Rotational Kinetic Energy

As you read, pay attention to the diagram of a spinning disk in Figure 10.15. For the disk to spin, work must be done on the disk. The force acting upon the disk must be perpendicular to the radius of the disk, which we know is torque. We also know torque is related to moment of inertia. We can relate the work done on the disk to moment of inertia using the equation $W=\tau\theta=I\alpha\theta$ .

Example 10.8 shows how to calculate the net work for a rotating disk using this work equation. In the second part of the example, the rotational velocity is determined using the equation for rotational acceleration and moment of inertia. Lastly, it uses this equation to calculate the rotational kinetic energy.

How Thick Is the Soup? Or Why Don't All Objects Roll Downhill at the Same Rate?

One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest?

The easiest way to answer these questions is to consider energy. Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy $\mathrm{PE}_{\text {grav }}$ , which is converted entirely to $KE$ , provided each rolls without slipping. $KE$ , however, can take the form of $\mathrm{KE}_{\text {trans }}$ or $\mathrm{KE}_{\text {rot }}$ , and total $KE$ is the sum of the two.

If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can's original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure 10.19.

The figure shows a flat surface inclined at a height of h from the surface level, with three cans of soup of different densities numbered as one, two, and three rolling along it.

Figure 10.19 Three cans of soup with identical masses race down an incline. The first can has a low friction coating and does not roll but just slides down the incline. It wins because it converts its entire PE into translational KE. The second and third cans both roll down the incline without slipping. The second can contains thin soup and comes in second because part of its initial PE goes into rotating the can (but not the thin soup). The third can contains thick soup. It comes in third because the soup rotates along with the can, taking even more of the initial PE for rotational KE, leaving less for translational KE.

Assuming no losses due to friction, there is only one force doing work – gravity. Therefore the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives

$\mathrm{PE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}$ .

More specifically,

$\mathrm{PE}_{\text {grav }}=\mathrm{KE}_{\text {trans }}+\mathrm{KE}_{\text {rot }}$

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$ .

So, the initial $m g h$ is divided between translational kinetic energy and rotational kinetic energy; and the greater $I$ is, the less energy goes into translation. If the can slides down without friction, then $\omega=0$ and all the energy goes into translation; thus, the can goes faster.

Take-Home Experiment

Locate several cans each containing different types of food. First, predict which can will win the race down an inclined plane and explain why. See if your prediction is correct. You could also do this experiment by collecting several empty cylindrical containers of the same size and filling them with different materials such as wet or dry sand.

Example 10.10 Calculating the Speed of a Cylinder Rolling Down an Incline

Calculate the final speed of a solid cylinder that rolls down a 2.00-m-high incline. The cylinder starts from rest, has a mass of 0.750 kg, and has a radius of 4.00 cm.

Strategy

We can solve for the final velocity using conservation of energy, but we must first express rotational quantities in terms of translational quantities to end up with $v$ as the only unknown.

Solution

Conservation of energy for this situation is written as described above:

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}$ .

Before we can solve for $v$ , we must get an expression for $I$ from Figure 10.12. Because $v$ and $\omega$ are related (note here that the cylinder is rolling without slipping), we must also substitute the relationship $\omega=v / R$ into the expression. These substitutions yield

$m g h=\frac{1}{2} m v^{2}+\frac{1}{2}\left(\frac{1}{2} m R^{2}\right)\left(\frac{v^{2}}{R^{2}}\right)$ .

Interestingly, the cylinder's radius $R$ and mass $m$ cancel, yielding

$g h=\frac{1}{2} v^{2}+\frac{1}{4} v^{2}=\frac{3}{4} v^{2}$ ,

Solving algebraically, the equation for the final velocity $v$ gives

$v=\left(\frac{4 g h}{3}\right)^{1 / 2}$ .

Substituting known values into the resulting expression yields

$v=\left[\frac{4\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(2.00 \mathrm{~m})}{3}\right]^{1 / 2}=5.11 \mathrm{~m} / \mathrm{s}$ .

Discussion

Because $m$ and $R$ cancel, the result $v=\left(\frac{4}{3} g h\right)^{1 / 2}$ is valid for any solid cylinder, implying that all solid cylinders will roll down an incline at the same rate independent of their masses and sizes. (Rolling cylinders down inclines is what Galileo actually did to show that objects fall at the same rate independent of mass.)

Note that if the cylinder slid without friction down the incline without rolling, then the entire gravitational potential energy would go into translational kinetic energy. Thus, $\frac{1}{2} m v^{2}=m g h$ and $v=(2 g h)^{1 / 2}$ , which is 22% greater than $(4 g h / 3)^{1 / 2}$ . That is, the cylinder would go faster at the bottom.