive angular velocities as larger than negative angular velocities. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these graphs on the basis of the angular acceleration of the object. Rank positive angular accelerations as larger than negative angular accelerations. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Below are angular velocity vs. time graphs for six different objects. a. Rank these graphs on the basis of the angular acceleration of the object. Rank positive angular accelerations as larger than negative angular accelerations. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these graphs on the basis of the angular displacement of the object over the time interval shown. Rank positive angular displacements as larger than negative angular displacements. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: An artificial satellite circles a space station at constant speed. The satellite passes through the six labeled points. For all questions below, use the indicated coordinate system. a. Rank the x-velocity of the satellite at each of the labeled points. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. b. Rank the y-velocity of the satellite at each of the labeled points. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. c. Rank the tangential velocity of the satellite at each of the labeled points. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. d. Rank the angular velocity of the satellite at each of the labeled points. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Six artificial satellites of identical mass circle a space station with constant period T. The satellites are located a distance R from the space station. R T A 5000 m 160 hrs B 2500 m 40 hrs C 2500 m 80 hrs D 10000 m 160 hrs E 5000 m 120 hrs F 10000 m 80 hrs a. Rank these satellites on the basis of their angular velocity. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. b. Rank these satellites on the basis of their tangential velocity. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your rankings: Six artificial satellites of identical mass circle a space station at constant speed v. The satellites are located a distance R from the space station. R v A 5000 m 160 m/s B 2500 m 40 m/s C 2500 m 80 m/s D 10000 m 160 m/s E 5000 m 120 m/s F 10000 m 80 m/s a. Rank these satellites on the basis of their angular acceleration. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. b. Rank these satellites on the basis of their radial acceleration. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. c. Rank these satellites on the basis of their tangential acceleration. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your rankings: The pulley below has the outer radius and inner radius indicated. In all cases, the horizontal rope is pulled to the right at the same, constant speed. Router Rinner A 0.4 m 0.2 m B 0.4 m 0.3 m C 0.8 m 0.4 m D 0.6 m 0.5 m E 0.2 m 0.1 m F 0.6 m 0.2 m a. Rank these scenarios on the basis of the angular speed of the pulley. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these scenarios on the basis of the speed of the block. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The two wheels, with radii indicated below, are linked together by an elastic conveyor belt that does not slip on either wheel. In all cases, the small wheel is turning at the same, constant angular speed. Rlarge Rsmall A 0.4 m 0.2 m B 0.4 m 0.3 m C 0.8 m 0.4 m D 0.6 m 0.5 m E 0.2 m 0.1 m F 0.6 m 0.2 m a. Rank the scenarios on the basis of the speed of the belt. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank the scenarios on the basis of the angular speed of the large wheel. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The two gears, with radii indicated below, have equal teeth spacing and are well-meshed. In all cases, the large gear is turning at the same, constant angular speed. Rlarge Rsmall A 0.4 m 0.2 m B 0.4 m 0.3 m C 0.8 m 0.4 m D 0.6 m 0.5 m E 0.2 m 0.1 m F 0.6 m 0.2 m a. Rank the scenarios on the basis of the speed of the teeth on the small gear. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank the scenarios on the basis of the angular speed of the small gear. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A pulley system is illustrated below. The horizontal rope is pulled to the right at constant speed. Each letter designates a point on either the pulley or on one of the two ropes. Neither rope slips in its contact with the pulley. a. Rank the designated points on the basis of their speed. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank the designated points on the basis of the magnitude of their acceleration. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: At the instant shown, the pulley below has the outer radius, inner radius, angular velocity, and angular acceleration indicated. Positive angular quantities are counterclockwise. Router Rinner   A 0.4 m 0.2 m 10 rad/s 0 rad/s2 B 0.4 m 0.3 m 10 rad/s 1 rad/s2 C 0.8 m 0.4 m 10 rad/s 0 rad/s2 D 0.6 m 0.5 m 5 rad/s -1 rad/s2 E 0.2 m 0.1 m 20 rad/s -4 rad/s2 F 0.6 m 0.2 m 15 rad/s 2 rad/s2 a. Rank these scenarios on the basis of the speed of the block. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these scenarios on the basis of the magnitude of the acceleration of the block. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: At the instant shown, the pulley below has the outer radius, inner radius, angular velocity, and angular acceleration indicated. Positive angular quantities are counterclockwise. Router Rinner   A 0.4 m 0.2 m 10 rad/s 0 rad/s2 B 0.4 m 0.3 m 10 rad/s 0 rad/s2 C 0.8 m 0.4 m 10 rad/s 0 rad/s2 D 0.6 m 0.5 m 5 rad/s 0 rad/s2 E 0.2 m 0.1 m 20 rad/s 0 rad/s2 F 0.6 m 0.2 m 15 rad/s 0 rad/s2 a. Rank these scenarios on the basis of the magnitude of the acceleration of a point on the inner rim of the pulley. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these scenarios on the basis of the magnitude of the acceleration of the block. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: After being turned on, a record player (an antique device used to listen to music) reaches its rated angular speed of 45 rpm (4.71 rad/s) in 1.5 s. Motion Information Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Mathematical Analysis A clothes dryer spins clothes at an angular speed of 6.8 rad/s. Starting from rest, the drier reaches its operating speed with an average angular acceleration of 7.0 rad/s2. Motion Information Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Mathematical Analysis In a secret Las Vegas research laboratory, a roulette wheels is undergoing extensive testing. The wheel is spun at 25 rad/s. After 6.4 s, the wheel has rotated through 100 rad. Motion Information Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Mathematical Analysis A rider on a merry-go-round, 3.0 m from the axis, is traveling at 4.0 m/s. The merry-go-round slows, and the rider reaches a speed of 0.50 m/s in 11 seconds. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = A rider on a merry-go-round, 3.0 m from the axis, is traveling at 4.0 m/s. The merry-go-round slows to rest over three complete revolutions. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = A rider on a 15 m diameter Ferris Wheel is initially at rest. The angular speed of the Ferris Wheel is increased to 1.5 rad/s over a time interval of 3.5 s. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = An automobile enters a constant 90 m radius of curvature turn traveling at 25 m/s north and exits the curve traveling at 35 m/s east. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = An automobile enters a constant 90 m radius of curvature turn traveling at 25 m/s north and exits the curve traveling east. The car completes the turn in 5.4 s. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = In a device built to acclimate astronauts to large accelerations, astronauts are strapped into a pod that is swung in a 6.0 m radius circle at high speed. The angular speed of the pod is increased from rest to an angular speed of 2.8 rad/s in a time interval of 20 seconds. The device is then linearly slowed to rest over a time interval of 40 seconds. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = a23 = 23 = Event 3: t3 = r3R = 3 = v3 = 3 = a3R = In a device built to acclimate astronauts to large accelerations, astronauts are strapped into a pod that is swung in a 6.0 m radius circle at high speed. The linear speed of the pod is increased from rest to a speed of 17 m/s after three complete revolutions of the pod. The device is then linearly slowed to rest over a time interval of 35 seconds. Motion Information Mathematical Analysis Event 1: t1 = r1R = 1 = v1 = 1 = a1R = a12 = 12 = Event 2: t2 = r2R = 2 = v2 = 2 = a2R = a23 = 23 = Event 3: t3 = r3R = 3 = v3 = 3 = a3R = A diver rotating at approximately constant angular velocity completes four revolutions before hitting the water. She jumped vertically upward with an initial velocity of 5 m/s from a diving board 4 m above the water. Motion Information Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis A baton twirler throws a spinning baton directly upward. As it goes up and returns to the twirler's hand, it turns through four complete revolutions. The constant angular speed of the baton while in the air is 10 rad/s. Motion Information Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis A quarterback throws a pass that is a perfect spiral, spinning at 50 rad/s. The ball is thrown at 19 m/s at an angle of 350 above the horizontal. The ball leaves the quarterback's hand 2.0 m above the Astroturf and is caught just before it hits the turf. (The opposing coach thinks the ball was trapped. We are still waiting for the result of the challenge.) Motion Information Event 1: t1 = r1x = r1y = 1 = v1x = v1y = 1 = a12x = a12y = 12 = Event 2: t2 = r2x = r2y = 2 = v2x = v2y = 2 = Mathematical Analysis A car, with 0.75 m diameter tires, slows from 35 m/s to 15 m/s over a distance of 70 m. The car’s tires do not slip in their contact with the road. Motion Information Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis A bowling ball of diameter 21.6 cm is rolled down the alley at 4.7 m/s. The ball slows with an acceleration of 0.2 m/s2 until it strikes the pins. The pins are located 18.3 m from the release point of the ball. Motion Information Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis The man at right is trapped inside a section of large pipe. If that’s not bad enough, the pipe begins to roll from rest down a 35 m long, 180 incline! The pipe has a diameter of 1.2 m. The pipe (and very dizzy man) reach the bottom of the incline after 6.32 s. Motion Information Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis The unlucky man is falling at 20 m/s, 75 m above the crocodile-infested waters below! In an attempt to save him, the brake shoe is pressed against the spinning pulley. The action of the brake shoe gives the pulley an angular acceleration of 7.5 rad/s2. The man is saved! (Barely.) Motion Information Lucky Man Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Pulley Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Mathematical Analysis Tired of walking up the stairs, an engineering student designs an ingenious device for reaching her third floor dorm room. An block is attached to a rope that passes over the outer diameter of a 0.7 m outer diameter, disk-shaped compound pulley. The student holds a second rope, wrapped around the inner 0.35 m diameter of the pulley. When the block is released, the student is pulled up to her dorm room, 8 m off the ground, in 11.2 s. Motion Information Student Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Pulley Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Block Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Mathematical Analysis The device at right is used to lift a heavy load. The free rope is attached to a truck that, from rest, accelerates to the right with a constant acceleration. The block is lifted 25 m in 45 s. The inner and outer pulley diameters are 0.40 m and 0.90 m, respectively. Motion Information Truck Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Pulley Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Block Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Mathematical Analysis The device at right is used to lift a load quickly. The free rope is attached to a truck that, from rest, accelerates to the right with a constant acceleration. The block is lifted 25 m in 15 s. The inner and outer pulley diameters are 0.40 m and 0.90 m, respectively. Motion Information Truck Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Pulley Event 1: t1 = 1 = 1 = 12 = Event 2: t2 = 2 = 2 = Block Event 1: t1 = r1 = v1 = a12 = Event 2: t2 = r2 = v2 = Mathematical Analysis Dynamics Concepts and Principles To study the dynamics of an arbitrary rigid body we will break the motion down into a pure translation of the CM and a pure rotation about the CM. We will use particle dynamics, i.e., Newton’s second law applied to the CM of the object, to study the translational portion of the motion. The study of the rotational portion of the motion requires a pair of new concepts. We will “invent” these concepts through the use of an analogy with linear dynamics. In linear dynamics, Newton’s second law states that the linear acceleration of an object is proportional to the total force acting on the object and inversely proportional to the mass, or inertia, of the object It would seem plausible that the angular acceleration of an object would depend on analogous concepts in the same manner. We will replace the concept of force, often thought of as the push or pull applied to an object, with a quantity measuring the twist applied to an object. We will call this new quantity torque, symbolized . We will replace the concept of mass, the measure of the resistance of the object to changes in its linear velocity, with a quantity measuring the resistance of the object to changes in its angular velocity. We will call this new quantity rotational inertia, symbolized I. In summary, Before we go any further, however, let’s define these new concepts more clearly. Torque In simple English, torque measures the twist applied to an object. The question remains, however, how do we quantify twist? Let’s examine a common device used to generate twist, a wrench. The magnitude, location, and orientation of the force applied to the wrench by the person’s hand are indicated. Each of these three parameters effects the amount of twist the person delivers to the wrench (and therefore to the bolt). If you’ve turned many bolts in your life, two things about this person’s bolt-turning technique should grab you. First, why is this person applying the force at such a silly angle? She would generate much more twist if she applied the same magnitude force perpendicular to the wrench, rather than at an angle far from 900. Second, why is she not applying the force at the far edge of the wrench? She would generate far more twist if she applied the same magnitude force at the far edge of the wrench. If the preceding paragraph makes sense to you, you understand how to quantify torque. To maximize torque, you should: 1. Apply the force far from the axis of rotation (the bolt). 2. Apply the force perpendicular to the position vector between the axis of rotation and the force. 3. Apply a large magnitude force. Mathematically, this is summarized by: Note that this function has a maximum when r is large, F is large, and  = 900. Torque will also be assigned a direction, either clockwise or counterclockwise, depending upon the direction of the twist applied to the object. Rotational Inertia We have constructed a rotational analogy to Newton’s second law, Our next task is to better define what we mean by I, the rotational inertia. The rotational inertia is a measure of the resistance of the object to changes in its angular velocity. Imagine applying the same torque to two objects, initially at rest. After applying the torques for some set amount of time, measure the angular velocity of the objects. The object with the smaller angular velocity has the larger rotational inertia, because it has the larger resistance to angular acceleration. Since more massive objects are harder to get moving linearly, it seems plausible that rotational inertia should depend on the mass of the object. It also seems plausible that rotational inertia should depend on the shape of the object. For example, it would be easier to get the smaller dumbbell spinning than the larger dumbbell, even though they have the same total mass. Let’s try to get more quantitative. Examining we can see that the units of I must be the units of torque (N.m) divided by the units of angular acceleration (s-2). Remembering that a Newton is equivalent to kg.m.s-2 leads to the units of rotational inertia being kg.m2. Thus, rotational inertia must be the product of a mass and a distance squared. It seems plausible (there’s that word again!) that the distance that is squared in the relationship for rotational inertia is the distance from the rotation axis. The farther a piece of mass is from the rotation axis, the more difficult it is to give the object an angular acceleration. Enough with the “plausibilities”, let’s finally just define the rotational inertia to be: Imagine the object of interest divided into a large number of small chunks of mass, each with mass m. Each chunk of mass is a distance r from the rotation axis. If you take the product of the mass of each chunk and the distance of the chunk from the rotation axis, squared, and sum this quantity over all the chunks of mass, you have the rotational inertia of the object of interest. This summation has been done for a number of common objects and the results are tabulated in the next section. In summary, we now have quantitative relationships for measuring torque and rotational inertia, and the hypothesis that these two quantities are related in a manner analogous to Newton’s second law, Rotational Inertia of Common, Uniform Solids Hoop about axis through CM Cylinder about axis through CM Solid sphere about axis through CM Hollow sphere about axis through CM Thin rod about axis through CM Rectangular plate about axis through CM Dynamics Analysis Tools Applying Newton's Second Law to Circular Motion Investigate the situation below, in which an object travels on a circular path. (Note that this situation does not require the new concepts of torque and rotational inertia.) During a high speed auto chase in San Francisco, a 1745 kg police cruiser traveling at constant speed loses contact with the road as it goes over the crest of a hill with radius of curvature 80 m. A free-body diagram for the car at the instant it’s at the crest of the hill is sketched below. Since the car is traveling along a circular path, it’s best to analyze the situation using polar coordinates. Remember, in polar coordinates the radial direction points away from the center of the circular path. In general, you should apply Newton’s second law independently in the radial (R) and tangential () directions, although since the cruiser is moving at constant speed, there is no acceleration in the tangential direction and therefore no net force in the tangential direction. Rdirection Since the cruiser loses contact with the road, Froad = 0 N. and therefore For the car to lose contact with the road, it must be traveling at a speed of 28 m/s or greater when it reaches the crest of the hill. At lower speeds, the car would remain in contact with the road. Since it is often useful to directly relate the radial acceleration of an object to its velocity, let’s construct a direct relationship between these two variables: This relationship, although mathematically equivalent to , is in a more “useful” form. Another Circular Motion Scenario A 950 kg car, traveling at a constant 30 m/s, safely makes a left hand-turn with radius of curvature 75 m. First, let's draw a pair of free-body diagrams for the car, a side-view (on the left) and a rear-view (on the right) . The free-body diagram on the left is a side-view of the car. Notice that the upward direction is the y-direction, and the forward direction, tangent to the turn, is the direction. The free-body diagram on the right is a rear-view of the car. This is what you would see if you stood directly behind the car. Notice that the upward direction is still the y-direction, and the horizontal direction, perpendicular to the direction of travel and hence directed radially outward, is the R-direction. The frictional force indicted is perpendicular to the tread on the tire. This force causes the car to accelerate toward the center of the turn. Remember, if the car is going to travel along a circular path, it must have an acceleration directed toward the center of the circle. Something has to be supplying the force that creates this acceleration. This something is the static friction between the tire and the road that acts to prevent the car from sliding out of the turn. Since the car has no velocity in the radial direction, the frictional force that points in this direction must be static! Now that we have all that straightened out (maybe), let's apply Newton's Second Law. ydirection Rdirection Thus, to safely make this turn requires at least 11400 N of static friction. Using this value, I should be able to compute the minimum coefficient of friction necessary for the car to safely round this turn at this speed. Although this is a large value for the coefficient of static friction, it is an attainable value for a sports car with performance tires. Applying Newton's Second Law in Translational and Rotational Form - I Investigate the scenario described below. The robotic arm at right consists of a pair of hydraulic jacks, each attached 10 cm from the pin “elbow". The elbow is 15 cm from the extreme edge of the 40 kg, 0.90 m long “forearm”. The forearm and attached 90 kg load are held stationary at an angle of 20° below horizontal. Since the jacks primarily exert forces through extension, the front, or "biceps", jack is exerting negligible force. To study the dynamics of this situation, we will first need a free-body diagram of the forearm (and the attached load). The forces due to the jack, the rope, and gravity should not need much explanation. However, the forces due to the pin elbow may need some explanation. First, the free-body diagram drawn at left is of the forearm, not including the pin that serves as the elbow. Thus, the pin is external to the object of interest, and thus its interactions with the object are forces that must be indicated on the diagram. If you were to include the pin as part of the forearm, then the upper part of the arm would be external to the object of interest (pin plus forearm) and its interactions with the “pin plus forearm” would have to be indicated as forces on the diagram. Both of these approaches are completely valid. Second, the direction of the force of the pin on the forearm may not be obvious. Does the pin push straight down on the forearm, pull up on the elbow, or push at some unspecified angle? The only thing that’s clear is that this force is directed somewhere in the xy-plane. If the force is in the xy-plane, then it must have components along both the x- and y-axis (although one of these components may turn out to have a magnitude of zero). Thus, to handle the generality of the situation you should include both an x- and y-component for the force of the pin on the forearm. For convenience, I’ll draw these forces as pointing in the positive x and y directions. (If my guess is wrong, the mathematics will tell me.) Now that we have a free-body diagram, we can apply Newton’s second law, both the linear form (in the x- and y-directions) and the rotational form (in the -direction). x-direction y-direction Since the forearm is held stationary, both ax and ay are equal to zero. Also, Frope is equal to the force of gravity on the load, 882 N. Obviously, we need another equation in order to determine the two unknown forces. The obvious choice is Newton’s second law in rotational form. To use the rotational form of Newton’s second law, we must determine the torque due to each force. (Remember, the angle, , in the relation for torque is the angle between r (oriented along the forearm) and F.) Pin, Y Jack Gravity Rope Finally, let’s apply Newton’s second law in rotational form, paying careful attention to the algebraic sign of each torque. Note that our coordinate system indicates that all torques acting counterclockwise are positive. Therefore, all torques acting clockwise are negative. Since the forearm is held stationary,  is equal to zero. The jack must exert a force of 7796 N to hold the forearm stationary. Plugging this value into our y-equation yields: The pin must exert a force of 9070 N upwards, since the mathematics determined that the algebraic sign of FPin,y was positive. Applying Newton's Second Law in Translational and Rotational Form - II A 75 kg man is attached to a rope wrapped around a 35 kg disk-shaped pulley, with inner and outer diameters 0.60 m and 0.90 m, respectively. The man is initially at rest. The brake shoe is pressed against the pulley with a force of 500 N. The coefficient of friction between the brake shoe and the pulley is (0.9, 0.8). First, we have to determine whether the force applied to the brake shoe is sufficient to hold the man stationary. If it's not, we have to find the man's acceleration. To accomplish this, we need free-body diagrams for both the man and the pulley. We’ll use counterclockwise as the positive -direction and down as the positive y-direction. The pin, or axle, at the center of the pulley exerts forces in the x- and y-direction, although the exact directions may be unknown. We’ll just label these forces as Fpin,x and Fpin,y. The frictional force that acts on the right edge of the pulley acts in a way to prevent the man from falling (or, if he does fall, to slow down the man’s fall). To begin my analysis, I'll assume that the man is stationary, solve for the value of the frictional force required to keep him stationary, and then determine whether this frictional force is possible given the applied force of 500 N and the coefficient of static friction. Let’s apply Newton's second law to the man in the y-direction. (Remember, we are assuming he is not falling.) Now look at Newton's second law applied to the pulley in the -direction. Notice that all of the other forces acting on the pulley do not exert torques on the pulley. They are either located at r = 0 m or have angular orientations of  = 180°. Therefore, to hold the man stationary requires 490 N of friction. However, Therefore, the man must accelerate downward. Now that we know the man must fall, let's write the same two equations as before, although this time the accelerations (both angular for the pulley and linear for the man) are not zero. For the pulley: This assumes that the rotational inertia of a "compound" pulley, one with more than one location where a rope can be wrapped, is the same as a regular pulley, and that the outermost radius of the pulley determines the inertia. Since the man is falling, the frictional force is now kinetic, and Thus our “pulley” equation becomes: Examining Newton's second law for the man, We now have two equations with three variables. However, the angular acceleration of the pulley and the linear acceleration of the man are directly related. Since the rope that the man is attached to is wrapped 0.3 m from the center of the pulley, the man accelerates at the same rate as the tangential acceleration of a point on the pulley 0.3 m from the center. Thus, Using this relationship allows me to rewrite the two Newton's second law equations as: This pair of equations has the solution The man falls with this acceleration and the rope exerts this force (upward on the man and downward on the pulley). Dynamics Activities Six artificial satellites of mass, m, circle a space station at constant speed, v. m v A 200 kg 160 m/s B 100 kg 160 m/s C 400 kg 80 m/s D 800 kg 40 m/s E 200 kg 120 m/s F 300 kg 80 m/s a. If the distance between the space station and the satellites is the same for all satellites, rank the magnitude of the net force acting on each satellite. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. If the period is the same for all satellites, rank the magnitude of the net force acting on each satellite. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Six artificial satellites of mass, m, circle a space station at distance, R. m R A 200 kg 5000 m B 100 kg 10000 m C 400 kg 2500 m D 800 kg 5000 m E 100 kg 2500 m F 300 kg 7500 m a. If the tangential speed is the same for all satellites, rank the magnitude of the net force acting on each satellite. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. If the period is the same for all satellites, rank the magnitude of the net force acting on each satellite. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Six rollercoaster carts pass over semi-circular “bumps”. The mass of each cart (including passenger), M, and the force of the track on the cart at the apex of each bump, F, are given below. M F A 200 kg 400 N B 100 kg 400 N C 400 kg 200 N D 800 kg 800 N E 100 kg 800 N F 300 kg 300 N a. If the radius of each bump is the same, rank the speed of the cart as it passes over the apex of the bump. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. If the speed of each cart is the same at the apex, rank the radius of each bump. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A rollercoaster track has six semi-circular “dips” with different radii of curvature, R. The rollercoaster cart rides through each dip at a different speed, v. R v A 30 m 16 m/s B 60 m 16 m/s C 15 m 8 m/s D 30 m 4 m/s E 15 m 12 m/s F 45 m 4m/s Rank the magnitude of the force of the rollercoaster track on the cart at the bottom of each dip. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A rollercoaster track has a semi-circular “bump” of radius of curvature R. A rollercoaster cart (including passenger) of total mass M rides over the bump. M R A 200 kg 30 m B 100 kg 60 m C 400 kg 15 m D 800 kg 30 m E 100 kg 15 m F 300 kg 45 m Rank the minimum velocity necessary for the cart to momentarily lose contact with the track at the top of the bump. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The wrench below has six equal magnitude forces acting on it. Rank these forces on the basis of the magnitude of the torque they apply to the wrench, measured about an axis centered on the bolt. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Two identical spheres are attached together by a thin rod. The rod lies on a line connecting the centers-of-mass of the two spheres. The length of the rod is equal to the diameter of each sphere. The object has six equal magnitude forces acting on it at the locations shown. a. Rank these forces on the basis of the torque they apply to the object, measured about axis 1. Let the counterclockwise direction be positive and rank positive torques as larger than negative torques. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank these forces on the basis of the torque they apply to the object, measured about axis 2. Let the counterclockwise direction be positive and rank positive torques as larger than negative torques. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: An artist constructs the mobile shown below. The highest two crossbars are 3 units long, and are hung from a point ⅓ along their length. The lower three crossbars are 2 units long, and are hung from their midpoint. In this configuration, the mobile is perfectly balanced. Rank the masses of the six hanging shapes. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: An artist constructs the mobile shown below. The crossbars are either 3 units long, and hung from a point ⅓ along their length, or are 2 units long, and hung from their midpoint. In this configuration, the mobile is perfectly balanced. Rank the masses of the six hanging shapes. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The six platforms below are initially at rest in deep space. The indicated forces act on the platforms at their endpoints, quarter-points, or midpoints. All forces have the same magnitude. Rank these platforms on the basis of the torque that acts on them, measured about their CM. Let the counterclockwise direction be positive and rank positive torques as larger than negative torques. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The six platforms below are initially at rest in deep space. The indicated forces act on the platforms at their endpoints, quarter-points, or midpoints. All forces have the same magnitude. Rank these platforms on the basis of the torque that acts on them, measured about their CM. Let the counterclockwise direction be positive and rank positive torques as larger than negative torques. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The six platforms below are initially at rest in deep space. The indicated forces act on the platforms at their endpoints, quarter-points, or midpoints. All forces have the same magnitude. Rank these platforms on the basis of the force that must be applied to them, at their left-edge, to keep them from rotating. Let the positive direction be upward and rank positive forces as larger than negative forces. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A roadside sign is to be hung from the end of a thin pole, and the pole supported by a single cable. Your design firm brainstorms the following six scenarios. In scenarios A, B, and D, the rope is attached to the pole ¾ of the distance between the hinge and the sign. In C, the rope is attached to the mid-point of the pole. Rank the design scenarios on the basis of the tension in the supporting cable. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Two identical, uniform, solid spheres are attached together by a solid, uniform thin rod. The rod lies on a line connecting the centers-of-mass of the two spheres. The axes A, B, C, and D are in the plane of the page (which contains the centers-of-mass of the spheres and the rod), while axes E and F are perpendicular to the page. Rank the rotational inertia of this object about the axes indicated. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Below are top views of six identical turntables, upon which are fastened different masses. The distance from the center of the turntable to the center of the mass is either zero, one-half the radius of the turntable, or the radius of the turntable. Rank the rotational inertia of these turntable-mass systems. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Below are top views of six identical turntables, upon which are fastened different masses. The distance from the center of the turntable to the center of the mass is either zero, one-half the radius of the turntable, or the radius of the turntable. Rank the rotational inertia of these turntable-mass systems. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A block of mass M is attached to the inner radius (Rinner) of the pulley shown below. A second rope is attached to outer radius (Router) of the pulley. Assume friction in the pulley mount is very small. M Rinner Router A 10 kg 10 cm 20 cm B 5 kg 20 cm 40 cm C 10 kg 5 cm 20 cm D 10 kg 5 cm 10 cm E 20 kg 5 cm 10 cm F 20 kg 20 cm 30 cm Rank these scenarios on the basis of the magnitude of the force that must be applied to the free rope to hold the block stationary. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A 10 kg block is attached to the inner radius (Rinner) of the pulley shown below. A second rope, attached to outer radius (Router) of the pulley, is used to raise the block at constant speed v. Assume friction in the pulley mount is very small. v Rinner Router A 10 m/s 10 cm 20 cm B 20 m/s 20 cm 40 cm C 5 m/s 5 cm 20 cm D 10 m/s 5 cm 10 cm E 5 m/s 5 cm 10 cm F 10 m/s 20 cm 30 cm Rank these scenarios on the basis of the magnitude of the force that must be applied to the free rope to raise the block at the speed indicated. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A rollercoaster cart travels through a semicircular “bump” of radius 20 m followed by a “dip” of radius 25 m. At the apex of the bump, the 50 kg passenger momentarily loses contact with her seat, and at the bottom of the dip the bathroom scale she’s sitting on reads 1020 N. a. How fast is the cart moving at the apex of the bump? Free-Body Diagram Mathematical Analysis b. How fast is the cart moving at the bottom of the dip? Free-Body Diagram Mathematical Analysis A 60 kg woman rides in a Ferris wheel of radius 16 m. In order to better understand physics, she takes along a bathroom scale and sits on it. When at the top, the scale reads 540 N. The Ferris wheel maintains the same speed throughout its motion. She checks the scale again at the bottom of the motion. a. What is the angular speed of the Ferris wheel? Free-Body Diagram Mathematical Analysis b. What does the bathroom scale read at the bottom of the motion? Free-Body Diagram Mathematical Analysis A 1.5 kg pendulum bob swings from the end of a 2 m long string. The maximum angle from vertical reached by the pendulum is 15°, and the bob reaches a speed of 1.16 m/s as it passes through its lowest point. a. What is the force exerted on the bob by the string at the maximum angle? Free-Body Diagram Mathematical Analysis b. What is the force exerted on the bob by the string at the lowest point? Free-Body Diagram Mathematical Analysis The 65 kg strange man at right is skateboarding in a half-pipe. The half-pipe is a half-circle with a radius of 5 m. At the instant shown he is 2 m from the bottom of the half-pipe, measured vertically, and is moving at 7.5 m/s. He reaches a speed of 9.5 m/s at the bottom. a. What is the force exerted by the skateboard on the man at the instant shown? Free-Body Diagram Mathematical Analysis b. What is the force exerted by the skateboard on the man at the bottom of the half-pipe? Free-Body Diagram Mathematical Analysis A 2.0 kg tether ball is attached to a vertical pole by a 1.5 m long rope. The ball is swung around the pole at an angle of 300 from vertical. Free-Body Diagram Mathematical Analysis y-direction -direction A 2.0 kg tether ball swings around a vertical pole attached to two 1.5 m long ropes, each at an angle of 300 from vertical. One rope is attached to the top of the ball and the top of the pole, the other rope is attached to the bottom of the ball and the bottom of the pole. The ball is traveling at 3.5 m/s. Free-Body Diagram Mathematical Analysis In the hammer throw, a 7.3 kg steel ball at the end of a 1.22 m wire is swung in an approximately circular path around a thrower’s head. Assume the ball is traveling at a constant speed of 7 m/s. Free-Body Diagram Mathematical Analysis A 1000 kg car rounds a 50 m radius curve at constant speed without slipping. The coefficient of friction between the car's tires and the road is (0.6,0.5). Free-Body Diagram Mathematical Analysis rear view An 400 m radius, banked, highway curve is designed to allow cars to drive through the curve at a speed of 20 m/s and not slip, even when the road is extremely icy. Free-Body Diagram Mathematical Analysis rear view A 100 m radius, 80 banked, highway curve has just been built. You decide to find maximum velocity with which you can drive your 1200 kg pick-up truck through this curve without sliding up or down the incline. The coefficient of friction between your enormous tires and the road is (0.7, 0.6) . Free-Body Diagram Mathematical Analysis rear view You enter an 80 m radius, 60 banked, highway curve at 30 m/s. The coefficient of friction between your tires and the road is (0.9, 0.8) . Free-Body Diagram Mathematical Analysis rear view In an amusement park ride called the Rotor, a circular, 4.0 m radius room is spun around at high speed and then the floor is removed. The people riding the Rotor feel that they are being pressed against the wall with such a large force that they do not slide down the wall to the floor. (Obviously they do not understand physics.) The coefficient of friction between the riders and the wall is (0.6,0.5). Free-Body Diagram Mathematical Analysis The roadside sign at right has a mass of 22 kg. It hangs from the end of a 1.6 m long, 14 kg support pole hinged to the wall. A support cable is attached to the pole 1.1 m from the wall. Free-Body Diagram Mathematical Analysis The roadside sign at right has a mass of 22 kg. It hangs from a 1.6 m long, 14 kg support pole hinged to the wall. A support cable is attached to the pole at both 0.4 m and 1.2 m from the wall. The sign is hung directly below the outer support cable. Free-Body Diagram Mathematical Analysis The roadside sign at right has a mass of 22 kg. It hangs from the end of a 1.6 m long, 14 kg support pole hinged to the wall at an angle of 250 above horizontal. A support cable is attached to the pole at a point 1.1 m from the wall along the pole. Free-Body Diagram Mathematical Analysis The 70 kg strange man has built a simple scaffold by placing a 6 m long, 35 kg board on top of two piles of bricks. The brick supports are 2 m apart and centered on the CM of the board. The man’s 10 kg toolbox is 1 m from the right edge of the board. The man is 1.1 m from the left edge of the board when he stops and realizes he probably can’t walk all the way to the left edge of the board. Free-Body Diagram Mathematical Analysis The two 70 kg strange twins have built a simple scaffold by placing a 6 m long, 35 kg board on top of two piles of bricks. The brick supports are 2 m apart and centered on the CM of the board. One twin is at the extreme right edge of the board while the other is 1.2 m from the left edge. Free-Body Diagram Mathematical Analysis The 70 kg strange man has climbed three-quarters of the way up the 20 kg, 10 m long ladder when he stops and realizes that 50° may not be a very safe angle for a ladder. The coefficient of friction between the base of the ladder and the ground is (0.6, 0.5). The friction between the ladder and the wall is negligible. Free-Body Diagram Mathematical Analysis The 75 kg man is falling at 20 m/s, 75 m above the crocodile-infested waters below! In an attempt to save him, the brake shoe is pressed against the spinning pulley. The coefficient of friction between the brake shoe and the pulley is (0.9, 0.8), and the 35 kg disk-shaped pulley has inner and outer diameters of 0.60 m and 0.90 m, respectively. Motion Information Free-Body Diagrams Object: Event 1: t1 = r1 = 1 = v1 = 1 = a1 = 1 = Event 2: t2 = r2 = 2 = v2 = 2 = a2 = 2 = Mathematical Analysis Tired of walking up the stairs, an 80 kg engineering student designs an ingenious device for reaching his third floor dorm room. An 84 kg block is attached to a rope that passes over a 0.7 m diameter, 15 kg, disk-shaped pulley. The student holds the other end of the rope. When the 84 kg block is released, the student is pulled up to his dorm room, 8 m off the ground. Motion Information Free-Body Diagrams Object: Event 1: t1 = r1 = 1 = v1 = 1 = a1 = 1 = Event 2: t2 = r2 = 2 = v2 = 2 = a2 = 2 = Mathematical Analysis Tired of walking up the stairs, an 80 kg engineering student designs an ingenious device for reaching her third floor dorm room. An 42 kg block is attached to a rope that passes over the outer diameter of a 0.7 m outer diameter, 15 kg, disk-shaped compound pulley. The student holds a second rope, wrapped around the inner 0.35 m diameter of the pulley. When the 42 kg block is released, the student is pulled up to her dorm room, 8 m off the ground. Motion Information Free-Body Diagrams Object: Event 1: t1 = r1 = 1 = v1 = 1 = a1 = 1 = Event 2: t2 = r2 = 2 = v2 = 2 = a2 = 2 = Mathematical Analysis A 60 kg student lifts herself from rest to a speed of 1.5 m/s in 2.1 s. The boson’s chair has a mass of 35 kg. The 20 kg, disk-shaped pulley has a diameter of 1.1 m. Motion Information Free-Body Diagrams Object: Event 1: t1 = r1 = 1 = v1 = 1 = a1 = 1 = Event 2: t2 = r2 = 2 = v2 = 2 = a2 = 2 = Mathematical Analysis The 65 kg man at right is trapped inside a section of large pipe. If that’s not bad enough, the pipe begins to roll, from rest, down a 35 m long, 180 incline! The pipe has a mass of 180 kg and a diameter of 1.2 m. (Assume the man’s presence inside the pipe has a negligible effect on the pipe’s rotational inertia.) The coefficient of friction between the pipe and the ground is (0.5, 0.4). Motion Information Free-Body Diagram Event 1: t1 = r1 = 1 = v1 = 1 = a12 = 12 = Event 2: t2 = r2 = 2 = v2 = 2 = Mathematical Analysis Conservation Laws Concepts and Principles The Angular Impulse-Angular Momentum Relation We’ve already used the impulse-momentum relation to analyze situations involving translations through three-dimensional space. The relation is typically applied in its component form: Arguing by analogy, if the change in momentum in the x-, y-, and z-directions is equal to the impulse applied to the object in each direction doesn’t it seem plausible that a similar relation would hold in the “-direction”? If so, this relation should be constructed between the torques acting on an object, the time interval over which the torques act, and the change in angular velocity of an object. The relation should be: The product of rotational inertia and angular velocity is termed the angular momentum of the object, typically denoted L, and the product of torque and the time interval over which it acts is termed the angular impulse applied to the object. Thus, if no angular impulse is applied to an object, its angular momentum will remain constant. This special case is referred to as angular momentum conservation. However, if an angular impulse is applied to the object, the angular momentum will change by an amount exactly equal to the angular impulse applied. Angular momentum is changed through angular impulse. Incorporating Rotation in the Work-Energy Relation Our previous encounter with the work-energy relation resulted in: Recall from our previous discussion of work-energy that this is not a vector equation, meaning it is not applied independently in each of the coordinate directions. Generalizing this equation to include rotation will involving adding terms to this equation, not creating a separate “angular” energy equation. Recall that we model the motion of an arbitrary rigid body as a superposition of a pure translation of the CM and a pure rotation about the CM. Let’s investigate the effect of this model on our calculation of the kinetic energy of the object. The translation portion of the motion is easy. We envision the object as a point particle, localized at the CM of the real object, traveling with the velocity of the CM of the object. Thus, this portion of the motion contributes a kinetic energy, What about the kinetic energy due to the pure rotation of the object about the axis through the CM? Every small chunk of the object, dm, moves in a circle around the CM. Each of these pieces of mass has a velocity magnitude given by Thus, the kinetic energy of each piece is Therefore, the total kinetic energy due to rotation of all the little pieces is: Combining the kinetic energy due to rotation with the kinetic energy due to translation leads to a total kinetic energy of: and a work-energy relation of: Conservation Laws Analysis Tools Applying the Work-Energy Relation including Rotation - I A mischievous child releases his mother’s bowling ball from the top of the family’s 25 m long, 150 above horizontal driveway. The ball rolls without slipping down the driveway and at the bottom plows into the mailbox. The 6.4 kg ball has a diameter of 24 cm. Let’s determine the speed of the ball when it hits the mailbox. To determine this value, we can apply the work-energy relation to the ball between: Event 1: The instant the ball is released. Event 2: The instant the ball hits the mailbox. For these two events, work-energy looks like this: where the bottom of the driveway is the zero for gravitational potential energy and the rotational inertia of the bowling ball is taken to be that of a sphere. Now we must carefully determine which, if any, of the forces on the bowling ball do work on the bowling ball. First, we don’t have to worry about the force of gravity. The gravitational potential energy function was developed to automatically incorporate the work done by the force of gravity. Second, the contact force can do no work on the ball because the contact force is always perpendicular to the motion of the ball. Finally, what about the force of friction? It does appear, at first glance, that the force of friction is applied over the entire motion of the ball down the driveway. However, let’s pay closer attention to the actual point at which the force acts. The frictional force is static in nature, because since the ball rolls without slipping down the driveway the bottom of the ball is always in static contact with the ground (i.e., there is no relative velocity between the bottom of the ball and the ground). If the bottom of the ball has a velocity of zero, then the force that acts on the bottom of the ball (static friction) can act through no distance. During the instant at which the frictional force acts on a particular point on the bottom of the ball, that point is not moving. That point on the ball only moves when it is no longer in contact with the ground, but by that time the frictional force is acting on a different point. To summarize (and stop saying the same thing over and over), the force of static friction can do no work because it acts on a point that does not move. Thus, our equation simplifies to: The linear and angular velocity of the ball must be related. Since the ball rolls without slipping, the bottom of the ball has no linear velocity. Since the velocity of the bottom of the ball is the sum of the velocity due to translation of the CM and the velocity due to rotation about the CM, the velocity due to rotation must be equal in magnitude to the velocity of the CM: Substituting this into our equation yields: Applying the Work-Energy Relation including Rotation - II The 60 kg crate is descending at 2.0 m/s when a brake shoe is applied to the 10 kg, 20 cm radius disk-shaped pulley. The coefficient of friction between the shoe and the pulley is (0.9, 0.7). The crate comes to rest after descending 3 m. To determine this function, let’s apply the work-energy relation to both the crate and the pulley between: Event 1: The instant the brake shoe is applied. Event 2: The instant the crate comes to rest. In addition to defining the two instants of interest, we’ll need free-body diagrams for both the crate and the pulley. Crate where the final position of the crate is the zero for gravitational potential energy. Pulley To find the work on the pulley, note that Fcontact and Fpin do no work since the distance over which these forces act is zero. However, both Frope and Ffriction do work. If the crate falls 3.0 m, both of these forces act over a displacement of 3.0 m. However, note that this displacement is in the same direction as Frope ( = 0) on the right side of the pulley but in the opposite direction on the other side of the pulley (opposite to Ffriction and therefore  = 180.) Now note that since the brake shoe does not accelerate, the external force applied to the shoe, F, is the same magnitude as the contact force between the shoe and the pulley. The brake shoe must have been pressed against the pulley with a force of 902 N in order to stop the crate in 3 m. Applying the Impulse-Momentum Relations (Linear and Angular) A 30 kg child running at 4.0 m/s leaps onto the outer edge of an initially stationary merry-go-round. The merry-go-round is a flat disk of radius 2.4 m and mass 70 kg. Ignore the frictional torque in the bearings of the merry-go-round. Let’s imagine we’re interested in determining the final angular velocity of the merry-go-round after the child has safely come to rest on its surface. To determine this angular velocity, we can apply the impulse-momentum relations to both the child and the merry-go-round between: Event 1: The instant before the child lands on the merry-go-round Event 2: The instant after the child comes to rest on the merry-go-round. The child and the merry-go-round interact via some unknown magnitude force. (Contrary to the free-body diagram, this force does not necessarily act on the child’s disembodied head.) There are also other forces due to gravity and supporting structures that act on both the child and the merry-go-round, however, these forces are in the vertical direction and supply no torque. Child x-direction Merry-Go-Round -direction The final velocity of the child and final angular velocity of the merry-go-round are related because at this instant the child is at rest (hanging on to) the merry-go-round. Therefore, Substituting this into our angular equation yields: By Newton’s Third Law, the force of the child on the merry-go-round and the force of the merry-go-round on the child must be equal in magnitude. Therefore our two equations can be summed and the impulses cancel. This gives: The child is slowed from 4 m/s to 1.85 m/s by jumping onto the merry-go-round. The merry-go-round, however, is accelerated from rest to: Conservation Laws Activities For each of the scenarios described below, indicate the amount of linear kinetic energy, rotational kinetic energy, and gravitational potential energy in the system at each of the events listed. Use a consistent scale for each motion. Set the lowest point of each motion as the zero-point of gravitational potential energy. a. A mischievous child releases his mother’s bowling ball from the top of the family’s 15 m long, 80 above horizontal driveway. The ball rolls without slipping down the driveway and at the bottom plows into the mailbox. b. A yo-yo of inner diameter 0.70 cm and outer diameter 8.0 cm is released from rest. The string is 0.80 m long. Assume the string does not slip on the inner diameter of the yo-yo as the yo-yo falls.. For each of the scenarios described below, indicate the amount of linear kinetic energy, rotational kinetic energy, and gravitational potential energy of each object at each of the events listed. Use a consistent scale for each motion. Set the initial positions of the objects as the zero-points of gravitational potential energy. a. Tired of walking up the stairs, an 80 kg engineering student (S) designs an ingenious device for reaching his third floor dorm room. An 84 kg block (B) is attached to a rope that passes over a 0.70 m diameter, 15 kg, disk-shaped pulley (P). The student holds the other end of the rope. When the block is released, the student is pulled up to his dorm room. b. A 75 kg man (M) is falling at 20 m/s, 75 m above crocodile infested waters! He holds a rope attached to a 15 kg simple pulley (P). In an attempt to save him, a brake shoe is pressed against the spinning pulley. The man is saved, but barely. For each of the scenarios described below, indicate the amount of linear kinetic energy, rotational kinetic energy, and gravitational potential energy of each object at each of the events listed. Use a consistent scale throughout both motions. Set the initial positions of the objects as the zero-points of gravitational potential energy a. In a horizontal skiing device, the skier begins from rest 35 m from the end of the skiing run. The skier (S) has a mass of 75 kg, the block (B) has a mass of 50 kg, and the simple pulley (P) has a mass of 15 kg. The coefficient of friction is extremely small. b. In an inclined skiing device, the skier begins from rest 35 m from the end of the 200 above horizontal inclined skiing run. The skier (S) has a mass of 75 kg, the block (B) has a mass of 50 kg, and the simple pulley (P) has a mass of 15 kg. The coefficient of friction is extremely small. A disk-shaped pulley of mass M and radius R is rotating at angular velocity . The friction in its bearings is so small that it can be ignored. A brake shoe is pressed against the pulley in order to stop it. In all cases, the brake shoe is pressed against the pulley with the same force and the coefficient of friction between the brake shoe and the pulley is the same. M R  A 10 kg 0.8 m 12 rad/s B 10 kg 0.4 m 24 rad/s C 20 kg 0.4 m 6 rad/s D 40 kg 0.2 m 3 rad/s E 20 kg 0.2 m 3 rad/s F 30 kg 0.6 m 8 rad/s a. Rank the scenarios below on the amount of time it takes to stop the pulley. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank the scenarios below on the angle through which the pulley rotates before stopping. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A disk-shaped pulley of mass M and radius R is rotating at angular velocity . The friction in its bearings is constant and ultimately causes the pulley to stop rotating. M R  A 10 kg 0.8 m 12 rad/s B 10 kg 0.4 m 24 rad/s C 20 kg 0.4 m 6 rad/s D 40 kg 0.2 m 3 rad/s E 20 kg 0.2 m 3 rad/s F 30 kg 0.6 m 8 rad/s a. Rank the scenarios below on the amount of time it takes to stop the pulley. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: b. Rank the scenarios below on the angle through which the pulley rotates before stopping. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A sphere of mass m and radius R is released from rest at the top of an incline and rolls without slipping down the incline. All spheres are released from rest from the same location on the incline. m R A 1 kg 30 cm B 2 kg 60 cm C 3 kg 10 cm D 2 kg 15 cm E 1 kg 15 cm F 2 kg 30 cm Rank the velocity of the sphere at the bottom of the incline. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: Six different equal-mass objects are released from rest from the same location at the top of an incline and roll without slipping down the incline. Object R A solid sphere 30 cm B hollow sphere 60 cm C solid disk 10 cm D hoop 10 cm E solid cylinder 20 cm Rank the velocity of the objects at the bottom of the incline. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: A merry-go-round of radius R is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round. m r A 10 kg 0.50 R B 10 kg 0.25 R C 20 kg 0.25 R D 40 kg 0.25 R E 10 kg 1.00 R F 15 kg 0.75 R Rank the scenarios on the basis of the angular speed of the merry-go-round after the sandbag “sticks” to the merry-go-round. Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest _____ The ranking cannot be determined based on the information provided. Explain the reason for your ranking: The 75 kg man is falling at 20 m/s, 75 m above the crocodile infested waters below! In an attempt to save him, the brake shoe is pressed against the spinning pulley. The coefficient of friction between the brake shoe and the pulley is (0.9, 0.8), and the 35 kg disk-shaped pulley has inner and outer diameters of 0.60 m and 0.90 m, respectively. The man is saved, but barely. Free-Body Diagrams Mathematical Analysis Event 1: Event 2: The 75 kg man is falling at 15 m/s! In an attempt to stop his fall, the brake shoe is pressed against the spinning pulley with a force of 1000 N. The coefficient of friction between the brake shoe and the pulley is (0.9, 0.8), and the 35 kg disk-shaped pulley has inner and outer diameters of 0.60 m and 0.90 m, respectively. a. How long does it take to stop the man’s fall? b. How far does the man fall? Free-Body Diagrams Mathematical Analysis Event 1: Event 2: Tired of walking up the stairs, an 80 kg engineering student designs an ingenious device for reaching his third floor dorm room. An 84 kg block is attached to a rope that passes over a 0.70 m diameter, 15 kg, disk-shaped pulley. The student holds the other end of the rope. When the 84 kg block is released, the student is pulled up to his dorm room in 5.3 s. Free-Body Diagrams Mathematical Analysis Event 1: Event 2: Tired of walking up the stairs, an 80 kg engineering student designs an ingenious device for reaching her third floor dorm room. An 42 kg block is attached to a rope that passes over the outer diameter of a 0.70 m outer diameter, 15 kg, disk-shaped compound pulley. The student holds a second rope, wrapped around the inner 0.35 m diameter of the pulley. When the 42 kg block is released, the student is pulled up to her dorm room, 8.0 m off the ground. Free-Body Diagrams Mathematical Analysis Event 1: Event 2: A 60 kg student lifts herself from rest to a speed of 1.5 m/s in 2.1 s. The boson’s chair has a mass of 35 kg. The 20 kg, disk-shaped pulley has a diameter of 1.1 m. Free-Body Diagrams Mathematical Analysis Event 1: Event 2: A mischievous child releases his mother’s bowling ball from the top of the family’s 15 m long, 80 above horizontal driveway. The ball rolls without slipping down the driveway and at the bottom plows into the mailbox. The 6.4 kg ball has a diameter of 21.6 cm. Free-Body Diagram Mathematical Analysis Event 1: Event 2: The 65 kg strange man at right is trapped inside a section of large pipe. If that’s not bad enough, the pipe begins to roll from rest down a 35 m long, 180 incline! The pipe rolls down the hill without slipping. The pipe has a mass of 180 kg and a diameter of 1.2 m. (Assume the man’s presence inside the pipe has a negligible effect on the pipe’s rotational inertia.) Free-Body Diagram Mathematical Analysis Event 1: Event 2: A 30 kg child running at 3.0 m/s leaps onto the outer edge of an initially stationary merry-go-round. The merry-go-round is a flat disk of radius 2.4 m and mass 50 kg. Ignore the frictional torque in the bearings of the merry-go-round. Free-Body Diagrams Mathematical Analysis child (top view) merry-go-round (top view) Event 1: Event 2: A 40 kg child running at 4.0 m/s leaps onto the outer edge of a merry-go-round initially rotating in the opposite direction. The merry-go-round is brought to rest. The merry-go-round is a flat disk of radius 2.4 m and mass 50 kg. Ignore the frictional torque in the bearings of the merry-go-round. Free-Body Diagrams Mathematical Analysis child (top view) merry-go-round (top view) Event 1: Event 2: A 30 kg bag of sand is dropped onto a merry-go-round 1.6 m from its center. The merry-go-round was rotating at 3.2 rad/s before the bag was dropped. The merry-go-round is a flat disk of radius 2.4 m and mass 50 kg. Ignore the frictional torque in the bearings of the merry-go-round. Free-Body Diagrams Mathematical Analysis sandbag (top view) merry-go-round (top view) Event 1: Event 2: A 30 kg bag of sand is dropped onto a merry-go-round. The merry-go-round was rotating at 2.2 rad/s before the bag was dropped, and 1.3 rad/s after the bag comes to rest. The merry-go-round is a flat disk of radius 2.4 m and mass 60 kg. Ignore the frictional torque in the bearings of the merry-go-round. Free-Body Diagrams Mathematical Analysis sandbag (top view) merry-go-round (top view) Event 1: Event 2: Selected Answers