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0 votes
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in Physics by (84.3k points)
A uniform disc of mass M and radius R is rotating freely about its central vertical axis with angular speed w0. Another disc of mass m and radius r is free to rotate about a horizontal rod AB. Length of the rod AB is`L (lt 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 figure. Find the time after which the slipping between the two discs will cease. Assume that normal reaction between the two discs is equal to mg. Coefficient of friction between the two discs is `mu`.
image

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2 Answers

0 votes
by (82.1k points)
Correct Answer - `(MR^(2) L omega_(0))/(2mu g [MR^(2) + mL^(2)])`
+1 vote
by (20 points)

Consider the Free Body Diagram of the system at any general time.

Now assume after time t slipping ceases and pure rolling starts, and at that time angular velocity of horizontal disc is \( \omega \) 

and the angular velocity of vertical disc is \( \omega' \)

So we have

(1) \( r\omega ' = L\omega \)

(2) \(\omega = \omega _0 - \frac{{\mu mgL}}{{MR^2 /2}}t \)

(3) \( \omega ' = \frac{{L\omega }}{r} = 0 - \frac{{\mu mgr}}{{Mr^2 /2}}t \)

from (2) and (3) 

\( \frac{L}{r}\left\{ {\omega _0 - \frac{{\mu mg2L}}{{MR^2 }}t} \right\} = \frac{{\mu mgr2}}{{mr^2 }}t \)

\( \Rightarrow t = \frac{{MR^2 L\omega }}{{2\mu g\left( {MR^2 + mL^2 } \right)}} \)

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