Solving for Angular Velocity in Trapeze Performance

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In summary, the problem involves a trapeze performer swinging in a vertical circle, which can be modeled as a rotating bar pivoted at one end. The mass of the bar is given as 70 kg with the center of mass located 1.2 m from the axis of rotation. Using the principle of conservation of energy, the angular velocity at the bottom of the swing can be found by considering the change in gravitational potential energy and using the equation for rotational kinetic energy. The length of the object is determined to be 2.4 m, resulting in a correct answer of 5.79 rad/s for the angular velocity at the bottom of the swing.
  • #1
BrainMan
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2

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.
 
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  • #2
ImageUploadedByPhysics Forums1406862427.606407.jpg
 
  • #3
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
Nathanael said:
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
BrainMan said:
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
Nathanael said:
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
You used the equation for linear kinetic energy.

Are you familiar with the equation for rotational kinetic energy?
 
  • #8
Nathanael said:
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
BrainMan said:
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
Nathanael said:
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
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
BrainMan said:
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
Nathanael said:
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
BrainMan said:
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
Nathanael said:
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!
 

1. How is angular velocity defined in trapeze performance?

Angular velocity is a measure of how quickly an object in motion is rotating around a fixed point. In trapeze performance, it refers to the rotational speed of the performer around the bar or other point of support.

2. What factors affect the angular velocity in trapeze performance?

The angular velocity in trapeze performance is affected by the length of the trapeze bar, the distance between the performer and the point of support, and the strength and technique of the performer.

3. How is angular velocity measured in trapeze performance?

Angular velocity in trapeze performance is typically measured in radians per second. This can be calculated by dividing the angle of rotation in radians by the time it takes for the rotation to occur.

4. How can angular velocity be increased in trapeze performance?

Angular velocity in trapeze performance can be increased by using a shorter trapeze bar, positioning the performer closer to the point of support, and by improving the performer's strength and technique.

5. Why is it important to solve for angular velocity in trapeze performance?

Solving for angular velocity in trapeze performance allows performers to control and manipulate their movements with precision and grace. It also helps in understanding the physical demands of the performance and can aid in preventing injuries.

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