# A case of Rolling

1. Nov 17, 2008

### ritwik06

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

A uniform cylinder of mass m of radius r(smaller) rolls on a cylindrical surface of radius R(larger). At a certain instant, the line OC has an angular velocity $$\omega$$ and angular acceleration $$\alpha$$.

Find
a) the acceleration of point of contact P (of the 2 cylinders) with respect to the surface.
b) the kinetic energy of the cylinder

3. The attempt at a solution
First of all since the line joining the centres goes through an angular velocity of $$\omega$$. The velocity of the centre of mass of the small cylinder with respect to the centre of fixed cylinder is $$\omega(R+r)$$. The angular velocity about its centre would be $$\frac{\omega(R+r)}{r}$$. I am stuck now.

For
a) I think the acceleration of this point= r * $$\beta$$ where $$\beta$$is the angular acceleration of the small cylinder about its centre.
$$\beta=\frac{d(\frac{\omega(R+r)}{r})}{dt}$$

Since it is given that d$$\omega$$/dt=$$\alpha$$
$$\beta=\frac{(R+r)\alpha}{r}$$

But the answer provide is not r * $$\beta$$. I am confused.

b) For the kinetic energy of the cylinder, I used
KE= $$0.5*m*v^{2}+0.5*I (\frac{\omega(R+r)}{r})^2$$

where I=mr*r
and v=$$\omega(R+r)$$

Last edited by a moderator: May 3, 2017
2. Nov 17, 2008

### tiny-tim

Hi ritwik06!
You're misunderstanding the question … "with respect to the surface" means with respect to the larger cylinder … in other words, with respect to the ground.

Try again!

3. Nov 17, 2008

### ritwik06

Hi tim,
I did not have a confusion regarding that. What I have tried to calculate is in accordance with ground frame. Is it not?

4. Nov 18, 2008

### tiny-tim

Hi ritwik06!

(have an alpha: α and a beta: β and an omega: ω and a squared: ² )
No, the acceleration of point of contact P (of the 2 cylinders) with respect to the surface (the ground frame) is rα.

(rβ is the acceleration with respect to the smaller cylinder)