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

Question

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 1.5 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? (c) Suppose that the turntable achieves its angular speed by starting from rest and undergoing a constant angular acceleration for 0.39 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Explanation / Answer

Part

Acceleration of seed is given by:

ac = v^2/r

v = w*r

ac = w^2*r

r = 1.5 cm = 0.015

w = 33 rev/min = 33*2*pi/60 = 3.45 rad/sec

So,

ac = 3.45^2*0.015 = 0.178 m/sec^2

Part B

Using force balance, if the seed is not to slip

Ff = Fnet

Ff = us*m*g

Fnet = m*ac

us*m*g = m*ac

us = coefficient of static friction

us = ac/g = 0.178/9.81 = 0.018

Part C

In this case net acceleration will be

a = sqrt (ac^2 + at^2)

ac = 0.178 m/sec^2

at = r*alpha = r*w/t

at = 0.015*3.45/0.39 = 0.133 m/sec^2

Now

a = sqrt (0.178^2 + 0.133^2) = 0.222 m/sec^2

So,

us = a/g = 0.222/9.81 = 0.023

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