First let us set the variables:
m = mass of the block
Initial angular speed wi = 2.5 rad/s
Initial distance from axis ri = 0.28 m
Coefficient of static friction u = 0.74
Say the smallest distance from the axis in which the block
can remain in place = rf
Using conservation of angular momentum:
m ri^2 wi = m rf^2 wf
Since m is constant, we can cancel that out:
ri^2 wi = rf^2 wf
wf = wi (ri/rf)^2 >>>>
(1)
At distance rf, the block just starts to slide, therefore
static friction force = centripetal force:
static friction force = u m g
centripetal force = m wf^2 * rf
Equating the 2 forces:
m wf^2 * rf = u m g
Cancelling m on both sides:
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)
Finally we get:
rf^(3/2) = wi ri^2/sqrt(u g)
Substituting given
values:
rf^(3/2) = 2.5 * 0.28^2/sqrt(0.74 * 9.8)
rf^(3/2) = 0.07278
rf = 0.07278^(2/3)
rf = 0.17 m
Ans: 0.17 m
