Solving Rolling Homework Statement: Acceleration & Kinetic Energy

  • Thread starter Thread starter ritwik06
  • Start date Start date
  • Tags Tags
    Rolling
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 4K views
ritwik06
Messages
577
Reaction score
0

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 [tex]\omega[/tex] and angular acceleration [tex]\alpha[/tex].
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 [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@!
 
Last edited by a moderator:
Physics news on Phys.org
Hi ritwik06! :smile:
ritwik06 said:
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].

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. :wink:

Try again! :smile:
 
tiny-tim said:
Hi ritwik06! :smile:


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. :wink:

Try again! :smile:

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?
 
Hi ritwik06! :smile:

(have an alpha: α and a beta: β and an omega: ω and a squared: ² :wink:)
ritwik06 said:
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. :wink:

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