# Rotational Inertia

1. Apr 27, 2007

### Weave

1. The problem statement, all variables and given/known data
A body radius R and mass m is rolling horizonttally without slipping with speed v. Then is rolls up a hill to maximum hieght h. If $$h=\frac{3v^2}{4g}$$
a) what is the body's rotational inertia?
b) What body might it be?

2. Relevant equations
$$h=\frac{3v^2}{4g}$$

3. The attempt at a solution
I have no clue!

2. Apr 27, 2007

### denverdoc

Show some effort, you just copied the problem under 2. Granted this is one of the tougher problems posted recently, but add some info re eqns for energy conservation in this instance.

3. Apr 27, 2007

### Weave

ok so
energy of conservation in this problem is $$K_{i} =U_{f}\longrightarrow\frac{1}{2}mv^2=mgh$$
and rotational inertia for a body is
$$I=\int r^2 dm$$

4. Apr 27, 2007

### Staff: Mentor

Call the rotational inertia I. Make use of the fact that it rolls without slipping. (Hint: The KE will be part translational and part rotational.)

5. Apr 27, 2007

### D H

Staff Emeritus
Some questions:
• What are the components of energy with which you need to be concerned? You have equations for translational kinetic energy and gravitational potential energy. Are you missing anything?
• What is the total energy of the body at the bottom of the hill (before it starts rolling up the hill)?
• What is the total energy of the body when it reaches the maximum height?

6. Apr 27, 2007

### Weave

so $$\frac{1}{2}I_{center of mass}w^2+\frac{1}{2}Mv_{center of mass}^2=Mgh$$ This is the potenial energy once it reaches its maximum hieght.
The intial KE is $$\frac{1}{2}I_{center of mass}w^2+\frac{1}{2}Mv_{center of mass}^2$$ since it starts from rolling
Components of energy are translational+rotational=rolling
and $$I=\int r^2 dm$$

Last edited: Apr 27, 2007
7. Apr 27, 2007

### D H

Staff Emeritus
The body is rolling without slipping. That means there is a relationship between the rotational velocity $\omega$ and the translational velocity $v$. You were also given a relationship between the height $h$ and the velocity. Use these relationships to simplify the equations.

8. Apr 27, 2007

### Weave

so you saying: $$v_{translational}=\omega R\longrightarrow\omega=\frac{v_{center of mass}}{R}$$
right?
I plugg that into my equation and the $$h=\frac{3v^2}{4g}$$

Last edited: Apr 27, 2007
9. Apr 27, 2007

### D H

Staff Emeritus
So what do you get?

10. Apr 27, 2007

### Weave

After simplifying every thing I got
$$I=\frac{M}{2}R^2$$
which happens to be the Rotational Inertia of a cylinder! AHH!

11. Apr 27, 2007

### D H

Staff Emeritus
Well done.