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

Solution

Physics sir · 3 steps

Friction is the same at every radius, so the coin slips wherever ω² r reaches μ g.

1. What holds the coin. Static friction is the only horizontal force on the coin, so it must supply the centripetal force mω² r. The most it can give is μ mg, and the coin just slips when the two are equal.

   μ m g = mω² r ⇒ ω² r = μ g = constant

2. Halve the speed. Since ω² r stays fixed, r ∝ 1/ω². Halving ω makes ω² one quarter, so r becomes 4 times larger: r₂ = 1 cm × 2² = 4 cm.
3. So. The coin now just slips at 4 cm from the centre, option (c).

The idea to keep. At the slipping point, μ g = ω² r. Mass drops out, and the slipping radius goes as 1/ω², so do not forget to square the change in ω.

Answer from the NTA final answer key. Solution by PrepWiser.