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.

Rotational Kinetic Energy: Work and Energy Revisited

In this module, we will learn about work and energy associated with rotational motion. Figure 10.14 shows a worker using an electric grindstone propelled by a motor. Sparks are flying, and noise and vibration are created as layers of steel are pared from the pole. The stone continues to turn even after the motor is turned off, but it is eventually brought to a stop by friction. Clearly, the motor had to work to get the stone spinning. This work went into heat, light, sound, vibration, and considerable rotational kinetic energy.

The figure shows a mechanic cutting metal with a metal grinder. The sparks are emerging from the point of contact and jumping off tangentially from the cutter.

Figure 10.14 The motor works in spinning the grindstone, giving it rotational kinetic energy. That energy is then converted to heat, light, sound, and vibration.


Work must be done to rotate objects such as grindstones or merry-go-rounds. Work was defined in Uniform Circular Motion and Gravitation for translational motion, and we can build on that knowledge when considering work done in rotational motion. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk (as shown in Figure 10.15) and remains perpendicular as the disk starts to rotate. The force is parallel to the displacement, and so the net work done is the product of the force times the arc length traveled:

\text { net } W=(\text { net } F) \Delta s.

To get torque and other rotational quantities into the equation, we multiply and divide the right-hand side of the equation by r, and gather terms:

\text { net } W=(r \text { net } F) \frac{\Delta s}{r}.

We recognize that r \text { net } F=\text { net } \tau and  \Delta s / r=\theta , so that

 \text { net } W=(\operatorname{net} \tau) \theta.

This equation is the expression for rotational work. It is very similar to the familiar definition of translational work as force multiplied by distance. Here, torque is analogous to force, and angle is analogous to distance. The equation n \text {
    net } W=(\text { net } \tau) \theta is valid in general, even though it was derived for a special case.

To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that net \text { net } \tau=I \alpha \text {, } so that

 \text { net } W=I \alpha \theta

The figure shows a circular disc of radius r. A net force F is applied perpendicular to the radius, rotating the disc in an anti-clockwise direction and producing a displacement equal to delta S, in a direction parallel to the direction of the force applied. The angle covered is theta.

Figure 10.15 The net force on this disk is kept perpendicular to its radius as the force causes the disk to rotate. The net work done is thus (\operatorname{net} F) \Delta s. The net work goes into rotational kinetic energy.

Making Connections

Work and energy in rotational motion are completely analogous to work and energy in translational motion, first presented in Uniform Circular Motion and Gravitation.

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

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

Next, we solve for \alpha \theta \text {: }

 \alpha \theta=\frac{\omega^{2}-\omega_{0}^{2}}{2} .

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

\text { net } W=\frac{1}{2} I \omega^{2}-\frac{1}{2} I \omega_{0}^{2}.

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) I \omega^{2}
    . to be rotational kinetic energy \mathrm{KE}_{\mathrm{rot}} for an object with a moment of inertia  I and an angular velocity \omega:

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

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 Figure 10.16.

The figure shows a bus carrying a large flywheel on its board in which rotational kinetic energy is stored.

Figure 10.16 Experimental vehicles, such as this bus, have been constructed in which rotational kinetic energy is stored in a large flywheel. When the bus goes down a hill, its transmission converts its gravitational potential energy into\mathrm{KE}_{\text {rot }}. It can also convert translational kinetic energy, when the bus stops, into \mathrm{KE}_{\text {rot }}. The flywheel's energy can then be used to accelerate, to go up another hill, or to keep the bus from slowing down due to friction.

 


Source: Rice University, https://openstax.org/books/college-physics/pages/10-4-rotational-kinetic-energy-work-and-energy-revisited
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