Trapeze performer

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  • #1
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Homework Statement

A trapeze performer swinging in a vertical circle can be modeled as a rotating bar pivoted at one end, as shown in figure 8.26. Let the mass of the bar be 70 kg with the center of mass located 1.2 m form the axis of rotation. If the angular velocity at the top of the swing is 3 rad/s, what is the value of w at the bottom of the swing? (hint use the principle of conservation of energy.)



Homework Equations






The Attempt at a Solution


I am not really sure where to begin with this problem. I was thinking that I could use kinetic and potential energy to solve this problem and basically convert the potential energy at the top to kinetic energy to find the velocity but it doesn't say how high the bar is.
 

Answers and Replies

  • #3
Nathanael
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It tells you that "the center of mass located 1.2 m from the axis of rotation"

So how much does the height (of the center of mass) change from the top of the swing to the bottom?
 
  • #4
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It tells you that "the center of mass located 1.2 m from the axis of rotation"

So how much does the height (of the center of mass) change from the top of the swing to the bottom?
2.2 meters
 
  • #5
Nathanael
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2.2 meters
Well no, it would be 2.4 meters, but I think that you just made a typo.

So you can now use the (change in) gravitational potential energy (along with conservation of energy) to solve the problem.
 
  • #6
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Well no, it would be 2.4 meters, but I think that you just made a typo.

So you can now use the (change in) gravitational potential energy (along with conservation of energy) to solve the problem.
OK well I did that and I am getting very close to the right answer. So
PE = mgy
PE = 70(9.8)(2.4)
PE = 1646.4

1646.4 = 1/2mv^2
v = 6.85857
v = rw
w = 5.72 rad/s
The correct answer is supposedly 5.79 rad/s
 
  • #7
Nathanael
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You used the equation for linear kinetic energy.

Are you familiar with the equation for rotational kinetic energy?
 
  • #8
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You used the equation for linear kinetic energy.

Are you familiar with the equation for rotational kinetic energy?
Yes but how do I find the moment of inertia of a bar?
 
  • #9
Nathanael
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Yes but how do I find the moment of inertia of a bar?
Two options, Calculus or Google :)

The moment of inertia is [itex]\frac{1}{3}MR^2[/itex]

EDIT: By "R" I meant "L" as in the length of the bar
 
Last edited:
  • #10
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Two options, Calculus or Google :)

The moment of inertia is [itex]\frac{1}{3}MR^2[/itex]

EDIT: By "R" I meant "L" as in the length of the bar
using that I got 8.4 which is still the wrong answer
1646.4 = 1/3IW^2
W = the sqaure root of 70.56
W = 8.2
 
  • #11
Nathanael
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If you want to do it by calculus, it would be like something like this:

[itex]I=\frac{M}{L}(^{L}_{0}∫x^2.dx)=\frac{1}{3}ML^2[/itex]

If you really really want, I can explain that equation, but I'm not great at explaining.
(I've barely just learned calculus, so I'm not the best at explaining it :tongue:)
 
  • #12
Nathanael
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1646.4 = 1/3IW^2
No, it should be [itex]\frac{1}{2}I\omega ^2=\frac{1}{2}\frac{ML^2}{3}\omega ^2=\frac{1}{6}ML^2\omega ^2[/itex]


But another thing, you forgot to include the rotational energy that it already had at the top of the swing (the problem said it was moving at 3 rad/s)
 
  • #13
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No, it should be [itex]\frac{1}{2}I\omega ^2=\frac{1}{2}\frac{ML^2}{3}\omega ^2=\frac{1}{6}ML^2\omega ^2[/itex]


But another thing, you forgot to include the rotational energy that it already had at the top of the swing (the problem said it was moving at 3 rad/s)
OK now I got the total energy to be 1797.6
so
1797.6 = 1/6MR^2w^2
1797.6 = 1/6(70)(1.2^2)w^2
w = 10.344 which is not correct
 
  • #14
Nathanael
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OK now I got the total energy to be 1797.6
so
1797.6 = 1/6MR^2w^2
1797.6 = 1/6(70)(1.2^2)w^2
w = 10.344 which is not correct
I think you're very close, you just made a mistake about the length of the object. The length of the object is not 1.2m

The problem does not tell you the length of the object, but if you assume that the object is uniform (uniform in mass) then you can deduce that the length of the object is 2.4m (because the center of mass would be at the midpoint, which was 1.2m)

So the energy is not 1796.6 (that would be the energy if the length was 1.2m)

Use the correct length of the object and see if you get the correct answer
 
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  • #15
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I think you're very close, you just made a mistake about the length of the object. The length of the object is not 1.2m

The problem does not tell you the length of the object, but if you assume that the object is uniform (uniform in mass) then you can deduce that the length of the object is 2.4m (because the center of mass would be at the midpoint, which was 1.2m)

So the energy is not 1796.6 (that would be the energy if the length was 1.2m)

Use the correct length of the object and see if you get the correct answer
OK thanks! I got it right!
 

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