# Rotational Motion & Energy Equation (Intro Physics)

1. Apr 5, 2010

### carlodelmundo

1. The problem statement, all variables and given/known data

[PLAIN]http://carlodm.com/phys/1.png [Broken]

2. Relevant equations

Energy Equation

3. The attempt at a solution

As shown in the scan.

Is this the correct approach? I have a feeling that the velocity, v, needs to be used. I used the Energy Equation for this problem but wasn't sure if it was the right equation to address kinetic energy, potential energy, and rotational motion.

Thanks

Last edited by a moderator: May 4, 2017
2. Apr 6, 2010

### Stonebridge

This assumes there is no energy lost due to friction, of course.
The initial energy is all potential in the mass M
At the end, the potential energy lost by the falling mass (Mgh) has appeared as kinetic energy in both the mass and the rotating table.
So you do need to consider the velocity v of the mass.

3. Apr 6, 2010

### carlodelmundo

Hi StoneBridge,

Thanks for your response. How do I account for the kinetic energy of the linear 'falling of the mass' when I am only given the velocity?

For the right hand side of the equation, do I just include another kinetic energy term (... KE = (1/2)Mv^2)

Would that account for the velocity?

Thanks

4. Apr 6, 2010

### Stonebridge

The question just asks that you express everything in terms of what's given in the question. You have the mass M and the velocity v of the mass as given.
So yes, ½Mv2 is the k.e. of the falling mass, and in your notes I see the expression for the k.e. of the turntable and the p.e. the mass has at the start.
So it's just a case of including all 3 terms and remembering that energy is conserved (no friction here!)

5. Apr 6, 2010

### carlodelmundo

Thanks a lot! I appreciate it!