# Homework Help: Friction challange

1. Jul 29, 2007

### gandharva_23

consider a funnel rotating about its axis with a constant angular speed w . A small particle remains at rest w.r.t. the wall of the funnel . If we solve the problem in particles reference frame we have a centrifugal force mw^2 r whose component will be tending to move the particle up the funnel's wall . Hence i can say that for certain values of w frictional force will be acting in the downward direction . Now i want to solve the problem in the groung frame . If i analyse the problem in ground frame there is no force that is tending to take the particle up the wall so friction force can never act down the wall ....

2. Jul 29, 2007

### mgb_phys

A couple of important points, the laws of physics don't depend on your frame of reference and centripetal force is tricky sometimes.

3. Jul 29, 2007

### gandharva_23

"laws of physics don't depend on your frame of reference "
thats what i want to ask .... in ground frame which force is trying to drag the particle up the plane ? How can i obtain a range of frictional force for which the particle remains in equilibrium w.r.t funnel by solving the problem in ground frame . we know that for larger values of w frictional force will be acting in the downward direction . If we are observing the particle in ground frame then which force is trying to drag the particle up the funnel if we are saying that friction is acting in down the funnel wall ?

4. Jul 29, 2007

### Staff: Mentor

Real forces--like friction--exist independent of reference frame. In the ground frame the particle is accelerating--and the only forces able to provide that acceleration are the normal force and friction. Hint: Consider vertical and horizontal force components; apply Newton's 2nd law.

In the accelerated frame the particle is at rest and, as you know, you must add a "fictitious" centrifugal force acting on the particle to produce equilibrium. (In an accelerated frame you must modify Newton's laws by adding such fictitious forces.)

Either way, you can solve for the range of friction force required.