Finding the Velocity of a Mass in Circular Motion

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Homework Help Overview

The discussion revolves around finding the velocity of a mass in circular motion, specifically analyzing the dynamics of a mass (M) and a bob (m) in a system involving centripetal force and conservation of momentum. Participants are examining the conditions under which the mass moves through a circular path and the implications of forces acting on it.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between momentum and centripetal force, questioning the assumptions made about the forces acting on the mass as it moves through the circular path. There are attempts to reconcile different equations and clarify the conditions for the mass to maintain its motion.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations of the problem. Some have identified potential errors in reasoning and are considering the implications of energy conservation in the context of variable centripetal force. There is no explicit consensus yet, but the dialogue is productive.

Contextual Notes

Participants are working with a diagram and specific equations related to momentum and centripetal force, but there are indications of missing information or assumptions that need to be clarified. The original poster expresses confusion about a factor discrepancy in their calculations.

barryj
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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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barryj said:
Set V = V
Won't M slow down as it moves from the bottom to the top?
 
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?
 
barryj said:
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.
 
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I think I see the error. I will fix the error and re post tomorrow.
 
barryj said:

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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