A platform 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 ax

First let us set the variables:

m = mass of the block <span>
Initial angular speed wi = 2.5 rad/s </span>
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 

<span>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 
<span>wf = wi (ri/rf)^2              >>>> (1)

<span>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 
<span>Equating the 2 forces:
m wf^2 * rf = u m g 
Cancelling m on both sides: 
wf^2 * rf = u g 
<span>wf = sqrt(u g/rf)                         >>>> (2) </span>

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) </span></span></span></span>

Substituting given values:<span>
rf^(3/2) = 2.5 * 0.28^2/sqrt(0.74 * 9.8) 
<span>rf^(3/2) = 0.07278
rf = 0.07278^(2/3) 
rf = 0.17 m 

Ans: 0.17 m<span> </span></span></span>