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:
[tex]Ff = m*a[/tex] where Ff = μ * N and [tex]a = \frac{V^2}{r}=\omega^2*r[/tex]
N - m*g = 0 So, N = m*g. Replacing everything on the original equation:
[tex]\mu*m*g = m*\omega^2*r[/tex] (eq2)
Solving for r:
[tex]r = \frac{\mu*g}{\omega^2}=1.41m[/tex]
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.
