A coin rests 12.5 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.330. The turntable starts from rest at t = 0 and rotates with a constant angular acceleration of 0.580 rad/s2.

(a) Once the turntable starts to rotate, what force causes the centripetal acceleration when the coin is stationary relative to the turntable? Under what condition does the coin begin to move relative to the turntable?
(b) After what period of time will the coin start to slip on the turntable?

Explanation:

Radius of circular path of coin R = 12.5  x 10⁻²,

coefficient of static friction μs = .33

In order that the coin rotates in circular path , it requires centripetal force which is provided by friction. As speed of rotation increases , force of friction also increases to provide it required centripetal force. When the speed of rotation becomes too high so that frictional force can not compensate the increase in centripetal force then coin will start slipping or it starts moving with respect to turntable.

b ) At this point of time

centripetal force = limiting force of friction

mω² R = μs mg  , m is mass of the coin , ω is angular velocity ,

ω² R = μs g

ω² x 12.5 x 10⁻², = .33 x 9.8

ω² = .33 x 9.8 /  12.5 x 10⁻²,

= 25.87

ω = 5.08 rad / s

2π / T = 5.08

T = 2π / 5.08

= 1.23 s .

where are the vectors ?

f=ma newton

f=100x2.5=250n