Aplatform is rotating at an angular speed of 2.5 rad/s. a block is resting on this platform at a distance of 0.28 m from the axis. the coefficient of static friction between the block and the platform is 0.74. 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

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 

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explanation:

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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