A coin rests 16.0 cm from the center of a turntable. The coefficient of static f

Question

A coin rests 16.0 cm from the center of a turntable. The coefficient of static friction between the coin and turntable surface is 0.320. The turntable starts from rest at

t = 0

and rotates with a constant angular acceleration of 0.620 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 / Answer

(a)
Answer is Fricton Force.
Coin begin to move relative to the turntable when the Centripetal Force exceeds the Frictional Force between the Coin and Table.

(b)
Now, Max Angular Velocity achieved by coin before sliiping.
m*v^2/r = u*m*g
Where we know, v = r*w
r*^2 = u*g
= sqrt(u*g/r)
= sqrt((0.320 * 9.8)/0.16)
= 4.43 rad/s

Time taken to reach this speed ,
= + t
4.43 = 0 + 0.620 *t
t = 7.15 s
Period of time after which the coin will start to slip on the turntable, t = 7.15 s

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