A coin rests 15.0 cm from the center of a turntable. The coefficient of static f

Question

A coin rests 15.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.350. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.730 rad/s2. Once the turntable starts to rotate, the coin tends to move outward from the center of the turntable. However, the static frictional force provides the centripetal force to keep the coin at rest relative to the turntable. After what period of time will the coin start to slip on the turntable?

Explanation / Answer

1. the force is the friction between the coin and surface if mv^2/r or mw^2 r (where w is angular velocity) exceeds u mg, then the coin will slip 2. w(t)=w0+alpha t = 0+0.730t we find when m w^2 r> u mg w^2 r = (0.730t)^2 r > u g t^2>u g/[(0.730)^2 r] solve the complete equation for answer!!!

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