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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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Just use conversion factors. One revolution is $2\pi\times50\text{ cm}$ (why?), and there are sixty seconds in a minute... – J. M. Dec 14 '11 at 10:14
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. – TonyK Dec 14 '11 at 10:18

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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it would be total distance covered by a point of rim but how much distance would that be per second. Since, angle is described per second equals (Θ) = 50π/60 = 5π/6. S = r x Θ = 50cms x 5π/6 = 130.9cms. – Nikhil Mulley Dec 14 '11 at 10:39
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 $\theta={5\over 12}\cdot 2\pi= {5\pi\over6}$.) – David Mitra Dec 14 '11 at 10:55

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