11.1.11.1: Illustrations

Illustration 1: Cross Product

We talk about the magnitude of the torque as the amount of force perpendicular to the radius arm on which it acts. No radius arm, no torque. Torque is positive (out of the page) if \(F\) acts to rotate the object counterclockwise via the right-hand rule (RHR) and negative if \(F\) acts to rotate the object clockwise (again, via the RHR). Restart.

In order to mathematically describe torque, we must use the mathematical construction of the vector or cross product. Torque is the vector product of the radius vector and the force vector, \(\mathbf{r}\times\mathbf{F}\). The magnitude of the torque is \(r F \sin(\theta )\), and the direction of the torque is determined by the RHR. \(\theta\) is the angle between the two vectors, and \(A\) and \(B\) are the magnitudes of the vectors \(\mathbf{r}\) and \(\mathbf{F}\), respectively. Drag the tip of either arrow (position is given in meters). The red arrow is \(\mathbf{r}\) and the green arrow is \(\mathbf{F}\). The magnitude of each arrow is calculated as well as the cross product.

The direction of the torque, \(\mathbf{r}\times\mathbf{F}\), is determined by the RHR (point your fingers toward \(\mathbf{r}\), curl them into the direction of \(\mathbf{F}\), and the direction that your thumb points is the direction of the torque. Therefore,

\(\tau =\mathbf{r}\times\mathbf{F} = r F \sin(\theta )\) with the direction prescribed by the RHR,

where \(\mathbf{r}\) is the moment arm on which the force acts and \(\mathbf{F}\) is the force.

Illustration 2: Rolling Motion

Many everyday objects roll without slipping (position is given in centimeters and time is given in seconds). Restart. This motion is a mixture of a pure rotation and a pure translation. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. So how might we combine the two motions together so that the disk rolls without slipping?

First consider the various points on the surface of the rotating wheel. Since it has a constant angular velocity, every point has the same speed but a different velocity. Consider three special points: the top of the wheel, the wheel's hub, and the bottom of the wheel. The top of the wheel has a velocity \(v =\omega R\), and the velocity points to the right. The hub has zero velocity. And the bottom of the wheel has a velocity \(v =\omega R\) to the left.

Now consider the pure translation. Every point on the wheel has a velocity \(v\) to the right.

So how do we combine the two motions together to get rolling without slipping? If the velocity of the point at the bottom of the wheel—the point that touches the ground—has a velocity of zero with respect to the ground, the wheel will not slip.

Consider the three special points again: the top of the wheel, the wheel's hub, and the bottom of the wheel. We will add the translational velocity to the rotational velocity and see what we get. The top of the wheel has a rotational velocity \(v = \omega R\) to the right, which when combined with the translational velocity of \(v\) to the right gives us \(2v\) to the right. The hub has zero rotational velocity, which when combined with the translational velocity of \(v\) to the right gives us \(v\) to the right. And finally, the bottom of the wheel has a velocity \(v =\omega R\) to the left, which when combined with the translational velocity of \(v\) to the right gives us \(0\)!

Therefore, as long as the angular velocity gives us a \(v\) that is the same \(v\) as the translation, we have rolling without slipping as in Animation 3.

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

Illustration 3: Translational and Rotational Kinetic Energy

How do we describe rolling without slipping from an energy standpoint? We already know how to represent the kinetic energy of translation: \((1/2) mv^{2}\). We also know how to represent the kinetic energy of rotation: \((1/2) I\omega^{2}\). But what if we have both? Restart.

As the ball rolls down the incline, the gravitational potential gets transformed into kinetic energy, but how much of each? With rolling without slipping, we found that there is a relationship between the linear velocity and the angular velocity: \(v =\omega R\). Given this relationship we know that \(KE_{\text{trans}} = (1/2) mv^{2}\), while \(KE_{\text{rot}} = (1/2) I (v^{2}/R^{2})\). But the moment of inertia always looks like \(CmR^{2}\), so we find that \(KE_{\text{rot}} = (1/2) Cmv^{2}\). Therefore, we find that \(KE_{\text{total}} = (1+C) (1/2) mv^{2}\). The gravitational potential energy gets transformed into the total kinetic energy, and what fraction goes into \(KE_{\text{trans}}\) or \(KE_{\text{rot}}\) is determined by the constant \(C\). Specifically,

\[KE_{\text{trans}}/KE_{\text{total}}=1/(1+C)\quad\text{and}\quad KE_{\text{rot}}/KE_{\text{total}}=C/(1+C)\nonumber\]

A ball of radius \(1.0\text{ m}\) and a mass of \(0.25\text{ kg}\) rolls down an incline, as shown (position is given in meters and time is given in seconds). The incline makes an angle \(\theta = 20^{\circ}\) with the horizontal. Watch the graph of gravitational potential energy and rotational and translational kinetic energy vs. time or position.

Why do you think that the energy vs. time graphs curve, while the energy vs. position graphs are straight lines?

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

Illustration 4: Angular Momentum and Area

One of the strangest ideas regarding angular momentum is that an object moving in a straight line can have angular momentum. Angular momentum for a particle is given by the cross product \(\mathbf{L}=\mathbf{r}\times\mathbf{p}\). Given this, we see that the origin matters for the calculation of angular momentum for a particle.

In the absence of a net external torque acting on a system, a particle's angular momentum remains constant. For this discussion, the particle is free, so angular momentum should be conserved. Is there a different way to state the concept of angular momentum conservation? There may be. Consider the statement, Does a particle sweep out equal areas in equal times (with respect to any origin)?

Specifically, in this Illustration does a free particle moving in a straight line sweep out equal areas in equal times?

Press "start" to begin the animation, and let the show begin: A black dot will move freely from left to right. Different colors show the area the particle sweeps out with respect to some fixed point (the origin). Do all the areas have the same size? Click within each area and see what will happen. Certainly from mathematical equations we know the area of a triangle \(=\) width \(\ast\) height\(/2\). All of the areas have the same height and the same width (\(= v_{x}\ast dt\)).

Note

Kepler's second law (see Chapter 12 on gravitation for more details) states, during equal time intervals, the radius vector from the sun to a planet sweeps out equal areas. What does this tell you about the angular momentum of the planets?

Illustration authored by Fu-Kwun Hwang and Mario Belloni.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.

Illustration 5: Conservation of Angular Momentum

A red mass (\(1\text{ kg}\)) is incident on an identical black mass (\(1\text{ kg}\)) that is attached to a massless rigid string so that it can rotate around the origin as shown (position is given in meters and time is given in seconds). At \(t = 2.6\text{ s}\) the red mass undergoes a completely elastic collision with the black mass. Restart.

Consider the beginning part of the animation in which a \(1\text{-kg}\) red ball is incident on a \(1\text{-kg}\) black ball that is constrained to move in a circle. From which point should we measure the red ball's angular momentum? The best place, given the collision with the pendulum, is the point \((0, 0)\), the pivot. This is because we can easily measure the angular momentum of the pendulum about this point. What then is the angular momentum of the red ball before the collision? Certainly it must be changing, since \(\mathbf{r}\) changes. No! The angular momentum for a particle is given by the cross product: \(\mathbf{L}=\mathbf{r}\times\mathbf{p}\), which means we can consider the part of \(\mathbf{r}\) that is perpendicular to \(\mathbf{p}\) (\(rp \sin\theta\), where \(\theta\) is the angle between \(\mathbf{r}\) and \(\mathbf{p}\)). Since \(\mathbf{p}\) is in the negative \(x\) direction, the part of \(\mathbf{r}\) that is perpendicular to \(\mathbf{p}\) is \(y\). Therefore, \(|L| = 50\text{ kg}\cdot\text{m}^{2}/\text{s}\). Note that even though \(\mathbf{r}\) changes, \(y\) does not. The direction of the angular momentum is found with the RHR and is into the page (which is the negative \(z\) direction).

Now what happens to the angular momentum after the first collision? Given that only the black ball moves, we find that \(|L| = mvr = I\omega = 50\text{ kg}\cdot\text{m}^{2}/\text{s}\) (again, into the page). The angular momentum is the same as before the collision. Given that there are no external torques (The pendulum string does not create a torque. Why?), angular momentum is conserved.

What about after the second collision? Well, this is a bit harder. The radius vector \(\mathbf{r}\) changes (before the first collision the radius changed, but the part of the radius perpendicular to the momentum was constant). We must use a better definition of the magnitude of \(\mathbf{r}\times\mathbf{p}\) than \(rp \sin\theta\). In general we get for the \(z\) component of the angular momentum: \(L_{z} = (xp_{y} - yp_{x})\). At \(t = 16\) seconds, we have \((-5.06) (-1.73) - (-12.51) (-4.69) = -50 = 50\text{ kg}\cdot\text{m}^{2}/\text{s}\) (again, into the page).

Note

Note that in general, \(\mathbf{A}\times\mathbf{B} = (A_{y}B_{z} - A_{z}B_{y})\mathbf{i} + (A_{z}B_{x} - A_{x}B_{z})\mathbf{j} + (A_{x}B_{y} - A_{y}B_{x})\mathbf{k}\).

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