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Torque and angular momentum of a swing

  • Thread starter dimpledur
  • Start date
  • #1
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Homework Statement


A gymnas is performing a giant swing on the high bar. IN a simplified model of the giant swing, assume that the gymnast keeps his arms and body straight. Also assume that the gymnast does no work. With what angular speed should he be moving at, at the bottom of the swing, to ensure that he can make it all the way around? The distance from the bar to his feet is 2m and his centre of gravity is 1m in.



The Attempt at a Solution



I'm not sure if I am supposed to apply conservation of energy for this question, or if there is a simpler way?

I was going to say:

Fnet=mv^2/r
Fnet*r/m = v^2
sqrt (Fnet*r/m) = v

and then convert v into angular velocity.

However, I dunno how to solve for Fnet etc.

I'm not sure how to approach this problem at all. Any help is greatly appreciated.
 

Answers and Replies

  • #2
194
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Actually, I may have figured it out. Correct me if I am wrong, please.

Since the question states that no work is done by the man swinging, then there is no translational kinetic energy. However, there is rotational kinetic energy.

Therefore, we can state:

Rotational kinetic energy at the bottom = potential energy at the top
Krot = Ut
1/2Iw^2 = mgh
w2 = 2mgh/I

We can have the man swinging assume the shape of a rod, where I = 1/3ML^2

so
w^2 = (2mgh)/(1/3ML^2)
w^2 = 6gh/L^2
w = sqrt (6gh) / L

now im not sure if L is the distance to the centre of gravity, or if it is the entire length of the rod. However, I assumed it was the distance of the entire rod.

so, w = sqrt (6*9.8*4) / 2 = 7.67rad /s


?????????
 
  • #3
194
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perhaps the question is unclear?
 
  • #4
43
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Well, it seems we are in the very same class and are having the very same problem. I came up with the same solution as yourself, and am having a similar problem completing the question.

I am unsure of what values to use for h and L
 

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