What is the Time for a Wave to Travel on a String in Circular Motion?

In summary, the Homework Statement states that it takes a wave on the string to travel from the center of the circle to the ball. It takes 3.36 seconds for the wave to travel this distance.
  • #1
Mugen Prospec
42
0

Homework Statement


The drawing shows a 16.9 kg ball being whirled in a circular path on the end of a string. The motion occurs on a frictionless, horizontal table. The angular speed of the ball is ω = 12 rad/s. The string has a mass of 0.013 kg. How much time does it take a wave on the string to travel from the center of the circle to the ball?



Homework Equations





The Attempt at a Solution

 
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  • #2
I have attempted this but I am just stuck I thought it has something to do with finding the tension but I don't know how to since there is no radii given.
 
  • #3
I'm going to have to agree with you there. I don't see how you can get anywhere without knowing the radius. It's the only length scale in the problem and the velocity definitely needs something with dimension length. No hidden clues in the diagram?
 
  • #4
Nope nothing ot indicate length

Although a similar one in my book has the answer as 3.36X10^-3]

Edit: ITs in second the answer is that in second but I do not see how it doesn't explain
 
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  • #5
m/sec, I would assume. Then are you supposed to carry some knowledge of a previous problem over into this one? You DEFINITELY need a radius. Otherwise, you just have to write the answer as radius*something.
 
  • #6
Mugen Prospec said:
Nope nothing ot indicate length

Although a similar one in my book has the answer as 3.36X10^-3]

Edit: ITs in second the answer is that in second but I do not see how it doesn't explain

Oh, my fault. I was thinking you needed a velocity, and you don't. You need a TIME. Sorry again. Put the radius equal to r and work everything out. In the end, the r will cancel.
 
  • #7
ok so what equations? do i set w=v/r to v=sqrt (f/(m/l))
 
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  • #8
Use F=Ma to get the tension T, use rho=m/r to get the density. Velocity=sqrt(T/rho), right? You have to know some these formulas. Just put them all together and put the unknown radius equal to r.
 
  • #9
It wants us to use the formula v=√(F/(M⁄L))
 
  • #10
It's not just about formulas, it's about understanding. F is the string tension, M is the string mass and L is the string length. How is F related to the mass at the end of the string, the angular speed and the length of the string? You have to do this. I can't do it for you.
 
  • #11
Yes I get that, I am not familiar with the Rho=m/r part haven't seen that in my 3 years of physics.
I also can't figure out how to change angular velocity to centripetal acceleration
 
  • #12
rho is linear density. You haven't seen that? kg/m. The main issue here is what is F, the tension. You've got have have seen that in three years of physics. What tension is required in the string? It's uniform circular motion.
 
  • #13
Mugen Prospec said:
I also can't figure out how to change angular velocity to centripetal acceleration

You have to put in the unknown radius to do that. Does rw^2 ring a bell?
 
  • #14
Im sorry I am just not getting it. This was not really for homework point I was just trying to get it for a final in a few days. Can you tell me your answer so I can back track it possibly
 
  • #15
A number isn't going to tell you anything. What's the tension in the string using the given mass at the end of the string, the angular velocity and a radius r? Yes, use the centripetal acceleration.
 
  • #16
Ok my answer I got was 0.00231 seconds.
In fact I used the answer from the previous problem I posted. I back tracked from that answer and I figured it out. I appreciate you attempts to steer me but I must also inform you that telling me the answer would have helped me much more. I understood the problem and you did helped me. I thank you for that. However I would have more quickly understood it had you given me the answer I requested. In other cases I may not have had a chance to make use of a similar problem and answer and may not have been able to work through it.
 
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  • #17
Judging by the amount of work and effort you showed in this thread, I'm going to have to guess somebody else gave you that answer until I'm proved wrong. Or can you post your work and show you actually know how to get it? Giving the answer simply isn't the way things are supposed to work here. You're welcome.
 
  • #18
Hey Dick quit being one. I genuinely came to here for help and you really wrote me off early. I don't understand what you stood to loose by giving me the answer or a detailed solution. Very few of us in the world want to become a Physicist and thanks to pompous people like you it discourages many young people for affirming that wish. Its this kind of behavior that encourages a self castration of our field, preventing future generations from even attempting the profession. This is a small community that should be close-knit I take every opportunity to help people in lower classes and when I ask for help because a very specific problem was giving me difficulty I am met with adversity and even worse on my first post. I am far from lazy so you can dismiss that belief and soon I will be posting my solution after I scan it into my computer. You have true soured my first impression of this forum.
-Anthony

Edit:
Hey Dick quit being one. I genuinely came to here for help and you really wrote me off early. I don't understand what you stood to loose by giving me the answer or a detailed solution. Very few of us in the world want to become a Physicist and thanks to pompous people like you it discourages many young people for affirming that wish. Its this kind of behavior that encourages a self castration of our field, preventing future generations from even attempting the profession. This is a small community that should be close-knit I take every opportunity to help people in lower classes and when I ask for help because a very specific problem was giving me difficulty I am met with adversity, and even worse it was on my first post. I am far from lazy so you can dismiss that belief and soon I will be posting my solution after I scan it into my computer. You have true soured my first impression of this forum.
-Anthony
 
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  • #19
"adversity and even worse"?. Oh come on, Mugen Prospec. Look back on the thread and try to figure out when and where you actually tried to work out the problem using the advice I was trying to give you. If you are going to get all ripped because I wouldn't give you the numerical answer, then that's your problem.
 
  • #20
I see you have all these awards for being a great help and you did help me on the canceling the radius part. However for someone to go from helpful to antagonistic is unnecesary. If you look at my original post I knew I need the tension and I knew the angular velocity caused the tension on the rope. All of this was already know. I reached an impasse and you showed me the hidden path then it was late so I had to sleep on it. I am not all "ripped up" but look what you did, you exhibited the worst traits in teaching: you became frustrated early and became scathing.
 
  • #21
Mugen Prospec, here are your options: you can say thank you to the person trying to help you (regardless of actual help given) or you can pray on getting help the next time around.
 
  • #22
You seem to be a lot more interested in complaining about the help offered by actual posted line count than you were in solving the problem. That's about all I have to say.
 
  • #23
Mugen Prospec said:
Ok my answer I got was 0.00231 seconds.
In fact I used the answer from the previous problem I posted. I back tracked from that answer and I figured it out. I appreciate you attempts to steer me but I must also inform you that telling me the answer would have helped me much more. I understood the problem and you did helped me. I thank you for that. However I would have more quickly understood it had you given me the answer I requested. In other cases I may not have had a chance to make use of a similar problem and answer and may not have been able to work through it.

I did in fact thank you but I also pointed out your true to your name, if we were in person I would shake your hand but the fact remains you were not very personable.
 
  • #24
Dick said:
You seem to be a lot more interested in complaining about the help offered by actual posted line count than you were in solving the problem. That's about all I have to say.

If I am confronted I will surly respond. I came here and asked for help, promptly I was answered with the same response I had. Then you had a break through, passed it on to me. I then had a ah ha moment, but soon I came to an impasse again. You became frustrated, then you insulted me, the gull to make such an assertion that I am ill-equipped to finish the problem without showing someone doing it for me. You had no reason to get frustrated as you obviously have. I even apologized because I was missing it. I am sure you time is valuable but the fact remains you got feed up and then closed to door to giving me more insight. If I was in the same situation and gotten feed up I would have posted the answer and ended it.
 

What is the speed of a wave on a string?

The speed of a wave on a string is determined by two factors: the tension of the string and the mass per unit length of the string. It can be calculated using the formula v = √(T/μ), where v is the speed of the wave, T is the tension of the string, and μ is the mass per unit length of the string.

How does the tension of the string affect the speed of the wave?

The higher the tension of the string, the faster the speed of the wave. This is because a higher tension results in a higher restoring force, which allows the wave to travel faster along the string.

How does the mass per unit length of the string affect the speed of the wave?

The higher the mass per unit length of the string, the slower the speed of the wave. This is because a higher mass means a greater inertia, which makes it more difficult for the wave to propagate through the string.

How does the wavelength of the wave affect its speed on a string?

The wavelength of the wave does not directly affect its speed on a string. However, as the wavelength increases, the frequency decreases and therefore the speed of the wave also decreases, since frequency and wavelength are inversely proportional.

Can the speed of a wave on a string be changed?

Yes, the speed of a wave on a string can be changed by altering the tension or mass per unit length of the string. It can also be changed by changing the properties of the medium through which the wave is traveling, such as temperature or density.

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