Solving Monkey Swing Physics Problem

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The discussion revolves around a physics problem involving a monkey swinging from two vines. The monkey initially hangs at rest, and upon releasing one vine, it swings on the other. Participants suggest using conservation of energy to determine the speed at the lowest point of the swing and discuss the need to calculate the height difference for potential energy. Tension in the remaining vine is also a key focus, with participants confirming the inclusion of the monkey's weight in calculations. The conversation emphasizes the importance of energy conservation principles in solving the problem effectively.
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Homework Statement


http://img256.imageshack.us/my.php?image=b911om5.jpg

sin30= .50 sin60= .87
cos30= .87 cos60= .50
tan30= .58 tan60= 1.73

A 5.0 kilogram monkey hangs initially at rest from two vines. A and B, as shown above. Each of the vines has length 10 meters and negligible mass.

For parts c and d:
The monkey releases vine A and swings on vine B. Neglect air resistance.
c. Determine the speed of the money as it pass through the lowest point of it's first swing.

d. Determine the tension in vine B as the monkey passes through the lowest point of its first swing.

Homework Equations


Centripetal force? m(v^2/r)


The Attempt at a Solution



Completed parts a and b. Force diagram and tensions in each vine before release. T1= 9N and T2= 5.13N

I'm not sure as what I am to do... I'm thinking that it has something to do with centripetal force but not sure cause he is swinging.
 
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Part C - the easiest way is probably conservation of energy.
Calculate the difference in height between it's initial position and the position when the vine is at it's lowest point. PE = m g h and KE = 1/2 m v^2

Part D - don't forget you also have the weight of the monkey.
 
that makes sense, thanks for the tip. Never would have thought to use conservation of energy.
 
Hothot330 said:
Never would have thought to use conservation of energy.
It's always a good bet - it can't necessarily answer all questions (part D for instance) but when it's right it's right! And is generally simple, you don't have to worry about angles and vectors etc.
 
ahh but how do you calculate the height? Are you sure the tension in both a and b are correct hothot?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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