Max/min values of v before m slips

[holezch]

Homework Statement

A very small cube of mass m is placed on the inside of a funnel rotating about a vertical axis at a constant rate of v revs/sec. The wall of the funnel makes an angle theta with the horizontal. If the coefficient of the static friction between the cube and the funnel is Mu, and the centre of the cube is a distance r from the axis of rotation, what are the a) largest and b) smallest values of v for which the cube will not move with respect to the funnel?

Homework Equations

friction static = MuN
F = Ma

The Attempt at a Solution

So, I found the min value of V.. I rotated my axis by theta and broke down my force vectors F = m v^2/r and W = mg. Then MuN = mgsintheta + m(v^2/r)costheta and then you can solve for v.. that's the minimum value of v.. I think I understand the question conceptually (the min value would be the friction keeping the block from falling, and the max value would be the value when which the block's frictional force overcomes everything else and starts sliding up). But how do I find the max value? thanks!

 

[holezch]

okay, I think I got it.. since it will be v^2/r, +/- the appropriate V will be the solutions..
is this right? thanks

 

[kerimek]

The maximum value of v can be found by setting the frictional force equal to the maximum possible static frictional force, which is given by MuN. This maximum value occurs when the cube is just about to slip, so the normal force N will be equal to the weight of the cube, mg. Substituting this into the equation for frictional force, we get:

MuN = mgsintheta + m(v^2/r)costheta

Solving for v, we get:

v = sqrt((Mu*g*sintheta*r)/(costheta - Mu))

So the maximum value of v is given by this equation.