Rotational Kinetic Energy

Example 10.8 Calculating the Work and Energy for Spinning a Grindstone

Consider a person who spins a large grindstone by placing her hand on its edge and exerting a force through part of a revolution as shown in Figure 10.17.

In this example, we verify that the work done by the torque she exerts equals the change in rotational energy. (a) How much work is done if she exerts a force of 200 N through a rotation of $1.00 \operatorname{rad}\left(57.3^{\circ}\right)$? The force is kept perpendicular to the grindstone's 0.320-m radius at the point of application, and the effects of friction are negligible. (b) What is the final angular velocity if the grindstone has a mass of 85.0 kg? (c) What is the final rotational kinetic energy? (It should equal the work.)

Strategy

To find the work, we can use the equation $\text { net } W=(\operatorname{net} \tau) \theta \text {. }$. We have enough information to calculate the torque and are given the rotation angle. In the second part, we can find the final angular velocity using one of the kinematic relationships. In the last part, we can calculate the rotational kinetic energy from its expression in $\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$.

Solution for (a)

The net work is expressed in the equation

$\text { net } W=(\text { net } \tau) \theta$,

where net $\tau$ is the applied force multiplied by the radius $(r F)$ because there is no retarding friction, and the force is perpendicular to $r$. The angle $\theta$ is given. Substituting the given values in the equation above yields

$\begin{aligned} \text { net } W &=r F \theta=(0.320 \mathrm{~m})(200 \mathrm{~N})(1.00 \mathrm{rad}) \\ &=64.0 \mathrm{~N} \cdot \mathrm{m} \end{aligned}$.

Noting that $1 \mathrm{~N} \cdot \mathrm{m}=1 \mathrm{~J}$,

$\text { net } W=64.0 \mathrm{~J} \text {. }$

Figure 10.17 A large grindstone is given a spin by a person grasping its outer edge.

Solution for (b)

To find $\omega$ from the given information requires more than one step. We start with the kinematic relationship in the equation

$\omega^{2}=\omega_{0}^{2}+2 \alpha \theta$.

Note that $\omega_{0}=0$ because we start from rest. Taking the square root of the resulting equation gives

$\omega=(2 \alpha \theta)^{1 / 2}$.

Now we need to find $\alpha$ . One possibility is

$\alpha=\frac{\text { net } \tau}{I}$.

where the torque is

$\text { net } \tau=r F=(0.320 \mathrm{~m})(200 \mathrm{~N})=64.0 \mathrm{~N} \cdot \mathrm{m}$

The formula for the moment of inertia for a disk is found in Figure 10.12:

$I=\frac{1}{2} M R^{2}=0.5(85.0 \mathrm{~kg})(0.320 \mathrm{~m})^{2}=4.352 \mathrm{~kg} \cdot \mathrm{m}^{2}$.

Substituting the values of torque and moment of inertia into the expression for $\alpha$, we obtain

$\alpha=\frac{64.0 \mathrm{~N} \cdot \mathrm{m}}{4.352 \mathrm{~kg} \cdot \mathrm{m}^{2}}=14.7 \frac{\mathrm{rad}}{\mathrm{s}^{2}}.$

Now, substitute this value and the given value for $\theta$ into the above expression for $\omega$:

$\omega=(2 \alpha \theta)^{1 / 2}=\left[2\left(14.7 \frac{\mathrm{rad}}{\mathrm{s}^{2}}\right)(1.00 \mathrm{rad})\right]^{1 / 2}=5.42 \frac{\mathrm{rad}}{\mathrm{s}}$.

Solution for (c)

The final rotational kinetic energy is

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$.

Both $I$ and $\omega$ were found above. Thus,

$\mathrm{KE}_{\mathrm{rot}}=(0.5)\left(4.352 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(5.42 \mathrm{rad} / \mathrm{s})^{2}=64.0 \mathrm{~J}$.

Discussion

The final rotational kinetic energy equals the work done by the torque, which confirms that the work done went into rotational kinetic energy. We could, in fact, have used an expression for energy instead of a kinematic relation to solve part (b). We will do this in later examples.

Example 10.9 Calculating Helicopter Energies

A typical small rescue helicopter, similar to the one in Figure 10.18, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length.

The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.

Solution for (a)

The rotational kinetic energy is

$\mathrm{KE}_{\mathrm{rot}}=\frac{1}{2} I \omega^{2}$.

We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find $\mathrm{KE}_{\text {rot }}$. The angular velocity$\omega$ is

$\omega=\frac{300 \mathrm{rev}}{1.00 \mathrm{~min}} \cdot \frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}} \cdot \frac{1.00 \mathrm{~min}}{60.0 \mathrm{~s}}=31.4 \frac{\mathrm{rad}}{\mathrm{s}}$

The moment of inertia of one blade will be that of a thin rod rotated about its end, found in Figure 10.12. The total $I$ is four times this moment of inertia, because there are four blades. Thus,

$I=4 \frac{M \ell^{2}}{3}=4 \times \frac{(50.0 \mathrm{~kg})(4.00 \mathrm{~m})^{2}}{3}=1067 \mathrm{~kg} \cdot \mathrm{m}^{2}$.

Entering $\omega$ and $I$ into the expression for rotational kinetic energy gives

$\begin{aligned} \mathrm{KE}_{\mathrm{rot}} &=0.5\left(1067 \mathrm{~kg} \cdot \mathrm{m}^{2}\right)(31.4 \mathrm{rad} / \mathrm{s})^{2} \\ &=5.26 \times 10^{5} \mathrm{~J} \end{aligned}$

Solution for (b)

Translational kinetic energy was defined in Uniform Circular Motion and Gravitation. Entering the given values of mass and velocity, we obtain

$\mathrm{KE}_{\text {trans }}=\frac{1}{2} m v^{2}=(0.5)(1000 \mathrm{~kg})(20.0 \mathrm{~m} / \mathrm{s})^{2}=2.00 \times 10^{5} \mathrm{~J}$.

To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

$\frac{2.00 \times 10^{5} \mathrm{~J}}{5.26 \times 10^{5} \mathrm{~J}}=0.380$.

Solution for (c)

At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:

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

or

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

We now solve for $h$ and substitute known values into the resulting equation

$h=\frac{\frac{1}{2} I \omega^{2}}{m g}=\frac{5.26 \times 10^{5} \mathrm{~J}}{(1000 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)}=53.7 \mathrm{~m}$.

Discussion

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades – something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.

Figure 10.18 The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades. The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown.

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.