physics question?

Question

A small block is placed on a rotating 45 degree funnel at a distance of 5 cm from the axis of rotation. The coefficient of friction of the block on the funnel surface is 0.5. Determine the minimum and maximum rotating rpms of the funnel such that the block does not slide up or down the funnel, but remains in the same fixed spot with respect to the funnel.

Answer 1

I assume that the funnel axis of rotation is vertical.
The body of mass m creates two forces: gravitational force mg and centrifugal force mrw^2, where w = 2*pi*f where f is revolution per sec. The vertical to the funnel surface total force = mg/sqrt(2) + mrw^2/sqrt(2). This force multiplied by friction coefficient gives the friction force parallel with funnel surface.
The driving force along the funnel surface that pulls the body down with small rps is mg/sqrt(2) - mrw^2/sqrt(2). The driving force that pulls the body up during high rps is mrw^2/sqrt(2) - mg/sqrt(2). (sqrt(2) comes from 45 degrees funnel slope).
For the body not to fall down the pulling force need to be smaller or equal friction force:
mg/sqrt(2) - rmw^2/sqrt(2) <= c(mg/sqrt(2) + mrw^2/sqrt(2)), where r is distance from axis, c is friction coefficient, and g is gravitationa constant 9.81 m/s^2. Therefore: rw^2(1+c) => g(1-c).
For the case of high rps pulling up force must be smaller or equal to the friction force:
rmw^2/sqrt(2) - mg/sqrt(2) <= c(mg/sqrt(2) + mrw^2/sqrt(2)), or
rw^2(1-c) <= g(1+c).
Therefore:
g(1-c)/(r(1+c)) <= w^2 <= g(1+c)/(r(1-c)), or
65.4 <= w^2 <= 588.6, or
8.087 <= w <= 24.261.
Therefore for the body not to slip either direction the revolutions per sec f must be:
1.288 <= f <= 3.863.
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