IIT JEE , AIEEE Forum - Physics , Mechanics

Tom_Bombadil

If V is the speed of the COM of the middle cylinder, and w is its angular speed:

V + w(R-r)/2 = w₂R - for point of contact with the outer cylinder

V - w(R-r)/2 = w₁r - for the inner cylinder

Solving these for w yeilds V = (w₁r + w₂R)/2.

Then, as in the previous solution, angular velocity about O becomes 2V/(R+r), or (w₁r + w₂R)/(R + r).

I am not confident that the above analysis is incorrect. Any input would be appreciated.

karthik2007

Yes, Tom is right. I too got option b).

karthik2007

@computer - you have got your constraints wrong dude. Check them again.

computer001

@ tom:" u shud be considerin the rel motion @ the bottom dude..
pl chk
n @karthik:im sure :\

madman

option a) is correct
here is my solution
i am going to solve the sum from the frame of the inner cylinder
velocity of the top most part of the outer cylinder with respect to the inner cylinder = Rw2 + rw1
therefore velocity of the centre of the sandwiched cylinder
=(Rw2 + rw1)/2 (pure rolling)
velocity of the surface of inner cylinder = wr1
velocity of centre of sandwiched cylinder from ground frame =
(Rw2 + rw1)/2 - rw1 = (Rw2 - rw1)/2
since they are asking its angular velocity about the centre
w = (Rw2 - rw1 )/2((R-r)/2 +r ) =(Rw2 - rw1 )/(r+R)

Ask & Discuss Questions with Community & Experts