Question

A record turntable is rotating at 33 1/2 rev/min. A watermelon seed is on the turntable 9.4 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 0.24 s. Calculate the minimum coefficient of static friction required for the seed not to slip during the acceleration period.

Answer 1

here,

the angular speed , w = 33.5 rev/min = 3.5 rad/s

r = 9.4 cm = 0.094 m

a)

the accelration of seed , ac = w^2 * r

ac = 3.5^2 * 0.094 m/s^2

ac = 1.16 m/s^2

b)

let the minimum value of the coefficient of static friction between the seed and the turntable be us

equating the forces

us * m * g = m * ac

us * 9.81 = 1.16

us = 0.12

c)

time taken , t = 0.24 s

let the angular accelration be alpha

w = 0 + alpha * t

3.5 = 0 = alpha * 0.24

alpha = 14.6 rad/s^2

let the minimum coefficient of static friction required for the seed not to slip during the acceleration period be us

equating the forces

us * m * g = m * ( w^2 * r)

us * 9.81 = 3.5^2 * 0.094

solving for us

us = 0.12