Swing Higher on Circus Trapeze: Solve Physics Problem

[mirandasatterley]

Any help with the following question would be appreciated. Please keep in mind that I'm not very good with physics.

"A circus trapeze consists of a bar suspended by two parallel ropes, each of length l, allowing performers to swing in a vertical circular arc. Suppose a performer with a mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle theta_i with respect to the vertical. Suppose the size of the performer's body is small compared to the length l, she does not pump the trapeze to swing higher, and air resistance is negigable.
A) Show that when the ropes make an angle theta with the vertical, the preformer must exert a force: mg(3 cos theta - 2 cos theta_1), so as to hang on.
B) Determine the angle theta_i for which the force needed to hang on at the bottom of the swing is twice the preformer's weight."

*I tried to attach a picture. Its not very good though, because I had to draw it*

I don't know how to approach the question but, my first question would be where is angle theta_i ?

 

[PhanthomJay]

Any help with the following question would be appreciated. Please keep in mind that I'm not very good with physics.

"A circus trapeze consists of a bar suspended by two parallel ropes, each of length l, allowing performers to swing in a vertical circular arc. Suppose a performer with a mass m holds the bar and steps off an elevated platform, starting from rest with the ropes at an angle theta_i with respect to the vertical. Suppose the size of the performer's body is small compared to the length l, she does not pump the trapeze to swing higher, and air resistance is negigable.
A) Show that when the ropes make an angle theta with the vertical, the preformer must exert a force: mg(3 cos theta - 2 cos theta_1), so as to hang on.
B) Determine the angle theta_i for which the force needed to hang on at the bottom of the swing is twice the preformer's weight."

*I tried to attach a picture. Its not very good though, because I had to draw it*

I don't know how to approach the question but, my first question would be where is angle theta_i ?

You posted this somewhere yesterday I think, and I tried to give you a couple of hints. The angle theta is somewhere in between beta and -beta. It is a 'given' in part a, and you've got to find it in part b. Maybe someone else will help because I'm not giving any more hints. If you're a beginner at centripetal force and conservation of energy equations, this is not the problem you want to tackle first.

 

[kyle8921]

Hi there,

I am happy to help you with this physics problem. The angle theta_i is the initial angle at which the performer steps off the elevated platform. Now, let's break down the problem and solve it step by step.

A) To show that the performer must exert a force of mg(3 cos theta - 2 cos theta_1), we need to use the law of conservation of energy. At the bottom of the swing, all of the potential energy is converted into kinetic energy. Therefore, we can equate the potential energy at the top of the swing to the kinetic energy at the bottom of the swing.

At the top of the swing, the potential energy is given by mg(2l - l cos theta). At the bottom of the swing, the kinetic energy is given by (1/2)mv^2, where v is the velocity of the performer at the bottom of the swing. Since the performer starts from rest, we can set the kinetic energy to zero. Equating these two energies, we get:

mg(2l - l cos theta) = 0

Solving for theta, we get:

cos theta = 2/3

But we also know that the angle theta_1 is given by cos theta_1 = l/2l = 1/2. Substituting this in the equation for theta, we get:

cos theta = 2 cos theta_1

Therefore, the force needed to hang on is mg(3 cos theta - 2 cos theta_1).

B) To determine the angle theta_i for which the force needed to hang on at the bottom of the swing is twice the performer's weight, we can use the same equation from part A and set it equal to 2mg (twice the weight):

mg(2l - l cos theta) = 2mg

Solving for theta, we get:

cos theta = 1/2

Therefore, the initial angle at which the performer steps off the platform should be theta_i = 60 degrees.

I hope this helps you understand the problem better. Remember, practice makes perfect when it comes to physics. Keep working at it and you will improve. Best of luck!