A uniform horizontal plank is resting symmetrically in a horizontal position on two cylindrical droms , which are apinning in in opposite direction about their horizontal axes with equal angular velocity . The distance between the axes with equal angular velocity. The distance between the axes is `2 L` and the coefficient of friction between the plank and cylender is `mu` . If the plank is displaced slightly from the equilibrium position along its length and released, show that it performs simple horizontal motion. Caculate also the time period of motion.

Explanation

When the plank is situated symmeetrically on the drums, the reations on the plank from the droms will be equal and so the force of friction will be equal in magnitude but opposite in direction and hence, the plank will be equilibrium along vertical as well as horizontal direction.

Now if plank is displacement by x to the right the reaction will not be equal. For vertical equilibrium of the plank.
`R_(A) + R_(B) = mg` (i)
And for rotational of plank, taking moment about center of mass, we have
`R_(A) (L + x) = R_(B) (L - x)` (ii)
Solving Eqs. (i) and (ii), we get `R_(A) = mg ((L - x)/(2 L))`
and `R_(B) = mg ((L - x)/(2 L))`
Now as `f = mu` R, so friction at B will be more than that at A and will bring the back, i.e. restoring force here is
`F = - (f_(B) - f_(A)) = - (f_(B) - f_(A)) = - mu (mg)/(L) x`
As the restoring force is linear, the motion will be simple harmonic motion with force constant `k = (mu mg)/(L)`
So `T = 2 pi sqrt((m)/(k)) = 2 pi sqrt((L)/(mu g))`