A block of mass m = 20 kg is kept at a distance R = 1 m from central axis of rotation of a round turn table table whose surface can rotate about central axis). Table starts from rest and rotates with

Question

A block of mass m = 20 kg is kept at a distance R = 1 m from central axis of rotation of a round turn table table whose surface can rotate about central axis). Table starts from rest and rotates with constant angular acceleration, α = 3 rad / sec 2 . The friction coefficient between block and table is μ = 0.5 . At time t = x /3 from starting of motion (i.e. t = 0 sec ) the block is just about to slip. Find the value of x .

TestHub · Accepted Answer

The block is about to slip when the net force equals the limiting friction: $m a_{t}^{2} + a_{r}^{2} = μ m g$ , where $a_{t} = α R$ and $a_{r} = ω^{2} R = (α t)^{2} R$ . Substituting these into the equation and solving for $t$ , we get $t = \frac{μ ^{2} g ^{2} + α ^{2} R ^{2} - α R}{α ^{2} R}$ . Given $t = x /3$ , we find $x = 3 t = 3 \frac{μ ^{2} g ^{2} + α ^{2} R ^{2} - α R}{α ^{2} R} = 2$ .

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