Question

A block is resting on a platform that is rotating at an angular speed of 2.4 rad/s. The coefficient of static friction between the block and the platform is 0.83. Determine the smallest distance from the axis at which the block can remain in place wothout skidding as the platform rotates.

Answer 1

Answer:

r = 0m is the Minimum distance from the axis at which the block can remain in place wothout skidding.

Explanation:

From a sum of forces:

$Ff = m*a$   where Ff = μ * N    and $a = \frac{V^2}{r}=\omega^2*r$

N - m*g = 0   So, N = m*g.   Replacing everything on the original equation:

$\mu*m*g = m*\omega^2*r$   (eq2)

Solving for r:

$r = \frac{\mu*g}{\omega^2}=1.41m$

If we analyze eq2 you can conclude that as r grows, the friction has to grow (assuming that ω is constant), so the smallest distance would be 0 and the greatest 1.41m. Beyond that distance, μ has to be greater than 0.83.