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

## 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

TSny
Homework Helper
Gold Member
Set V = V
Won't M slow down as it moves from the bottom to the top?

kuruman
Science Advisor
Homework Helper
Gold Member
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?

TSny
Homework Helper
Gold Member
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.

• scottdave
I think I see the error. I will fix the error and re post tomorrow.

## 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???

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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