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# Rolling Energy Problems

## Explores equations used to describe the energy of rolling or sliding objects.

Estimated3 minsto complete
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Progress
Practice Rolling Energy Problems
Progress
Estimated3 minsto complete
%
Rolling Energy Problems

Students will learn how to use energy conservation in situations that include rotating objects. Remember, conservation of energy still holds. A rolling object has two parts to its kinetic energy: the linear kinetic energy that you've already done, and this rotational kinetic energy. The method to calculate the rotational energy from the moment of inertia and rotational velocity is shown below.

### Key Equations

Einitial=EfinalThe total energy does not change in closed systems\begin{align*} \sum E_{\text{initial}} = \sum E_\text{final} \; \; \text{The total energy does not change in closed systems}\end{align*}

\begin{align*}KE_{rot} = \frac{1}{2} I\omega^2\end{align*}

Guidance
You can use energy conservation in rotation problems just like you did in the energy conservation lessons for linear motion. Note that rotating objects also have kinetic energy of rotation and this must be included when accounting for energy conservation.

#### Example 1

A hoop of mass 1 kg and radius .75 m, begins at rest at the top of a ramp, 3m above the ground. What is it's rotational velocity at the bottom of the ramp if the hoop rolls down the ramp without slipping?

##### Solution

To solve this problem, we'll apply energy conservation to the hoop. The hoop only has gravitational potential energy at the top of the ramp, and it has both rotational and linear kinetic energy at the bottom of the ramp.

\begin{align*} E_i&=E_f\\ PE&=KE + KE_{rot} && \text{start with conservation of energy}\\ mgh&=\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 && \text{substitute the proper value for each energy term}\\ mgh&=\frac{1}{2}mv^2 + \frac{1}{2}mr^2\omega^2 && \text{substitute in the moment of inertia of the hoop}\\ mgh&=\frac{1}{2}m(\omega r)^2 + \frac{1}{2}mr^2\omega^2 && \text{we can put } v \text{ in terms of omega because the hoop is rolling without sliding}\\ 2gh&=(\omega r)^2 + r^2\omega^2 && \text{now we will multiply by 2 and divide by m to simplify the equation}\\ 2gh&=2\omega^2r^2\\ \omega&=\frac{\sqrt{gh}}{r} && \text{solving for }\omega\\ \omega&=\frac{\sqrt{9.8\;\text{m/s}^2 * 3\;\text{m}}}{.75\;\text{m}} && \text{plug in the known values}\\ \omega&=7.2\;\text{rad/s}\\ \end{align*}

### Time for Practice

1. Your bike brakes went out! You put your feet on the wheel to slow it down. The rotational kinetic energy of the wheel begins to decrease. Where is this energy going?
2. The upper pulley has mass of 2kg and a radius of 30cm. The lower pulley has mass of 4kg and a radius of 1.0m. The 12 kg mass is 5m off the ground.
1. With what velocity will the 12kg mass hit the ground if μk=0?
2. With what velocity will the 12kg mass hit the ground if μk=0.2?

1. The energy goes into the heat generated by the friction between your foot and the tire.

2. a. 5.87m/s b. 4.79 m/s