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Question

since the tension is always perpendicular to the velocity vector, the work done by the tension force will be zero. Hence, according to the work energy

Vedclass · Accepted Answer

$2$

Vedclass · Answer

since the tension is always perpendicular to the velocity vector, the work done by the tension force will be zero. Hence, according to the work energy theorem, the kinetic energy or velocity of the disc will remain constant during it's motion. Hence, the sought time $t=\frac{s}{v_{0}},$ where $s$ is the total distance traversed by the small disc during it's motion.

Now, at an arbitary position $(Fig.)$ $d s=\left(l_{0}-R 	hetaight) d 	heta$

so, $s=\int_{0}^{{l/R}}\left(l_{0}-R 	hetaight) d 	heta$

$\mathbf{o r}$

$s=\frac{l_{0}^{2}}{R}-\frac{R l_{0}^{2}}{2 R^{2}}=\frac{l_{0}^{2}}{2 R}$

Hence, the required time, $t=\frac{l_{0}^{2}}{2 R v_{0}}$

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