A record turntable is rotating at 33 1/2 rev/min. A watermelon seed is on the tu
ID: 2034529 • Letter: A
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.
Explanation / Answer
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