4.1: Kinematics

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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.

Screen-Shot-2017-07-27-at-2.41.26-PM-298x300.png

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 vector

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. Moreover, 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 (w(t)) and the same angular acceleration (a(t)).

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 the angular speed and angular acceleration of some point on the object.

Given that every point on a spinning object undergoes circular motion, the results from our study of circular motion will be very important in analyzing spinning objects. Recall that with q(t) defined as the angular position of an arbitrary point on the rigid-body,

and

Moreover, the velocity and acceleration of any point on a spinning, rigid body can be related to the angular quantities:

and

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 Dr), 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 Dq), 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.

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

q1 = 0 rad

w1 =

a1 =

Event 2: The ceiling fan stops

t2 = 22 s

q2 = 18 (2p) = 113 rad

w2 = 0 rad/s

a2 =

Since no specific information concerning the angular acceleration of the fan is given, let’s assume the angular acceleration is constant. (Thus, a1 = a2 = a12.) Since the relationships between angular position, velocity, and acceleration are the same as the relationships between linear position, velocity and acceleration, the kinematic equations for constant linear acceleration must have direct analogies for constant angular acceleration. Thus,

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.

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

a1 = 1.5 m/s2

Event 2: Block raised 15 m

t2 =

r2 =

v2 =

a2 = 1.5 m/s2

Pulley

Event 1: Truck begins to move

t1 = 0 s

q1 = 0 rad

w1 = 0 rad/s

a1 =

Event 2: Block raised 15 m

t2 =

q2 =

w2 =

a2 =

Block

Event 1: Truck begins to move

t1 = 0 s

r1 = 0 m

v1 = 0 m/s

a1 =

Event 2: Block raised 15 m

t2 =

r2 = 15 m

v2 =

a2 =

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 s is (technically) the arc length over which a rope is wrapped around the pulley and q is the angular displacement of the pulley. Of course, the amount of rope wrapped around a pulley is exactly equal to the displacement of the object attached to the rope. 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 q direction to be the direction in which the tires rotate.

Event 1: The Cadillac begins to accelerate

t1 = 0 s

r1 = 0 m

q1 = 0 rad

v1 = 0 m/s

w1 = 0 rad/s

a1 =

Event 2: The Cadillac reaches 25 m/s

t2 = 6.2 s

r2 =

q2 =

v2 = 25 m/s

w2 =

a2 =

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^[1]. Thus, the velocity of any point on the tire is given by

where r is the distance between the point of interest and 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

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 w will allow the tire to roll without slipping.

From above,

At the bottom of the tire, , since motion tangent to the tire at the bottom of the tire is motion to the left. Thus,

Therefore, at the end of the acceleration, the tire has an angular speed of 62.5 rad/s.

I included all of the vector notation to ensure I did the calculation correctly, but hopefully we can understand the result without getting bogged down in notation. Basically, the bottom of the tire can’t be moving relative to the ground if the wheel rolls without slipping. Since the CM of the tire is moving forward at 25 m/s, the bottom of the tire must be moving backward at 25 m/s relative to the CM. From the CM frame of reference, the tire is just spinning, and I can apply v = r w to calculate the angular speed of the tire.

We can find the angular acceleration and angular position of the tire by using the linear variables determined above and the method described in the previous example, or by using the equations for constant angular acceleration.

Thus, the wheel rotates through 194 rad, or 30.8 revolutions, while accelerating.

Activities

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.