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Given a circular wheel that is rotating at the rate of $25$ revolutions per minute. If the radius of the wheel is $50 \space cm$, what could be the distance covered by a point on the rim in one second (Given that $\pi = 3.1416$)

Any takes? Thanks.

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    Is the wheel rolling along the ground? Or just rotating about a fixed axle? In the first case, the distance covered depends on the initial position of the point.2011-12-14

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The number of revolutions per second is $\def\textstyle{} 25{\textstyle {\text{ rev} \over \text{ min}}}\cdot\textstyle{1\over 60}\textstyle{\text{min}\over \text{sec} } ={25\over 60}{\text{rev}\over\text{sec}} ={5\over12}{\text{rev}\over\text{sec}}.$ The point travels $\pi\cdot2\cdot50=100\pi {\text{ cm}\over\text{rev}}$ . So, in one second, the distance would be $\underbrace{100\pi \textstyle {\text{ cm}\over\text{rev}}}_{\text{ dist per rev}}\cdot\underbrace{\textstyle {5\over12}\textstyle{\text{rev}\over\text{sec}}\cdot 1\text{sec}}_{ \text{number of revs.}}={125\pi\over3} \text{ cm}.$

See TonyK's comment. The above is for a wheel whose center remains stationary.

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    Yes, that's correct. The point travels $125\pi/3$ (which is the same as you have) centimeters per second. (In the above, the wheel is rotating at $5/12$ revs per second, which gives the angle as $\theta={5\over 12}\cdot 2\pi= {5\pi\over6}$.)2011-12-14