PHYS101: Introduction to Mechanics

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.