Revolutions of disk?

Question

A horizontal disk of radius 5.00 cm accelerates angularly from rest with an angular acceleration of 0.250 rad/s^2. A small pebble located half way out from the center has a coeffient of static friction with the disk of 0.15.

a) how long will the pebble take to slip on the disk surface?

b) how many revolutions will the disk have rotated through that time?

Answer 1

In uniform circular motion the net force 'F' provides an acceleration towards the center of rotation:

F = m * a = m * v^2 / r

The friction force just before the pebble slips is also 'F'

F = mu * m * g

where mu is the coefficient of static friction

solve for v: v = sqrt (mu * r * g)

v = sqrt (0.15 * 0.025 * 9.81) = 0.19 m/s

omega = v / r = 0.19 / 0.025 = 7.6 rad/s

the time to slip 't' = omega / alpha

alpha = angular acceleration

t = 7.6 / 0.25 = 30.4 seconds

the radians to slip = 1/2 alpha * t^2

number of radians = 115.5

number revolutions = 115.5 / 2 / pi = 18.4