Question

A platform is rotating at an angular speed of 1.62 rad/s. A block is resting on this platform at a distance of 0.437 m from the axis. The coefficient of static friction between the block and the platform is 0.720. Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates.

Answer 1

Solution-

Let us assume m be the mass of the block

Initial angular speed wi = 1.62 rad/s
Initial distance from axis ri = 0.437 m
Coefficient of static friction u =0.720

Let rf be the smallest distance from the axis at which the block can remain in place

By using the consevation of angular momentum,
m ri^2 wi = m rf^2 wf
Dividing both sides by m,
ri^2 wi = rf^2 wf
wf = wi (ri/rf)^2 --------(1)

At distance rf, the block just begins sliding.
Hence static friction force = u m g
This provides centripetal force = m wf^2 * rf
Therefore
m wf^2 * rf = u m g
Dividing both sides by m,
wf^2 * rf = u g
wf = sqrt(u g/rf) -----------(2)

From equations (1) and (2),
wi (ri/rf)^2 = sqrt(u g/rf)
wi ri^2/rf^2 = sqrt(u g) / sqrt(rf)
rf^2/sqrt(rf) = wi ri^2/sqrt(u g)

rf^(3/2) = wi ri^2/sqrt(u g)
= 1.62 * 0.437 ^2/sqrt(0.720 * 9.8)
rf^(3/2) =0.1165
rf = 0.1164^(2/3)
rf = 0.24 m