Asked by phil

A bicycle wheel of radius 0.4 m is spinning at a rate of 40 revolutions per minute.

What is the centripetal acceleration of a point on the edge of the wheel?

What is the period of the wheel's motion?

Answers

Answered by bobpursley
40rpm=80PI/60 radians/s =1 1/3 PI rad/s=w

centripetal acceleartion=w^2 r

Period=1/freq=2PI/w
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