Max/min values of v before m slips

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SUMMARY

The discussion focuses on determining the maximum and minimum values of rotational speed (v) for a cube of mass m placed inside a funnel at an angle theta. The static friction coefficient (Mu) plays a crucial role in this analysis. The minimum value of v is derived using the equation MuN = mgsintheta + m(v^2/r)costheta, where N is the normal force. The maximum value of v is identified as the point where the frictional force is overcome, leading to the cube sliding up the funnel.

PREREQUISITES
  • Understanding of static friction and its coefficient (Mu)
  • Knowledge of basic physics concepts including force vectors and rotational motion
  • Familiarity with Newton's second law (F = Ma)
  • Ability to manipulate trigonometric functions in physics equations
NEXT STEPS
  • Study the derivation of forces in rotating systems, particularly in conical shapes
  • Learn about the dynamics of friction in circular motion scenarios
  • Explore advanced applications of static friction in engineering contexts
  • Investigate the effects of varying angles (theta) on rotational stability
USEFUL FOR

This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the dynamics of rotating systems and frictional forces.

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

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