(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]\omega[/tex] and angular acceleration [tex]\alpha[/tex].

http://img82.imageshack.us/img82/5380/cylinerssm0.jpg [Broken]

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 [tex]\omega[/tex]. The velocity of the centre of mass of the small cylinder with respect to the centre of fixed cylinder is [tex]\omega(R+r)[/tex]. The angular velocity about its centre would be [tex]\frac{\omega(R+r)}{r}[/tex]. I am stuck now.

For

a) I think the acceleration of this point= r * [tex]\beta[/tex] where [tex]\beta[/tex]is the angular acceleration of the small cylinder about its centre.

[tex]\beta=\frac{d(\frac{\omega(R+r)}{r})}{dt}[/tex]

Since it is given that d[tex]\omega[/tex]/dt=[tex]\alpha[/tex]

[tex]\beta=\frac{(R+r)\alpha}{r}[/tex]

But the answer provide is not r * [tex]\beta[/tex]. I am confused.

b) For the kinetic energy of the cylinder, I used

KE= [tex]0.5*m*v^{2}+0.5*I (\frac{\omega(R+r)}{r})^2[/tex]

where I=mr*r

and v=[tex]\omega(R+r)[/tex]

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

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# Homework Help: A case of Rolling

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