Strange friction AP

  • #1
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A small block of mass m is on a horizontal frictionless surface as it travels around the inside of a hoop of radius R. The coefficient of friction between the block and the wall is mu; therefore, the speed v of the block decreases. In terms of m, R, mu, and v, find expressions for:

a. the frictional force on the block. I got this part, friction is the centripetal force or mv^2/R.

b. the block's tangential acceleration, dv/dt.

c. the time required to reduce the speed of the block from an initial value v0 to v0/3.

I haven't formally taken calc yet (this is an APC mechanics free response), but I think I have a good enough understanding of it. The thing that hangs me up is that friction is not proportional to weight as usual. The only friction is caused by the hoop. The force on the hoop at an instant is the velocity. Does this mean friction = mu x v?.
I think I'm on the right track, the sliding causes a friction force which lowers the velocity which in turn lowers the friction force, lowering the velocity, etc. But how do I express this mathematically and answer the last two parts?
 

Answers and Replies

  • #2
The friction will still be proportional to the normal force.
 
  • #3
312
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Thanks a lot, its not as hard as I thought then. So the opposite of friction (mu x mg) is ma and a=-mu x g??
 
  • #4
The normal force of a mass on a horizontal surface is often equal to mg, not always, though. I think that must be where you are getting your mu x mg from. But it doesn't apply in this case, because only fhe hoop has friction, so the weight of the mass (mg) doesn't enter into this problem at all. I don't see g as being relevant to solving this problem, since the hoop is horizontal.

Instead, focus on the force perpendicular to the wall of the hoop. That's the normal force, N. Then the frictional force will be mu*N as always, and that is what is creating the tangential acceleration. I think you are very close to solving it.

Dorothy
 

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