Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 7.8 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

Answer 1

Given data:

● Angular velocity of the turntable ω = 33 rev/min

Therefore,

ω = 33 rev/min

ω = 33rev/min × 2π rad/rev × 1 min/60 sec = 3.45 rad/s

● The distance of the watermelon seed from the axis of rotation r = 7.8 cm = 0.078 m

● μs = the coefficient of static friction

Section a:

The seed is undergoes circular motion and it is been effected by centripetal acceleration.

ac = rω^2

ac = 0.078 × 3.45^2

ac = 0.9284 m/s^2

Therefore,

the centripetal acceleration of the seed is 9.274 m/s^2

Section b:

If the seed is observed not to slip at the course of the circular motion, then the supplied frictional force given by the seed and surface of turntable would at least be equivalent to the centripetal force working on the seed.

Centripetal Force = Frictional Force

mrω^2 = μsmg

μs = rω^2 /g

μs = 0.078 × 3.45^2

------------

9.81

μs = 0.09464

Thus,

the coefficient of static friction is 0.09464.