# Angular mometume question

1. Feb 23, 2009

### physics_geek

1. The problem statement, all variables and given/known data
A solid sphere is released from height h from the top of an incline making an angle θ with the horizontal.
(a) Calculate the speed of the sphere when it reaches the bottom of the incline in the case that it rolls without slipping. (Use g,h, and theta for θ as necessary.)
(b) Calculate the speed of the sphere when it reaches the bottom of the incline in the case that it slides frictionlessly without rolling. (Use g,h, and theta for θ as necessary.)
(c) Compare the time intervals required to reach the bottom in cases (a) and (b).
rolling time/sliding time =

2. Relevant equations
no idea

3. The attempt at a solution

2. Feb 23, 2009

### Delphi51

You can do questions like this using forces and accelerations, but it is MUCH easier to do them with energy formulas. Basically,
Energy at top = energy at bottom
PE at top = KE at bottom + rotational energy at bottom

With that start, you can't go wrong - begin by putting in the detailed formula for each kind of energy.

3. Feb 23, 2009

### physics_geek

yea thanks a lot!
i think i got it

4. Feb 24, 2009

### Bob S

Let's answer b) first. The total kinetic energy at the bottom is mgh, independent of theta. When the sliding ball reaches the bottom, mgh = (0.5)mv^2, and v=sqr(2gh).
Let's answer a) now. Again the total energy is mgh. However now, the potential energy has been converted to both energy of translation (0.5)mv^2 AND energy of rotation. Because the moment of inertia of a uniformly dense sphere is (2/5)mr^2, the energy of rotation is
(1/5)m(r^2)(dtheta/dt)^2 = (1/5)mv^2. So when the ball rolls without slipping, the kinetic energy at the bottom is mgh = (1/2 + 1/5)mv^2 = (7/10)mv^2, and so now v=sqr(10gh/7). So now the translational velocity is lower.