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Kinetic / Potential Energy Problem

  • Thread starter x93
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  • #1
x93
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Homework Statement


A small block of mass m starts from rest and slides along a frictionless loop-the-loop as shown in the image. What should be the initial height h, so that m pushes against the top of the track (at a) with a force equal to its weight?

Ans. h=3R

prob17a.gif



Homework Equations


KE=(1/2)mv2
PE=mgh
Fc=mv2/r


The Attempt at a Solution


PEi=mgh
KEa=PEi-PEa=mgh-2mgR
(1/2)m(va)2=mgh-2mgR
va=√(2gh-4gR)
Fc=mg
mg=(2mgh-4mgR)/R
2h-4R=R
h=(5/2)R

I think I attempted the problem in the correct manner, but perhaps made a subtle algebraic mistake since the coefficient is only off by 1/2.
 

Answers and Replies

  • #2
PhanthomJay
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Your error is in the calculation of Fc, which is not equal to mg. The net centripetal force, Fc, is the sum of all forces acting on the block in the cenripetal direction.
 
  • #3
x93
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Okay, would the net centripetal force be 2mg? I'm not sure that I understand why. My guess is that the centripetal force is constant all the way around the loop, and at the bottom/start of the loop the centripetal force is equal to the normal force mg. Therefore, the magnitude of the net force at the top of the loop is the weight of the mass, mg, plus the normal force, mg. Is that right?
 
  • #4
PhanthomJay
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Okay, would the net centripetal force be 2mg?
yes...
I'm not sure that I understand why. My guess is that the centripetal force is constant all the way around the loop,
no, it is not constant, since v is changing as the object makes its way around the loop.
and at the bottom/start of the loop the centripetal force is equal to the normal force mg.
no-o.
Therefore, the magnitude of the net force at the top of the loop is the weight of the mass, mg, plus the normal force, mg. Is that right?
Yes, but you have the wrong reasoning. Draw free body diagrams. At the top of the loop, there are 2 forces acting downward (centripetally) on the object. Since it is given that the mass pushes (upward) on the track at the top of the loop with a (normal) force of mg, then by Newton's 3rd Law, the track normal force pushed downward on the object with a force of mg. The object's weight also acts downward on it. The net sum of these 2 forces is ___?___., which is the net centripetal force acting on the object.
 
  • #5
x93
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Ah, okay, I see. Thanks for your help.
 

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