A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis along the y- axis and free to rotate about its axis. The platform is given a motion in the x-direction given by x= Acos(ωt). There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is?
τ = Iα = F x R
Frictional force = μmg
The Attempt at a Solution
Differentiating the given equation 2 times, I found out that maximum acceleration = Aω²
Torque will be maximum when acceleration is maximum to prevent slipping.
If an observer is placed on the plank,
the cylinder will appear at rest to him.
Thus to balance forces, frictional force= Pseudo force = mAω²
⇒μmg = mAω²
⇒μ = Aω²/g
Now to find torque due to friction -
We take a small ring on the disk (bottom part of the cylinder) of thickness dx and radius x,
Mass of this ring = M/πr² (2πr dx)
⇒dτ = M/πr² (2πx dx) μg x
Integrating x from 0 to R
τ = 2/3 μMgR
Putting the value of μ
τ = 2/3 MRAω²
But the answer is τ = 1/3 MRAω²