Question

A block of mass m = 20 kg is kept at a distance R = 1m from central axis of rotation of a round turn table (A table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, $\alpha$ = 3 rad/sec². The friction coefficient between block and table is $\mu$ = 0.5. At time t = $\frac{x}{3}$ sec from starting of motion (i.e. t = 0 sec) the block is just about to slip. Find the value of x.

Answer 1

2

Answer 2

The block on the rotating turntable experiences both tangential and centripetal accelerations. The tangential acceleration is $a_t = \alpha R$, and the centripetal acceleration is $a_c = \omega^2 R = (\alpha t)^2 R$. Since these are perpendicular, the net acceleration required is $a_{net} = \sqrt{a_t^2 + a_c^2}$. This acceleration is provided by the static friction force. When the block is just about to slip, the required force $m a_{net}$ equals the maximum static friction force $\mu mg$. Equating $a_{net} = \mu g$ and substituting the expressions for accelerations in terms of $\alpha$, $R$, and $t$, we solve for $t$. Finally, comparing the calculated $t$ with the given $t = x/3$ yields the value of $x$.