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 0.25 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Explanation / Answer

I assume a steady rotation. Convert speed from RPM into an omega in rad/sec.

(A)
The acceleration of the watermellon seed is centripetal, since it is traveling in a circular path.
a_cent = omega^2 * R

(B)
Friction holds it in place. Friction is proportional to normal force, via roughness coefficient mu. Normal force is equal and opposite weight in order that it doesn't fall through turntable surface.

F = mu*m*g

mu*m*g = m*a_cent
mu*m*g = m*omega^2 * R

Solve for mu:
mu = omega^2*R/g

(C) The watermelon seed also undergoes tangential acceleration, a perpendicular component to centripetal acceleration.

Tangential acceleration:
a_tan = alpha*R

alpha = omega/?t

thus, a_tan = omega*R/?t

Combine accelerations in Pythagorean theorem combination:
a_net = sqrt(a_tan + a_cent)

a_net = sqrt(omega*R/?t + omega^2*R)

In order that the seed remains in place without slipping, this acceleration must be caused by friction.
mu*m*g = m*a_net

mu*g = sqrt(omega*R/?t + omega^2*R)

Result for mu:
mu = sqrt(omega*R/?t + omega^2*R)/g

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Results for question
Data:
omega := 10*Pi/9 rad/sec; R:=0.015 m; ?t:=0.5 sec; g:=9.8 N/kg;

Part A) a_cent = omega^2 * R
a_cent = 0.1827 m/sec^2

Part B) mu = omega^2*R/g
mu = 0.01865

Part C) mu = sqrt(omega*R/?t + omega^2*R)/g
mu = 0.1712

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