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Energy analysis of a particle moving in a shrinking circle

  • #1
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Homework Statement


A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ##r_0## with angular velocity ##w_0##, but I now pull the string down through the hole until a length r remains between the hole and the particle.

a) What is the particle's angular velocity now?

b) Assuming that I pull the string so slowly that we can approximate the particle's path by a circle of slowly shrinking radius, calculate the work I did pulling the string.

c)Compare your answer to part b with the particle's gain in kinetic energy.

Homework Equations


Rotational energy: ##E = \frac{1}{2}I\omega^2##


The Attempt at a Solution


The main assumption I made seems correct to me, but the question seems to imply it's wrong. I thought about the rotation and decided that because we assume the path is a circle of slowly decreasing radius, there is no work done as the force of the string is always tangential to the direction of motion. If we didn't assume this, then there would be a translational component to energy when we move and maybe that would lead to some work gain. If the assumption is true, the solution is the following:

a) Because no work is done, the energy is conservative and ##T_{in} = T_{fin}##. ##T_{in} = \frac{1}{2}MR_0^2\omega_0^2## and ## T_{fin} = \frac{1}{2}Mr^2\omega^2## after a bit of math, ##\omega = \frac{R_0\omega_0}{r}##

b)As argued above, the work Is zero.

c)My gain is zero
 

Answers and Replies

  • #2
Orodruin
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there is no work done as the force of the string is always tangential to the direction of motion.
This is incorrect. The approximation is only to ease the computation of the force in the radial direction. You will still have to move inwards to decrease the radius and this will result in work being done.
 
  • #3
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This is incorrect. The approximation is only to ease the computation of the force in the radial direction. You will still have to move inwards to decrease the radius and this will result in work being done.
You do, but I was thinking that an increase in angular rotation would make up for the decreased radius. I figured the assumption was wrong, but decided to roll with it.

I realize now that even though we simplify the computation as we did, the force is acting in a non tangential direction because the circle is shrinking in the radial direction. I'm just unsure how to calculate the work. Could I use an integral like ##\int_{r_0}^r F_r \ dr## and let ##F_r = ma?##
For part 1 I don't know how to do it without calculating the work since it seems you'd have to in some form to figure out how much energy was added to the system.
 
  • #4
tnich
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Homework Statement


A particle of mass m is moving on a frictionless horizontal table and is attached to a massless string, whose other end passes through a hole in the table, where I am holding it. Initially the particle is moving in a circle of radius ##r_0## with angular velocity ##w_0##, but I now pull the string down through the hole until a length r remains between the hole and the particle.

a) What is the particle's angular velocity now?

b) Assuming that I pull the string so slowly that we can approximate the particle's path by a circle of slowly shrinking radius, calculate the work I did pulling the string.

c)Compare your answer to part b with the particle's gain in kinetic energy.

Homework Equations


Rotational energy: ##E = \frac{1}{2}I\omega^2##


The Attempt at a Solution


The main assumption I made seems correct to me, but the question seems to imply it's wrong. I thought about the rotation and decided that because we assume the path is a circle of slowly decreasing radius, there is no work done as the force of the string is always tangential to the direction of motion. If we didn't assume this, then there would be a translational component to energy when we move and maybe that would lead to some work gain. If the assumption is true, the solution is the following:

a) Because no work is done, the energy is conservative and ##T_{in} = T_{fin}##. ##T_{in} = \frac{1}{2}MR_0^2\omega_0^2## and ## T_{fin} = \frac{1}{2}Mr^2\omega^2## after a bit of math, ##\omega = \frac{R_0\omega_0}{r}##

b)As argued above, the work Is zero.

c)My gain is zero
As @Orodruin says, you will have to perform work to pull the string downward, even if you do it slowly. So conservation of energy will not help you solve this problem. There is another conservation law you can use, though.
 
  • #5
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As @Orodruin says, you will have to perform work to pull the string downward, even if you do it slowly. So conservation of energy will not help you solve this problem. There is another conservation law you can use, though.
I'm really sad I forgot about conservation of angular momentum. Since there is no torque, angular momentum is conserved. I got ##mr_0^2\omega_0 = mr^2\omega## for the initial equation. Then ##\omega = \frac{r_0^2\omega_0}{r^2}##. My guess on how to calculate the work separately feels pretty wrong though.
 
  • #6
tnich
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I'm really sad I forgot about conservation of angular momentum. Since there is no torque, angular momentum is conserved. I got ##mr_0^2\omega_0 = mr^2\omega## for the initial equation. Then ##\omega = \frac{r_0^2\omega_0}{r^2}##. My guess on how to calculate the work separately feels pretty wrong though.
Take a shot at calculating the work and show us what you get.
 
  • #7
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Okay, so I calculated work by doing ##T_{fin} - T_{in}## because that's the smart and simple way to do it (and I had misread part c. to mean I had to calculate it in a unique way to compare the answers, not compare the gain in KE). I got my result to be ##W = T_{in}(\frac{r_0^2}{r^2}-1) ##which means that the increase in kinetic energy simply depends on the ratio between the initial and final radius. That makes sense. Time wouldn't be in the answer since the time it takes is irrelevant to the actual work performed. If I plug in ##\omega## from earlier into ##T_{fin}## I get ##T_{in}(\frac{r_0^2}{r^2})## which is what I would expect
 
  • #8
tnich
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Okay, so I calculated work by doing ##T_{fin} - T_{in}## because that's the smart and simple way to do it (and I had misread part c. to mean I had to calculate it in a unique way to compare the answers, not compare the gain in KE). I got my result to be ##W = T_{in}(\frac{r_0^2}{r^2}-1) ##which means that the increase in kinetic energy simply depends on the ratio between the initial and final radius. That makes sense. Time wouldn't be in the answer since the time it takes is irrelevant to the actual work performed. If I plug in ##\omega## from earlier into ##T_{fin}## I get ##T_{in}(\frac{r_0^2}{r^2})## which is what I would expect
I think you have it, but I can't quite tell. I assume ##T_{in}## represents initial kinetic energy and ##T_{fin}## is final kinetic energy. How did you calculate ##T_{fin} - T_{in}##? Did you use the slowly shrinking radius approach requested in the problem statement?
 
  • #9
haruspex
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Did you use the slowly shrinking radius approach requested in the problem statement?
I think it only specifies slowly so that KE due to radial velocity can be ignored.
 
  • #10
tnich
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I think it only specifies slowly so that KE due to radial velocity can be ignored.
I think that @DanielA has correctly calculated the difference in KE by subtracting the KE at radius ##r_0## from the KE at radius ##r##. But part c) of the problem does not make sense unless the energy difference is calculated in a different way in part b), specifically by finding the work required to move the mass from radius ##r_0## to radius ##r##.
As a hint to @DanielA, what would be the work required to change the radius from ##r_0## to ##r_0 - Δ## for a very small Δ?
 
  • #11
haruspex
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I think that @DanielA has correctly calculated the difference in KE by subtracting the KE at radius ##r_0## from the KE at radius ##r##. But part c) of the problem does not make sense unless the energy difference is calculated in a different way in part b), specifically by finding the work required to move the mass from radius ##r_0## to radius ##r##.
As a hint to @DanielA, what would be the work required to change the radius from ##r_0## to ##r_0 - Δ## for a very small Δ?
I was just meaning to clarify what "the approach" was. Sorry for any confusion.
 
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  • #12
Orodruin
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Could I use an integral like ##\int_{r_0}^r F_r \ dr##
Yes, this is the definition of work done. According to the work-energy theorem, this should be equal to the difference in energies, which is essentially what the problem is going to illustrate to you. The big question is now: How do you find ##F_r##?
 
  • #13
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Yes, this is the definition of work done. According to the work-energy theorem, this should be equal to the difference in energies, which is essentially what the problem is going to illustrate to you. The big question is now: How do you find ##F_r##?
I think I said it earlier, but I dropped the integral approach because with the info given, we can't really find the force. We might be able to do the integral by saying ##F_r = ma## and then claim a isn't in terms of r and so we can pull it out and evaluate the integral (I'm pretty sure this is invalid here, but it is a trick I've used in a similar way in other physics courses), but that would give me an acceleration factor which makes no sense because the time the work takes to accomplish is irrelevant so I'm confident at least the approach is wrong, but most likely the integral trick itself is wrong.

Instead, I just found the difference in initial and final energies by the difference in ##T_{fin}## and ##T_{in}## since there is no potential in the problem.
 
  • #14
haruspex
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we can't really find the force
Yes you can. This is why it says it is slowly reeled in. The force will be negligibly more (at any given radius) than is necessary to maintain the current radius.
 
  • #15
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Yes you can. This is why it says it is slowly reeled in. The force will be negligibly more (at any given radius) than is necessary to maintain the current radius.
Would that just be a little more than v^2/r then? Though really, it seems dealing with it is entirely unnecessary since the work can be calculated indirectly using conservation of angular momentum (since I can find ##\omega## from it and therefore do ##T_{fin}-T_{in}## to find work since the only energy at the start and ending is rotational kinetic energy) because the force and position vectors are pointing in opposite directions with the origin at the hole in the center so there is no torque.
 
  • #16
haruspex
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Would that just be a little more than v^2/r then?
Yes.
dealing with it is entirely unnecessary since the work can be calculated indirectly using conservation of angular momentum
True, but the question is partly an exercise in observing that the laws of physics and mathematics are consistent. In case you had not guessed.
 
  • #17
Orodruin
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Would that just be a little more than v^2/r then?
Yes.
Well, that is the acceleration. To get the force you need to multiply by the mass otherwise the dimensions wont match.

Though really, it seems dealing with it is entirely unnecessary since the work can be calculated indirectly using conservation of angular momentum (since I can find ##\omega## from it and therefore do ##T_{fin}-T_{in}## to find work since the only energy at the start and ending is rotational kinetic energy) because the force and position vectors are pointing in opposite directions with the origin at the hole in the center so there is no torque.
As haruspex said and as I alluded to in #12, the entire point of the exercise is to illustrate to you that the work-energy theorem holds, i.e., that you can get the difference in the energies by computing the work done or vice versa.
 
  • #18
tnich
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I think I said it earlier, but I dropped the integral approach because with the info given, we can't really find the force. We might be able to do the integral by saying ##F_r = ma## and then claim a isn't in terms of r and so we can pull it out and evaluate the integral (I'm pretty sure this is invalid here, but it is a trick I've used in a similar way in other physics courses), but that would give me an acceleration factor which makes no sense because the time the work takes to accomplish is irrelevant so I'm confident at least the approach is wrong, but most likely the integral trick itself is wrong.

Instead, I just found the difference in initial and final energies by the difference in ##T_{fin}## and ##T_{in}## since there is no potential in the problem.
The concept you need is centripetal force. I suggest that you look it up and find out how to use it here.
EDIT: Sorry, I see you have already discussed this.
 
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