11.1.10.1: Illustrations

Illustration 1: Coordinates for Circular Motion

How would you describe the motion of the object shown (position is given in meters and time is given in seconds)? Restart. The object is moving in a circle about \(x = 0\text{ m}\) and \(y = 0\text{ m}\), but the object's \(x\) and \(y\) coordinates vary with time. They vary in a special way such that \(x\) and \(y\) are always between \(-1\text{ m}\) and \(1\text{ m}\). To see this, look at Animation 2 and watch the \(x\) and \(y\) values change in the table. This is called the component form. We can also describe the motion in terms of the vector form. In this case, the radius vector, \(\mathbf{r}\), always has a magnitude of \(1\text{ m}\). but it changes direction. Look at Animation 3. We describe the direction of this vector in terms of the angle it makes with the positive \(x\) axis. Therefore the angle-when measured in degrees-varies from \(0\) to \(360\). It is often convenient to give the angle in a unit different than degrees. We call this unit a radian. The radian unit is defined as \(2\pi\) radians \(= 360^{\circ}\). Notice that both units are defined in terms of one full revolution. To see the angle given in radians look at Animation 4.

So why use radians? Well, it turns out that there is a really nice relationship between angle in radians (\(\theta\)), the radius (\(r\)), and the arc of the circle (\(s\)). This geometric relationship states that: \(\theta = s/r\). Why is this useful? It allows us to treat circular motion like one-dimensional motion. The arc is the linear distance traveled, which is \(s = vt\) when the motion is uniform. This means that \(\theta = (v/r) t\), since \(s = r\theta\). We call \(v/r\) by the name omega, \(\omega\), and it is the angular velocity. Therefore, \(\theta =\omega t\), for motion with a constant angular velocity. When there is a constant angular acceleration, we call it by the name alpha, \(\alpha\), and it is related to the tangential acceleration by \(a_{t}/r\). So when we are using radians we can use our one-dimensional kinematics formulas with \(x \to\theta\), \(v\to\omega\), and \(a\to\alpha\).

Illustration 2: Motion about a Fixed Axis

Many objects rotate (spin) about a fixed axis. Shown is a wheel of radius \(5\text{ cm}\) rotating about a fixed axis at a constant rate (position is given in centimeters and time is given in seconds). Restart.

Consider the various points on the surface of the rotating wheel. By watching the line rotate, we can see that the wheel is rotating at a constant rate. In fact, if we watch a point on the surface of the wheel (a radius of \(5\text{ cm}\)), we would say it has a constant speed. What about a point halfway out (a radius of \(2.5\text{ cm}\))? It also has a constant speed. But how does this speed compare to the speed of a point on the surface of the wheel?

We can determine this by first considering a quantity that is not related to the radius, the angular speed, \(\omega\). The angular speed is the angular displacement divided by the time interval (in this Illustration, since there is no acceleration, the average speed and the instantaneous speed are the same). So what is the wheel's angular speed? From one revolution of the wheel, the angular displacement is \(2\pi\) and the time interval (called the period, \(T\)) is \(5\) seconds. Therefore the angular speed is the angular displacement over the time interval (\(\omega = 2\pi /T\)) \(0.4\pi\text{ radians/s} = 1.256\text{ radians/s}\).

How do we relate the angular speed to the linear (tangential) speed of a point on the wheel? First consider the speed of a point on the surface of the wheel. It is again easiest to measure the speed by considering one rotation of the wheel. In this case, the distance traveled by that point is \(2\pi r\), and therefore the average (and in this case instantaneous) tangential speed is \(2\pi r/T = 2\pi\text{ cm/s} = 6.28\text{ cm/s}\). The relationship between the angular speed and the tangential speed must be \(\omega = v/r\). (Recall that we found above that \(\omega = 2\pi /T\).)

This works because there is a relationship between the angular displacement and the tangential displacement (the arc length \(s\)), namely that \(\Delta\theta =\Delta s/r\). This also must be the case for an angular displacement of one revolution: \(2\pi = 2\pi r/r\).

Since linear velocity is a vector we might expect that angular velocity is a vector as well. This is indeed the case. So in which direction does the angular velocity point for the rotating wheel? We use the right-hand rule (RHR) to determine the direction of the angular velocity. Using your right hand, curl your fingers in the direction of the rotation of the wheel: the direction your thumb points is the direction of the wheel's angular velocity. Here, \(\omega\) is into the page (computer screen). This may seem weird; after all you might say that the wheel is rotating clockwise. Clockwise is not a good description, as it does not imply a vector-like quantity and the description is not unique. Why is it not unique? If you were on the other side of the page or computer screen, you would say that the wheel is rotating counterclockwise instead!

Can you guess what the relationship between the angular acceleration and the tangential acceleration is? Well, given that acceleration is change in velocity for a given time, \(\Delta\mathbf{v}/\Delta t\), you have probably guessed that angular acceleration, called \(\alpha\), is the change in angular velocity for a given time, \(\Delta\)\(\omega\)\(/\Delta t\). Given how \(v\) and \(\omega\) are related to each other, it must be that \(a =\alpha r\).

Illustration authored by Mario Belloni.
Script authored by Steve Mellema, Chuck Niederriter, and Mario Belloni.

Illustration 3: Moment of Inertia, Rotation Energy, and Angular Momentum

Many objects rotate (spin) about a fixed axis. Shown is a wheel (a disk) of radius \(5\text{ cm}\) and a mass of \(200\) grams rotating about a fixed axis at a constant rate (position is given in centimeters and time is given in seconds). Restart.

In Illustration 10.2 we discussed how linear speed (velocity) was related to angular speed (velocity), and in the process how angular acceleration is related to the angular velocity (\(\alpha\) \(=\Delta\)\(\omega\)\(/\Delta t\)). In this Illustration we will discuss kinetic energy of rotation, \(KE_{\text{rot}}\), and angular momentum, \(\mathbf{L}\).

The easiest way to remember the forms for the kinetic energy of rotation and the angular momentum is by analogy with the kinetic energy of translation and the linear momentum. We recall that \(KE = 1/2 m v^{2}\) and \(\mathbf{p} = m\mathbf{v}\). Can you guess what the rotational kinetic energy and angular momentum will look like?

First, what will play the role of \(v\) and \(\mathbf{v}\) in the rotational expressions? If you said \(\omega\) and \(\omega\) you are right. Next we must consider what plays the role of \(m\), and we will be all set. The property of mass describes an object's resistance to linear motion. Therefore, what we are looking for is a property of objects that describes their resistance to rotational motion. This is called the moment of inertia. The moment of inertia depends on the mass of the object, its extent, and its mass distribution. It turns out that for most simple objects the moment of inertia looks like \(I =CmR^{2}\), where m is the object's mass, \(R\) is its extent (usually a radius or length), and \(C\) is a dimensionless constant that represents the mass distribution.

Therefore, we have that \(KE_{\text{rot}} = 1/2 I\omega^{2}\) and \(\mathbf{L} = I\)\(\omega\). What are this disk's \(KE_{\text{rot}}\) and \(\mathbf{L}\)? Well, from Illustration 10.2 we know that \(\omega = 1.256\) radians/s. Since the wheel is a disk, \(C=2\). Therefore, we can calculate the moment of inertia as: \(2.5\times 10^{-4}\text{ kg}\cdot\text{m}^{2}\). Finally, we have that \(KE_{\text{rot}} = 1.97\times 10^{-4}\text{ J}\) and \(\mathbf{L} = 3.14\times 10^{-4}\text{ J}\cdot\text{s}\) (into the page or computer screen). Note that these are small values because \(I\) for this disk is small. A \(1\text{-m}\) radius and \(2\text{-kg}\) mass disk would have a moment of inertia of \(1.0\text{ kg}\cdot\text{m}^{2}\).

Illustration authored by Mario Belloni.
Script authored by Steve Mellema, Chuck Niederriter, and Mario Belloni.

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