Question

A block is resting on a platform that is rotating at an angular speed of 1.9 rad/s. The coefficient of static friction between the block and the platform is 0.77. Determine the smallest distance from the axis at which the block can remain in place without skidding as the platform rotates.

Answer 1

IMPORTANT NOTE - THE QUESTION IS MISSING SOMETHING, PROBABLY THE ORIGINAL DISTANCE FROM AXIS OF ROTATION

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block will move when centripetal force exceeds force of static friction

so,

Let d be smallest distance and w be the final angular velocity

then

u_smg = mdw²

u_sg = dw²

If we treat the block as point mass

then

w = (D² / d² ) w_i

where D is original distance from center axis ( this should have been given in the question)

then

u_sg = d * (D⁴ / d⁴ ) w²_i

simplifying, we get

u_sg = (D⁴ / d³ )w_i

so

d³ = D⁴ * w_i² / u_sg

given data,

w_i = 1.9 rad/sec

u_s = 0.77 g = 9.8 m/s²

so, you just need value of D which was missing in question to solve for d.

d = ( D⁴ * w_i² / u_sg)^1/3