IIT JEE , AIEEE Forum - Physics , Mechanics

karthik2007

Friction is the force that retards the first disc and accelerates the second disc. Hence, to find the frictional torque on a disc, consider an annular ring of thickness dx. Its area = $\pi(x+dx)^2 - \pi x^2$ = $2\pi xdx$

Hence, mass of the disc = $\frac{2mxdx}{r^2}$ (Assuming areal distribution of mass, and assuming m is the mass of the disc)

Hencr frictional force = $\frac{2\mu mgxdx}{r^2}$

Torque of friction about this elemental ring about center = $\frac{2 \mu mgx^2dx}{r^2}$

Total torque of friction = $\int^r_0 \frac{2\mu mgx^2}{r^2} \, dx = \frac{2\mu mgr}{3}$

Now, |angular acceleration| for either disc is found by using:

$\frac{2\mu mgr}{3} = \frac{mr^2 \alpha}{2}$

$\alpha = \frac{4\mu g}{3r}$

For the discs to acquire a common angular velocity, we have:

$\omega - \frac{4\mu g}{3r} t = 0 + \frac{4\mu g}{3r} t$

Solving, we get t = $\frac{3r \omega}{8\mu g}$ .

(excuse me for calculation mistakes , if I have made any)

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