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A uniform disc of mass M and radius R is rotating freely about its stationary smooth central vertical axis with angular speed $ω_{0} .$ Another disc of mass m and radius r is free to rotate about a horizontal rod AB. Length of the rod AB is $L (< R) .$ And its end A is rigidly attached to the vertical axis of the first disc. The disc of mass m, initially at rest, is placed gently on the disc of mass M as shown in the figure. Find the time in seconds after which the slipping between the two discs will cease. Assume that normal reaction between the two disc is equal to mg, coefficient of friction between the two discs is $μ$ given M = 1 kg, m = 1 kg, L = 0.5 m, $μ = 0.1, ω_{0} = 5 r a d / \sec$

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Detailed Solution

$f = μ m g$

$T_{A B} = μ m g r$

$α = \frac{μ m g r}{m r^{2} / 2} = \frac{2 μ g}{r}$

$ω = α t = \frac{2 μ g}{r} . t$

T on larger disc = $μ m g L = \frac{m R^{2}}{2} . α_{0}$

$α_{0} = \frac{2 μ m g L}{M R^{2}}, ω_{L} = ω_{0} - α_{0} t$

No slipping, $ω_{L} L = ω r$

$t = \frac{M R^{2} ω_{0} L}{2 μ g (M R^{2} + m L^{2})}$

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