A coin of mass m is placed on top of a horizontal turntable a distance R from the axis of rotation. Gravity is acting vertically downward with magnitude mg and the normal force on the coin from the turntable is acting vertically upward. The coefficient of static friction between the coin and the turntable is µ. If at time t = 0 the turntable is at rest and accelerates with a constant angular acceleration α < µg/R, at what time T will the coin begin to slide on the turntable?

Question

Kelsie Andersen

Physics

23 June, 06:55

A coin of mass m is placed on top of a horizontal turntable a distance R from the axis of rotation. Gravity is acting vertically downward with magnitude mg and the normal force on the coin from the turntable is acting vertically upward. The coefficient of static friction between the coin and the turntable is µ. If at time t = 0 the turntable is at rest and accelerates with a constant angular acceleration α < µg/R, at what time T will the coin begin to slide on the turntable?

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Answers (1)

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Cade Ali · Answer 1

T = 6.28fR²/µg

Explanation:

a=µg/R

(v-vi) / (t-ti) = µg/R

v: linear velocity of turntable at which coin begins to slip

vi: initial velocity

t: time at which coin begins to slide

ti: time at which turntable started moving

(v-0) / T=µg/R

T=Rv/µg

to find v,

v = wR

here, w is angular velocity and is equal to 2πf where f is frequency of turntable rotation

v = 2πfR

so,

T = R (6.28fR) / µg

T = 6.28fR²/µg