A hollow cylinder of mass m and radius R is s | vedclass

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(A) For a hollow cylinder,the moment of inertia about its central axis is $I = mR^2$.When placed on the incline,the friction force $f$ acts up the pla

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$\omega_0 R / (g \sin 	heta)$

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(A) For a hollow cylinder,the moment of inertia about its central axis is $I = mR^2$.When placed on the incline,the friction force $f$ acts up the plane to oppose the slipping. The friction force is $f = \mu N = \mu mg \cos 	heta = (	an 	heta) mg \cos 	heta = mg \sin 	heta$.The torque $	au$ about the centre of mass is $	au = fR = (mg \sin 	heta) R$.Using $	au = I\alpha$,we get $(mg \sin 	heta) R = (mR^2) \alpha$,which gives the angular acceleration $\alpha = (g \sin 	heta) / R$.Since the cylinder is spinning clockwise,the friction force acts up the plane,causing an angular deceleration. The final angular velocity $\omega = \omega_0 - \alpha t$.Setting $\omega = 0$ to find the time when the cylinder stops spinning (and thus the centre of mass remains stationary until pure rolling begins),we have $0 = \omega_0 - (g \sin 	heta / R) t$.Solving for $t$,we get $t = \omega_0 R / (g \sin 	heta)$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

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