10.4 Rotational kinetic energy: work and energy revisited  (Page 2/9)

Now, we solve one of the rotational kinematics equations for $\text{αθ}$ . We start with the equation

Next, we solve for $\text{αθ}$ :

Substituting this into the equation for net $W$ and gathering terms yields

This equation is the work-energy theorem    for rotational motion only. As you may recall, net work changes the kinetic energy of a system. Through an analogy with translational motion, we define the term $\left(\frac{1}{2}\right){\mathrm{I\omega }}^2$ to be rotational kinetic energy     ${\text{KE}}_{\text{rot}}$ for an object with a moment of inertia $I$ and an angular velocity $\omega$ :

The expression for rotational kinetic energy is exactly analogous to translational kinetic energy, with $I$ being analogous to $m$ and $\omega$ to $v$ . Rotational kinetic energy has important effects. Flywheels, for example, can be used to store large amounts of rotational kinetic energy in a vehicle, as seen in [link] .

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 [link] . 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 $\text{1.00 rad}(\mathrm{57.3º})$ ? 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 $\text{net}W=\left(\text{net τ}\right)\theta$ . 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 ${\text{KE}}_{\text{rot}}=\frac{1}{2}{\mathrm{I\omega }}^2$ .

Solution for (a)

The net work is expressed in the equation

where net $\tau$ is the applied force multiplied by the radius $(\text{rF})$ 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

Noting that $\mathrm{1\; N}·\text{m}=\mathrm{1\; J}$ ,

Solution for (b)

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

Note that ${\omega }_0=0$ because we start from rest. Taking the square root of the resulting equation gives