(e) If the coin is placed at a radius of 10 cms and the turn table is rotating at a speed of 6 radians /sec, the inward force acting on the coin will be mω²r = 2(180)= 360 gm cm/sec² .

If the coefficient of friction is μ between the coin and the turn table, then force of Friction would be μR, where is R= 2g, where 2g is the weight of the coin in gm cm/sec². In this case g will be equal to 981cm/sec².

Hence μ (2)(981) = 360

Which gives μ= 360/ 2(981) = 0.1834

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