What is the angular velocity that the rod will rotate

In summary, the conversation discusses a homework problem involving two masses connected by a rod and rotating on a horizontal plane. The question asks for the angular velocity of the rod after a collision when a mass of 2M is added. The two methods used to calculate the angular momentum -I*ω and |r x p| - are discussed, with a clarification that the given omega is specified with an axis of rotation at the center. A summary of the solution attempted is also provided.
  • #1
MMS
148
4
Hello!
I'm pretty much translating this question word by word, so if you have problem understanding something off of it, please, let me know and I'll try explaining it better.

Homework Statement



Two masses of similar mass M are connected by a rod of length 2a. the rod is rotating on a horizontal plane in an angular velocity of ω, when its center of mass (point A) is at rest. The table is horizontal and friction is neglected. (the rod is not wired to the table).
At some given time t, we locate a mass of 2M so that it sticks to one of the masses.

Question:What is the angular velocity that the rod will rotate in around its new center of mass after the collision?
Z2Oxpgs.png

Homework Equations



The Attempt at a Solution


What I did was this:
1. I calculated the location of the new center of mass (the after frame) and it turns out to be a/2 towards the 3M.

2. I looked at the frame just a moment before collision so that the mass 2M is really close to the mass M (one of them) but yet to collide with one. I calculated the angular momentum before collision by I*ω around the new center of mass that I stated above:

[m*(3a/2)^2 + m*(a/2)^2]*ω = 2.5m*(a^2)*ω

3. Now, I looked at the frame just a moment after the collision so the masses pretty much made no movement and again I calculated the angular velocity after collision by I*ω around the new center of mass:

[m*(3a/2)^2 + m*(a/2)^2 + 2m*(a/2)^2]*ω' = 3m*(a^2)*ω'

4. Of course there is conservation of angular momentum, therefore I equalized the two expressions to get that the new angular velocity is ω' = (5/6)*ω and not (2/3)*ω as the correct answer is.

Here's the thing that's bothering me other than that answer is wrong. I tried applying RxMV on the before system with the location of the new center of mass and it turns out to be right!
I really don't understand where have I mistaken in the solution I've given and why in A LOT of cases when I do I*ω and RxMV in this type of questions (conservation of angular momentum) I get different results. I thought they're equal...

Anyway, I'd really appreciate it if you guys told me at which point of the solution I did an incorrect step and also answer my question in the last paragraph.

THANK YOU!
 
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  • #2
I agree with your answer.
Regarding cases where the two methods disagree, can you post an example?
 
  • #3
Hi and thanks for the reply.

What answer do you agree with? the ω'?
 
  • #4
mms said:
hi and thanks for the reply.

What answer do you agree with? The ω'?

5ω/6
 
  • #5
Oh alright. But it's wrong so... :P
 
  • #6
MMS said:
Oh alright. But it's wrong so... :P

The omega they give (the omega "before") is with respect to the center of mass located at the center (distance of "a" for both masses), not at the new position.
And using [itex] I \omega [/itex] or [itex] | \vec{r} \times \vec{p} | [/itex] has, of course, to give the same answer in any given problem. You should show us an example.
 
  • #7
nrqed said:
The omega they give (the omega "before") is with respect to the center of mass located at the center (distance of "a" for both masses), not at the new position.
A rotation rate is a rotation rate. It will take the same time to complete one revolution regardless of where you take the axis to be.
 
  • #8
haruspex said:
A rotation rate is a rotation rate. It will take the same time to complete one revolution regardless of where you take the axis to be.

But if you have an object rotating with respect to some axis and you change the axis, the value of omega will change.

I did not want to explain everything in details to let the OP do some work but since you missed my point, let me be more detailed: the calculation of the angular momentum depends on where the axis of rotation is. You won't get the same value of [itex] I \omega [/itex] if you use a different point. Since the [itex] \omega [/itex] they give is specified with an axis of rotation at the center, it is that axis of rotation that one must use to calculate the angular momentum.
 
  • #9
nrqed said:
But if you have an object rotating with respect to some axis and you change the axis, the value of omega will change.

I did not want to explain everything in details to let the OP do some work but since you missed my point, let me be more detailed: the calculation of the angular momentum depends on where the axis of rotation is. You won't get the same value of [itex] I \omega [/itex] if you use a different point. Since the [itex] \omega [/itex] they give is specified with an axis of rotation at the center, it is that axis of rotation that one must use to calculate the angular momentum.
The angular momentum changes, of course, but not the rotation rate.
 
  • #10
MMS said:
Hello!
I'm pretty much translating this question word by word, so if you have problem understanding something off of it, please, let me know and I'll try explaining it better.

Homework Statement



Two masses of similar mass M are connected by a rod of length 2a. the rod is rotating on a horizontal plane in an angular velocity of ω, when its center of mass (point A) is at rest. The table is horizontal and friction is neglected. (the rod is not wired to the table).
At some given time t, we locate a mass of 2M so that it sticks to one of the masses.

Question:What is the angular velocity that the rod will rotate in around its new center of mass after the collision?
Z2Oxpgs.png

Homework Equations






The Attempt at a Solution


What I did was this:
1. I calculated the location of the new center of mass (the after frame) and it turns out to be a/2 towards the 3M.

2. I looked at the frame just a moment before collision so that the mass 2M is really close to the mass M (one of them) but yet to collide with one. I calculated the angular momentum before collision by I*ω around the new center of mass that I stated above:

[m*(3a/2)^2 + m*(a/2)^2]*ω = 2.5m*(a^2)*ω

Like I mentioned in my previous post, since they say that the initial rotation is with respect to the center of the rod, you must use that information to find the initial angular momentum. Therefore it is

[itex] (m a^2 + m a^2) \omega = 2 m a ^2 \omega [/itex]

Set this equal to your final angular momentum and you will get the correct answer.
 
  • #11
haruspex said:
The angular momentum changes, of course, but not the rotation rate.

The total momentum is zero. Thus the angular momentum will be the same around any point. The momentum of inertia the OP is using would be fine if the object was fixed at that point, which is not the case. Seen from the CoM after the collision, the two masses would have different angular velocities at the collision time (they have the same velocity and one is further away). This does not mean the object´s rotation has a different angular frequency, just that the angular velocities seen from the post-collision CoM are different.
 
  • #12
nrqed said:
Like I mentioned in my previous post, since they say that the initial rotation is with respect to the center of the rod, you must use that information to find the initial angular momentum. Therefore it is

[itex] (m a^2 + m a^2) \omega = 2 m a ^2 \omega [/itex]

Set this equal to your final angular momentum and you will get the correct answer.


I have noticed that if I do it with respect to the center of the rod on the before angular momentum I get the right answer. But notice what you're doing here. You're actually applying conservation of momentum with respect to two different axes. The angular momentum before with respect to the center of the rod and the angular momentum after with respect to the new location of the center of mass which is completely invalid.

Also, I believe that the angular velocity is the same at any point on the rod.
 
  • #13
nrqed said:
The omega they give (the omega "before") is with respect to the center of mass located at the center (distance of "a" for both masses), not at the new position.
And using [itex] I \omega [/itex] or [itex] | \vec{r} \times \vec{p} | [/itex] has, of course, to give the same answer in any given problem. You should show us an example.

I really want to type it down but I'm on my phone and it's completely annoying.

Pleasw, try calculating Iw with respect to the location of the new center of mass and after that calculate it by rxmv and see the results for yourself.
 
  • #14
Orodruin said:
The total momentum is zero. Thus the angular momentum will be the same around any point. The momentum of inertia the OP is using would be fine if the object was fixed at that point, which is not the case. Seen from the CoM after the collision, the two masses would have different angular velocities at the collision time (they have the same velocity and one is further away). This does not mean the object´s rotation has a different angular frequency, just that the angular velocities seen from the post-collision CoM are different.

You've taken my post out of context. Nrqed was claiming the rate of rotation of an object depends on the axis with respect to which you measure it. My response was that the angular momentum (in general) depends on the choice of axis, but the rotation rate does not. You are right that in the present case the angular momentum does not depend on it either.
 
  • #15
MMS said:
I have noticed that if I do it with respect to the center of the rod on the before angular momentum I get the right answer. But notice what you're doing here. You're actually applying conservation of momentum with respect to two different axes. The angular momentum before with respect to the center of the rod and the angular momentum after with respect to the new location of the center of mass which is completely invalid.

Also, I believe that the angular velocity is the same at any point on the rod.

At last I see where we both went wrong. We calculated the prior angular momentum wrongly. We did it as though the rod were rotating about the system CoM. Instead, look at the linear velocities. Each mass has speed aω, so the moment is maω(3a/2 + a/2) = 2maω2.
 
  • #16
haruspex said:
At last I see where we both went wrong. We calculated the prior angular momentum wrongly. We did it as though the rod were rotating about the system CoM. Instead, look at the linear velocities. Each mass has speed aω, so the moment is maω(3a/2 + a/2) = 2maω2.

I don't believe it's right to say that each has a a linear velocity of what you said. The linear velocity of each mass is v=ωxr where r is the distance from the location of our new center of mass to where the body is and then you apply rxmv where r, again, is the distance mentioned above.
What you've done is calculate the linear velocity of each mass with respect to the point they're rotating around before colliding with the 3rd mass.
 
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  • #17
Nrqrd is right.

The initial linear velocities were v and -v, where v=aω, as the rod rotated about its centre.

The angular momentum of a point mass about any fixed point is mrxv. That of the system of the rotating equal masses, with respect to the centre of the rod, was 2mav=2ma2ω.
The angular momentum in the new frame of reference was (3/2 a ) v m + (1/2 a) v m= 2avm again as orodruin pointed out in Post #11.
The 2m mass was added to the system. Originally in rest, it had zero contribution to the angular momentum. After it stuck to the bottom mass, the angular momentum became IΩ where the moment of inertia I=m(3/2 a)2+3m (1/2 a)2= 3 a2m .

The angular momentum is conserved:3 a2m Ω =2ma2ω.
 

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  • #18
ehild said:
Nrqrd is right.

The initial linear velocities were v and -v, where v=aω, as the rod rotated about its centre.

The angular momentum of a point mass about any fixed point is mrxv. That of the system of the rotating equal masses, with respect to the centre of the rod, was 2mav=2ma2ω.
The angular momentum in the new frame of reference was (3/2 a ) v m + (1/2 a) v m= 2avm again.
The 2m mass was added to the system. Originally in rest, it had zero contribution to the angular momentum. After it stuck to the bottom mass, the angular momentum became IΩ where the moment of inertia I=m(3/2 a)2+3m (1/2 a)2= 3 a2m .

The angular momentum is conserved:3 a2m Ω =2ma2ω, The new angular speed is 2/3 of the initial one.

First off, thanks for the solution you've posted.

I've posted a question about the linear velocities. Can you please take a look at my last post (#16)?

Thanks.
 
  • #19
My post was the answer to your Post #16. Just before the collision, the linear velocities of the masses were v and -v, in any rest frame of reference. The velocities do not change if you change the point of reference. v=rω is valid if r is measured from the centre of rotation, and it was at the middle of the stick before the collision.

ehild
 
  • #20
haruspex said:
You've taken my post out of context. Nrqed was claiming the rate of rotation of an object depends on the axis with respect to which you measure it. My response was that the angular momentum (in general) depends on the choice of axis, but the rotation rate does not. You are right that in the present case the angular momentum does not depend on it either.

I don't even see what you mean by saying that a rate of rotation is independent of the axis of rotation. We can't even talk about rotation of an object without defining the axis around which it rotates. Let's take two masses M connected by a rigid rod of negligible mass. If I tell you that it is rotating at 20 rad/second, what does that mean?!? It does not mean anything unless you know around what axis it is rotating. If it does, then explain to me what this means without specifying an axis of rotation.

The rod could be held at the center, at one extremity or it could be some other rotation entirely. It is meaningless to give a rotation rate without specifying the axis of rotation (and it is even less meaningful to do any calculation without knowing about the axis of rotation).

Regards,

Patrick
 
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  • #21
nrqed said:
I don't even see what you mean by saying that a rate of rotation is independent of the axis of rotation. We can't even talk about rotation of an object without defining the axis around which it rotates. Let's take two masses M connected by a rigid rod of negligible mass. If I tell you that it is rotating at 20 rad/second, what does that mean?!? It does not mean anything unless you know around what axis it is rotating. If it does, then explain to me what this means without specifying an axis of rotation.

The rod could be held at the center, at one extremity or it could be some other rotation entirely. It is meaningless to give a rotation rate without specifying the axis of rotation (and it is even less meaningful to do any calculation without knowing about the axis of rotation).

Regards,

Patrick
Remember that our point of disagreement does not concern an actual change of axis, but the choice of axis for calculating angular momentum..
Any pair of points of the body defines a direction in space. If I rotate a body such that every such pair defines a line parallel to the line they originally defined then I have not changed the orientation of the body, but I may have changed its location.
If I rotate it steadily about a fixed axis, after some period of time T it will be back at its original orientation. Every line defined by points in the body will similarly be back in its original orientation. Thus the period of rotation (and hence the rate) is independent of which axis I consider the rotation to be about.
Another way to think of this is to consider a different inertial frame, moving at constant velocity wrt to the original frame. The axis of rotation will appear to be different, but the rate the same.
 
  • #22
haruspex said:
Remember that our point of disagreement does not concern an actual change of axis, but the choice of axis for calculating angular momentum..
Any pair of points of the body defines a direction in space. If I rotate a body such that every such pair defines a line parallel to the line they originally defined then I have not changed the orientation of the body, but I may have changed its location.
If I rotate it steadily about a fixed axis, after some period of time T it will be back at its original orientation. Every line defined by points in the body will similarly be back in its original orientation. Thus the period of rotation (and hence the rate) is independent of which axis I consider the rotation to be about.

But it is a mistake to say that if the period of rotation is the same, the rate is necessarily the same. This is obvious to see if you consider an explicit example.

Consider a rod rotating at constant ##\omega ## with respect to its center. Then the extremities have an angular velocity that is constant and equal to ## \omega ##. Now take another reference point, somewhere at a certain distance from the rod (not within the disk covered by the rod). The ##\omega## of of the masses will clearly not be the same as the ##\omega ## with respect to the center. The ##\omega## measured from this other point even changes sign!
 
  • #23
nrqed said:
But it is a mistake to say that if the period of rotation is the same, the rate is necessarily the same. This is obvious to see if you consider an explicit example.

Consider a rod rotating at constant ##\omega ## with respect to its center. Then the extremities have an angular velocity that is constant and equal to ## \omega ##. Now take another reference point, somewhere at a certain distance from the rod (not within the disk covered by the rod). The ##\omega## of of the masses will clearly not be the same as the ##\omega ## with respect to the center. The ##\omega## measured from this other point even changes sign!
Time to move this discussion to a private thread.
 
  • #24
And time to define ##\omega##.

ehild
 
  • #25
ehild said:
And time to define ##\omega##.

ehild

Yes. I just want to post a last time in order to make sure that there is no confusion for others who might happen to read this thread so let me again emphasize my point: the value of [itex] \omega [/itex] of a certain object ## DOES ## depend on what axis of rotation is it measured from.

Let me be very precise. Consider an object in rotation, for example the wheel of a car being driven at constant speed on a road. What is the ## \omega ## of the wheel? The first key point is that this question is ambiguous, from a physics point of view. Of course, the intuitive answer is that it has a unique and precise value but when people jump to this conclusion it is because they are ## implicitly ## thinking about the angular speed with respect to the center of the wheel and they are therefore already choosing an axis of rotation (even if they might not want to admit it ;-) )

So let's be very precise and careful. First, in order to define an ## \omega ##, we must pick a specific point mass. Let's pick a small piece of rubber at the edge of the wheel.

Now we want to calculate ## \omega ##. What is the definition of ## \omega ##? It is

[tex] \omega = \frac{ d \theta (t)}{dt} [/tex]

So we need to obtain the function ## \theta (t) ## giving the angular position of the piece of rubber at the edge of the wheel as a function of time. And now comes the obvious point: this function depends on where we set our axis of rotation! Let's say you set the axis of rotation at the center of the wheel and wait a time equal to half the period of rotation, then the angle changes by ## \pi ## radians in that time. Now use a different axis of rotation, fixed let's say at a point on the ground a certain distance from the car, then it is clear that in the same amount of time the point on the wheel will not cover the same angle. Therefore the function ## \theta (t) ## depends on the reference point used for the axis of rotation and ## \omega## also depends on the axis of rotation chosen.
 
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  • #26
haruspex said:
Remember that our point of disagreement does not concern an actual change of axis, but the choice of axis for calculating angular momentum..
Any pair of points of the body defines a direction in space. If I rotate a body such that every such pair defines a line parallel to the line they originally defined then I have not changed the orientation of the body, but I may have changed its location.
If I rotate it steadily about a fixed axis, after some period of time T it will be back at its original orientation. Every line defined by points in the body will similarly be back in its original orientation. Thus the period of rotation (and hence the rate) is independent of which axis I consider the rotation to be about.
Another way to think of this is to consider a different inertial frame, moving at constant velocity wrt to the original frame. The axis of rotation will appear to be different, but the rate the same.

Just to be clear. I do agree with you that with respect to any point on the object the rotation rate is the same. The point I made when I wrote my first post about the rotation rate ##\omega ## given in the original problem being given with respect to the center of mass was an indirect hint I was giving to make the OP realize that when calculating the angular momentum, he/she would have to work with respect to the COM. I did not mean to imply that the omega depends on what axis we pick ## on ## ## the ## ## object ##. My point was that given that the value of the omega given was with respect to a rotation relative to the COM, one would have to pay attention to this when calculating the angular momentum.

But I stand to my point that in general, the value of ## \omega ## does depend on where we put the axis of rotation as is clear by choosing a point outside of the object. Your point was that, as long as one uses a special subset of axis located on the object , the value of omega is the same. I agree with you on that point.

Regards,

Patrick
 
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  • #27
Patrick, your ω=dθ/dt seems to be the derivative of the angular coordinate in a polar coordinate system.

Here we speak about the angular velocity of the rotation of a rigid body. And it does not depend on the choice of the point of reference.

O is the point of reference in the inertial frame. O' is a fixed point and P is an arbitrary point of the rigid body. Its velocity is

##\vec v = \dot {\vec R} + \vec ω \times \vec {r'} =\vec V+\vec ω \times \vec {r'}##
If you shift the reference point in the body by ##\vec a ## then ##\vec {r'}= \vec a +\vec {r'' }##, so ##\vec v =\vec V+\vec ω \times \vec {a} + \vec ω \times \vec {r''}##. On the other hand, the point of reference can be placed anywhere in the rigid body, that does not change the velocity of P. So ##\vec v =\vec {V''}+\vec {ω'' }\times \vec {r''}##---> ##\vec{V''}=\vec V +\vec ω \times \vec a ## and ##\vec ω = \vec {ω''}##

The angular velocity ω is characteristic of the rotation of the rigid body. It does not depend on the choice of reference inside the body.

ehild
 

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  • #28
ehild said:
Patrick, your ω=dθ/dt seems to be the derivative of the angular coordinate in a polar coordinate system.

Here we speak about the angular velocity of the rotation of a rigid body. And it does not depend on the choice of the point of reference.

O is the point of reference in the inertial frame. O' is a fixed point and P is an arbitrary point of the rigid body. Its velocity is

##\vec v = \dot {\vec R} + \vec ω \times \vec {r'} =\vec V+\vec ω \times \vec {r'}##
If you shift the reference point in the body by ##\vec a ## then ##\vec {r'}= \vec a +\vec {r'' }##, so ##\vec v =\vec V+\vec ω \times \vec {a} + \vec ω \times \vec {r''}##. On the other hand, the point of reference can be placed anywhere in the rigid body, that does not change the velocity of P. So ##\vec v =\vec {V''}+\vec {ω'' }\times \vec {r''}##---> ##\vec{V''}=\vec V +\vec ω \times \vec a ## and ##\vec ω = \vec {ω''}##

The angular velocity ω is characteristic of the rotation of the rigid body. It does not depend on the choice of reference inside the body.

ehild

Thanks for your comment. I agree completely and I apologize if I gave the impression that I disagree with this point.
The first comment I made was to emphasize to the OP that the object was rotating with respect to its center, which was essential to calculate the correct angular momentum. But I see now that my comment could be misconstrued as saying that omega depends on the object. I apologize to you and to Haruspex. I agree that the omega of a rigid body is the same with respect to any point. My later comment was that the general definition of angular velocity of a point object does depend on the axis of rotation. I went to a case too general for the discussion, I guess.

Thanks again.

Regards,

Patrick
 

1. What is angular velocity?

Angular velocity is the rate at which an object rotates around a fixed point or axis. It is measured in radians per second (rad/s) or degrees per second (°/s).

2. How is angular velocity calculated?

Angular velocity is calculated by dividing the change in angle by the change in time. The formula for angular velocity is ω = Δθ/Δt, where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.

3. What factors affect the angular velocity of an object?

The angular velocity of an object can be affected by its moment of inertia, the applied torque, and any external forces acting on the object. The shape and mass distribution of the object also play a role in determining its angular velocity.

4. How does angular velocity differ from linear velocity?

Angular velocity is a measure of rotational speed, while linear velocity is a measure of straight-line speed. Angular velocity is expressed in radians per second or degrees per second, while linear velocity is expressed in meters per second or feet per second.

5. Can angular velocity change over time?

Yes, angular velocity can change over time if there is a change in the applied torque or if external forces act on the object. The moment of inertia can also change, affecting the angular velocity of an object.

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