A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the centre. If the angular velocity of the table is halved, it will just slip when placed at a distance of _____ from the centre:

Step-by-step solution

The coin will slip when μ m g = m ω 2 r ⇒ μ g = ω 2 × 1   cm       . . . ( i ) Now, ω ' = ω 2 μ g = ω 2 2 R       . . . ( i i ) Hence we can write from (i) and (ii) ω 2 × 1   cm = ω 2 4 R ⇒ R = 4   cm