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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 9.6 cm from the axis of rotation. (a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip

1 Answer

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Answer:

(A) Acceleration will be
1.145rad/sec^2

(B) Coefficient of static friction will be 0.116

Step-by-step explanation:

We have given angular speed


\omega =33rpm=(2\pi * 33)/(60)=3.454rad/sec

Distance from the axis r = 9.6 cm = 0.096 m

(a) Acceleration is equal to


a_c=\omega ^2r=3.454^2* 0.096=1.145rad/sec^2

(b) For seed is not to slip


ma=\mu mg


\mu =(a)/(g)


\mu =(1.145)/(9.8)=0.116

So coefficient of static friction will be 0.116

User Jonathan Hartley
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