Solving Rolling Homework Statement: Acceleration & Kinetic Energy

[ritwik06]

Homework Statement

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.
http://img82.imageshack.us/img82/5380/cylinerssm0.jpg

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

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 \betais 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)

But I get wrong results for both. Please Help@!

 

[tiny-tim]

Hi ritwik06!

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.

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!

 

[ritwik06]

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!

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?

 

[tiny-tim]

Hi ritwik06!

(have an alpha: α and a beta: β and an omega: ω and a squared: ² )

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?

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)