11.1.3.1: Illustrations

Illustration 1: Vector Decomposition

A red vector is shown on a coordinate grid, and several properties of that vector are given in a data table (position is given in meters). How can you represent this vector? There are two ways: component form and magnitude and direction form. Both ways of representing vectors are correct, but in different circumstances one way may be more convenient than the other. Restart.

You can drag the head of the vector by click-dragging the small circle at the vector's head.

Magnitude and Direction Form: When you think of a vector, such as the one depicted, you are thinking of the magnitude and direction form. We describe the magnitude as the size of the vector (depicted in the table as \(r\), which is always a positive number) and the direction as an angle (also depicted in the table and given in degrees). This angle is measured from the positive \(x\) axis to the direction that the vector is pointing.

Component Form: When you are solving problems in two dimensions, you often need to decompose a vector into the component form. So how do you do that? Look at the show components version of the animation. As you drag around the red vector, the maroon vectors show you the \(x\) and \(y\) components of the red vector (these are also shown in the table as \(x\) and \(y\)). Try to keep the length of the vector the same and change the angle. How did the components change with angle? As the angle gets smaller, the \(x\) component of the vector gets larger (it approaches the magnitude of the vector) and the \(y\) component of the vector gets smaller (it approaches zero). As the angle approaches \(90^{\circ}\) the \(x\) component of the vector gets smaller (it approaches zero) and the \(y\) component of the vector gets larger (it approaches the magnitude of the vector). Mathematically this is described by the statement that

\[x=r\cos(\theta )\quad\text{and}\quad y=r\sin(\theta)\nonumber\]

Once in component form, we can of course, go back to magnitude and direction form by using the relationships

\[r=(x^{2}+y^{2})^{1/2}\quad\text{and}\quad\theta = \tan^{-1}(y/x)\nonumber\]

Notice that the magnitude of the vector, here \(r\), must be positive as stated above.

Illustration 2: Motion on an Incline

Galileo was the first person to realize that a well-polished (a very slippery or frictionless) inclined plane could be used to reduce the effect of gravity. He realized that if you started with a vertical incline (angle of \(90^{\circ}\)), the scenario was equivalent to free fall. If the incline was horizontal (an angle of \(0^{\circ}\)), the object would not move at all. He therefore reasoned that as you decreased this angle from \(90^{\circ}\) the acceleration would decrease. He was able to measure this acceleration and, thereby, determine the acceleration due to gravity. Mathematically, this amounts to the realization that as a function of the incline's angle:

\[g_{\text{eff}}=g\sin (\theta)\nonumber\]

where \(g_{\text{eff}}\) is the acceleration down the incline. See Illustration 2.5 and Chapter 4 for more details.

By varying the type of object sliding down the slippery (frictionless) incline, he was able to show that all objects accelerate at the same rate. Try the experiment for yourself (time is given in seconds and distance is given in meters) with the above three animations.

Galileo started his objects from rest on an incline. What did he find from his experiments? Galileo's conclusion was that during successive equal-time intervals the objects' successive displacements increased as odd integers: \(1,\: 3,\: 5,\: 7,\:\ldots\). What does that really mean? Consider the chart below, which converts Galileo's data into data we can more easily understand (data shown for an incline whose angle yields an acceleration of \(2\text{ m/s}^{2}\)):

Elapsed time \((\text{s})\)	Displacement during the time interval \((\text{m})\)	Total displacement \((\text{m})\)
\(1\)	\(1\)	\(1\)
\(2\)	\(3\)	\(4\)
\(3\)	\(5\)	\(9\)
\(4\)	\(7\)	\(16\)

Table \(\PageIndex{1}\)

The third column is constructed by adding up all of the previous displacements that occurred in each time interval to get the net or total displacement that occurred so far. What is the relationship between displacement and time? The displacement is related to the square of the time elapsed. Does that idea look familiar? It should. We found in Chapter 2 that \(x = x_{0} + v_{0}\ast t + 0.5\ast a\ast t^{2}\) or that, for no initial velocity, \(x = x_{0} + 0.5\ast a\ast t^{2}\) or that \(\Delta x\) is proportional to \(t^{2}\).

Illustration 3: The Direction of Velocity and Acceleration Vectors

A putted golf ball rolls across the green as shown in the animation (position is given in meters and time is given in seconds). The animation represents the top view of the motion of the ball. Restart.

What is the direction of the velocity of the golf ball at any instant of time? View the velocity vector to check your answer (notice that a black line tangent to the object's path is also drawn). The direction of the velocity vector is determined by a fairly simple rule: It is always tangent to the path and in the direction of motion. The "direction of motion" is basically the direction of the object's displacement during a very small time interval. Since the displacement divided by this very small time interval approaches the instantaneous velocity (see Illustration 2.3), the instantaneous velocity must point in the direction of motion. This also directly follows from the definition of instantaneous velocity as the derivative of the position vector with respect to time.

What about the acceleration vector? It points in the direction of the change in velocity during any small time interval. This is again a result of the definition of acceleration. However, it also follows an interesting rule.

The acceleration vector can be resolved into two components, a component tangent to the path (called the tangential acceleration) and a component perpendicular to the path (called the radial component). View the velocity and acceleration vectors. The radial component of the acceleration is related to the change in direction of the velocity vector and points along the radius of curvature. The tangential component of the acceleration is related to the change in the magnitude of the velocity vector. In other words, it is related to the change in the speed. If the object is slowing down, then the tangential component of the acceleration is opposite to the velocity. If the object is speeding up, the tangential component of the acceleration is in the same direction as the velocity.

Click here to view the velocity vector (blue), acceleration vector (orange), and acceleration components (yellow for the tangential component and red for the radial component).

Illustration authored by Aaron Titus with support by the National Science Foundation under Grant No. DUE-9952323 and placed in the public domain.

Illustration 4: Projectile Motion

A purple ball undergoes projectile motion as shown in the animation (position is given in meters and time is given in seconds). The blue and red objects illustrate the \(x\) and \(y\) components of the ball's motion. Ghost images are placed on the screen every second. To understand projectile motion, you must first understand the ball's motion in the \(x\) and \(y\) directions separately (any multidimensional motion can be resolved into components). Restart.

Consider the \(x\) direction. Notice that the \(x\) coordinate of the projectile (purple) is identical to the \(x\) coordinate of the blue object at every instant. What do you notice about the spacing between blue images? You should notice that the displacement between successive images is constant. So what does this tell you about the \(x\) velocity of the projectile? What does it tell you about the \(x\) acceleration of the projectile? This should tell you that the object moves with a constant velocity in this direction (which is also depicted on the left graph).

Now consider the \(y\) direction. Notice that the \(y\) coordinate of the projectile (purple) is identical to the \(y\) coordinate of the red object at every instant. What do you notice about the spacing between successive images for the red object? You should notice that the displacement between successive images gets smaller as the object rises and gets larger as the object falls. This means that it has a downward acceleration. By studying the right-hand graph, we can also see that the \(y\) acceleration is constant.

A particularly important point to understand for the motion of a projectile is what happens at the peak. What is the velocity of the projectile at the peak? This is a tricky question because you have a good idea that the \(y\) velocity is zero. However, does this mean that the velocity is zero? Remember that velocity has two components, \(v_{x}\) and \(v_{y}\). At the peak, \(v_{x}\) is not zero. Therefore, the velocity at the peak is not zero. Click here to view the velocity and acceleration vectors.

Illustration authored by Aaron Titus with support by the National Science Foundation under Grant No. DUE-9952323 and placed in the public domain.

Illustration 5: Uniform Circular Motion and Acceleration

Uniform circular motion is an interesting mixture of one- and two-dimensional concepts. Restart. For uniform circular motion the speed of the object must be constant. This is the uniform in uniform circular motion. Show the ball undergoing uniform circular motion. So is an object moving in a circle with a constant speed accelerating?  Yes!  Why? The velocity is changing with time. Watch the animation (position is shown in meters and time is shown in seconds).  The animation depicts an object moving in a circle at a constant speed. To determine the acceleration, we need to consider the change in velocity for a change in time.

Since the speed does not change in time, what does change in time? It is the direction of the velocity that changes with time. Draw two velocity vectors corresponding to two different times to convince yourself that the direction of the velocity changes with time. Recall that velocity has both a direction (which always points tangent to the path, the so-called tangential direction) and a magnitude, and either or both can change with time. Where does the change in velocity point? Calculate the acceleration. The acceleration points toward the center of the circle. This direction—toward the center of the circle—is called the centripetal or center-seeking direction. It is often also called the radial direction, since the radius points from the center of the circle out to the object (the net acceleration points in the opposite direction).

Therefore, for uniform circular motion, the acceleration always points toward the center of the circle, no matter the cause. This is despite the fact that the velocity and the acceleration point in changing directions as time goes on. However, we get around this apparent difficulty in describing direction by defining the centripetal or radial direction and the tangential direction (the direction tangent to the circle). These directions change, but the velocity is always tangent to the circle and the acceleration is always pointing toward the center of the circle. The following animation shows velocity and acceleration as the object undergoes uniform circular motion.

Illustration 6: Circular and Noncircular Motion

A planet (green) orbits a star (yellow) as shown in the two animations. Restart. One animation depicts the uniform circular motion of a planet and the other one depicts the noncircular motion of a planet (position is given in \(10^{3}\mathbf{ km}\) and time is given in years). This Illustration will compare the two motions by focusing on the velocity and the acceleration of the planet in each of the animations.

Start the uniform circular motion animation and watch the planet's motion. How would you describe the motion of the planet (consider velocity and acceleration)? The speed of the planet is certainly a constant since the motion of the planet is uniform. But using our usual \(xy\) coordinates, the velocity certainly changes with time. Recall that the term velocity refers to both the magnitude and direction. However, if we use the radial and tangential directions to describe the motion of the planet, the velocity can be described as tangential and the acceleration can be described as being directed along the radius (the negative of the radial direction). Click here to view the velocity vector (blue) and the black line tangent to the path. Click here to view the acceleration vector (red), too. Notice that the acceleration vector points toward the star at the center of the circle.

Start the noncircular motion animation and watch the planet's motion. How would you now describe the motion of the planet (consider velocity and acceleration)? The speed of the planet is certainly no longer a constant since the motion of the planet is no longer uniform. Again using our usual \(xy\) coordinates, the velocity certainly changes with time since now both the direction and the magnitude change. However, if we use the radial and tangential directions to the path of the planet, the velocity can be described as tangential and the acceleration can be described as being directed along the radius. Click here to view the velocity vector (blue) and click here to view the acceleration vector (red), too. Notice that the velocity and the acceleration are no longer perpendicular for most of the orbit of the planet.

Notice that between points A and C the planet is speeding up, and between points C and A the planet is slowing down. This means that at points A and C the tangential component of acceleration is zero. It turns out that for a planet orbiting a star (if there are no other planets or stars nearby) the acceleration of the planet is directed exactly toward the star whether the motion of the planet is uniform or not.

Illustration authored by Aaron Titus and Mario Belloni.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.