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What is the velocity of the bullet to make the plumb bob swing to the top of the arc?

  • Thread starter barryj
  • Start date
  • #1
747
36

Homework Statement


See attached diagram and statement of problem

Homework Equations


Conservation of momentum ...
mv = MV + mv/2
mv/2 = MV
V = mv/(2M)

centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5

The Attempt at a Solution



Set V = V

mv/(2M) = (Lg)^.5
v = 2M(Lg)^.5/m

The answer is supposed to be v = 4M(Lg)^.5/m
I am off by a factor of 2 where did I go wrong???
 

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Answers and Replies

  • #2
TSny
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Set V = V
Won't M slow down as it moves from the bottom to the top?
 
  • #3
kuruman
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1. What is the speed at the top of the path when the bob just barely makes it?
2. Given you answer in 1, what is the centripetal acceleration?
 
  • #4
TSny
Homework Helper
Gold Member
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centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5
This would be valid if the stiff rod were replaced by a string. As @kuruman noted, you need to reconsider what the required speed of M must be in order to barely make it over the top.
 
  • #5
747
36
I think I see the error. I will fix the error and re post tomorrow.
 
  • #6
348
80

Homework Statement


See attached diagram and statement of problem

Homework Equations


Conservation of momentum ...
mv = MV + mv/2
mv/2 = MV
V = mv/(2M)

centripetal force...centripetal force equals weight at top of circle.
MV^2/L = Mg
V^2 = Lg
V = (Lg)^.5

The Attempt at a Solution



Set V = V

mv/(2M) = (Lg)^.5
v = 2M(Lg)^.5/m

The answer is supposed to be v = 4M(Lg)^.5/m
I am off by a factor of 2 where did I go wrong???
 
  • #7
348
80
Think, is the motion of the bob with a constant centripetal force or variable centripetal force. In all situations energy has to be conserved so why not try it out?
 

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