A small roller of diameter 20 cm has an axle | vedclass

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(C) Let $R$ be the radius of the roller and $r$ be the radius of the axle.Given: $2R = 20 \ cm \implies R = 10 \ cm$ and $2r = 10 \ cm \implies r = 5

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(C) Let $R$ be the radius of the roller and $r$ be the radius of the axle.Given: $2R = 20 \ cm \implies R = 10 \ cm$ and $2r = 10 \ cm \implies r = 5 \ cm$.For the roller to roll without slipping on the floor,the velocity of the center of the roller is $V_{	ext{center}} = \omega R$,where $\omega$ is the angular velocity.The velocity of the scale,which is on top of the axle,is $V_{	ext{scale}} = V_{	ext{center}} + \omega r$.Substituting $\omega = \frac{V_{	ext{center}}}{R}$,we get:$V_{	ext{scale}} = V_{	ext{center}} + \left(\frac{V_{	ext{center}}}{R}ight)r = V_{	ext{center}} \left(1 + \frac{r}{R}ight)$.Given $r = 5 \ cm$ and $R = 10 \ cm$,we have $V_{	ext{scale}} = V_{	ext{center}} \left(1 + \frac{5}{10}ight) = 1.5 V_{	ext{center}}$.If the roller moves a distance $d_{	ext{roller}} = 50 \ cm$ in time $t$,then $V_{	ext{center}} \cdot t = 50 \ cm$.The distance moved by the scale in the same time $t$ is $d_{	ext{scale}} = V_{	ext{scale}} \cdot t = 1.5 V_{	ext{center}} \cdot t = 1.5 	imes 50 \ cm = 75 \ cm$.Thus,the scale moves $75 \ cm$ relative to the ground,while the center of the roller moves $50 \ cm$ relative to the ground. The position of the scale relative to the center of the roller is $75 \ cm - 50 \ cm = 25 \ cm$ ahead of the center.

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