Question

An inclined plane of angle $\alpha$ is fixed onto a horizontal turntable with its line of greatest slope in same plane as a diameter of turntable. A small block is placed on the inclined plane a distance r from the axis of rotation of the turn-table and the coefficeint of friction is $\mu$. The turntable along with inclined plane spins about its axis with constant minimum angular velocity $\omega$.
a. Find an expression for the minimum angular velocity, ${\omega }_c$, to prevent the block from sliding down the plane, in terms of $g,r,\mu$ and the angle of the plane $\alpha$.
b. Now a block of same mass but having coefficient of friction (with inclined plane) $2\mu$ is kept instead of the original block. Find the ratio of friction force acting between block and inlined now to the friction force acting in part (a).

Answer 1

a. Net force along vertical axis is zero and along radial axis provides centripetal acceleration.
$\mu Nsin\alpha +Ncos\alpha -mg=0$
$Nsin\alpha -\mu Ncos\alpha =m{\omega }^{2}r$
$mg\frac{(sin\alpha -\mu cos\alpha )}{(\mu sin\alpha +cos\alpha )}=m{\omega }^{2}r$
$\omega =\sqrt{\frac{g(sin\alpha -\mu cos\alpha )}{r(\mu sin\alpha +cos\alpha )}}$
b. As block remains static at same height and radial distance, requirement of friction is same as in part (b).
Force of friction will remain unchanged. Hence ratio is 1.