Freezing of water inside a hollow sphere which is rolling

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Homework Statement


If we have a hollow ball completely filled with water which is rolling without slipping on a horizontal ground. If the water freezes which of the parameter will remain unchanged-
angular speed, angular momentum, linear momentum, kinetic energy, total energy

Homework Equations


$$\vec L=I_{cm}\omega + m\vec v x \vec r $$

The Attempt at a Solution


I am quite sure about the angular momentum as the water(presumably not rotating) and ball will have all its factors same that affect its angular momentum. Same goes for kinetic energy. But I have no clue regarding the angular speed and linear momentum.
The answer is given as angular momentum and total energy.
 

Answers and Replies

  • #2
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Please help me out.. This is screwing up my head.
 
  • #3
TSny
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I am quite sure about the angular momentum as the water(presumably not rotating) and ball will have all its factors same that affect its angular momentum.
What origin are you taking for the angular momentum? Is angular momentum going to be conserved about an arbitrary point?
You need to know the general condition that must be met in order for angular momentum about some specified point to be conserved.

Same goes for kinetic energy.
Why would KE necessarily be conserved?

It might help to consider the system shown below where the water is replaced by a rod.
upload_2018-4-14_15-7-29.png

This shows a hollow ball rolling without slipping. Inside the ball is a vertical rod that is free to rotate about its center. The rod is initially not rotating (it is just translating to the right with the ball). When the two little pegs simultaneously strike the ends of the rod, the rod sticks to the pegs so that the rod ends up rotating along with the ball.
 

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  • #4
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What origin are you taking for the angular momentum? Is angular momentum going to be conserved about an arbitrary point?
You need to know the general condition that must be met in order for angular momentum about some specified point to be conserved.

Why would KE necessarily be conserved?

It might help to consider the system shown below where the water is replaced by a rod.
View attachment 224056
This shows a hollow ball rolling without slipping. Inside the ball is a vertical rod that is free to rotate about its center. The rod is initially not rotating (it is just translating to the right with the ball). When the two little pegs simultaneously strike the ends of the rod, the rod sticks to the pegs so that the rod ends up rotating along with the ball.

Oh sorry. That was a huge blunder from my part... I just considered that after freezing the angular speed and linear speed will be same as before (totally wrong) . If that would have been the case, angular momentum will be conserved from any point. Any more views about the topic??? Thanks for letting me know my mistake. ..
 
  • #5
TSny
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Oh sorry. That was a huge blunder from my part... I just considered that after freezing the angular speed and linear speed will be same as before (totally wrong) . If that would have been the case, angular momentum will be conserved from any point. Any more views about the topic??? Thanks for letting me know my mistake. ..
Try to find a specific point in space about which the angular momentum of the system will be conserved. A free body diagram showing all external forces acting on the ball-water system will be very helpful.
 
  • #6
TSny
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@Jnan . I hope I haven't misinterpreted the problem statement. I assumed that initially the water is not rotating with the ball, but when the water freezes the ice does rotate with the ball. The results would be different if the water is initially rotating with the ball before it freezes. But, then, the problem doesn't seem to be very interesting.
 
  • #7
haruspex
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is rolling without slipping
I assume the question intends that it still does not slip during the freezing.
 
  • #8
TSny
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I assume the question intends that it still does not slip during the freezing.
Yes, I agree.
 
  • #10
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Try to find a specific point in space about which the angular momentum of the system will be conserved.

Okay this should be the point of contact of the ball with ground.
 
  • #11
TSny
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Okay this should be the point of contact of the ball with ground.
Well, I think it would be best to pick a fixed point in an inertial reference frame. The point of contact of the ball with the ground is moving relative to the ground (it's always vertically under the center of the ball), and this point might change its speed relative to the ground while the water freezes.
 
  • #12
rude man
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Water freezing means the ball has to expand, but if we ignore this (inflexible ball diameter) then seems to me there would be no change in either kinetic or rotational energy since neither external force nor torque was applied to the ball, and mass is unchanged of course. But not too sure about this. Othet comments welcome ...
 
  • #13
TSny
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Water freezing means the ball has to expand, but if we ignore this (inflexible ball diameter) then seems to me there would be no change in either kinetic or rotational energy since neither external force nor torque was applied to the ball, and mass is unchanged of course. But not too sure about this. Othet comments welcome ...
Hi, rude man. Are you assuming that the water is initially not rotating with the ball but the ice does rotate? If so, then I think KE will change if we assume the ball always rolls without slipping on the ground.
 
  • #14
haruspex
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Hi, rude man. Are you assuming that the water is initially not rotating with the ball but the ice does rotate? If so, then I think KE will change if we assume the ball always rolls without slipping on the ground.
I agree. It might seem strange that the KE changes, but it must.
 
  • #15
TSny
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I agree. It might seem strange that the KE changes, but it must.
Yes. It's not too hard to work it out explicitly. The behaviors of the angular velocity and the total angular momentum about the center of the sphere are also interesting.
 
  • #16
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If so, then I think KE will change if we assume the ball always rolls without slipping on the ground.

I didn't really get your statement. How can pure rolling have a effect on KE?
 
  • #17
haruspex
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But the question assumes that the ball is rolling without slipping. Hence the point of contact must be at rest relative to ground. It's definitely an inertial frame.
There is a difference between taking a fixed point, which happens to be the point of contact at some instant, as the reference, and taking a reference that moves with the current point of contact. The second is not an inertial frame, and will not give the right answer.
 
  • #18
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There is a difference between taking a fixed point, which happens to be the point of contact at some instant, as the reference, and taking a reference that moves with the current point of contact. The second is not an inertial frame, and will not give the right answer.
OH yeah.. I agree with that.. So with the centre of mass we can conserve angular momentum right....
 
  • #19
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Anyone who would like to add more about the angular momentum story..
 
  • #20
haruspex
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OH yeah.. I agree with that.. So with the centre of mass we can conserve angular momentum right....
Not with the centre of mass as axis, if that is what you meant.
For angular momentum of a system to be conserved about an axis, you must choose the axis so that the external forces have no net torque about it. What external forces are there? What choice of axis does that lead to?
 
  • #21
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Not with the centre of mass as axis, if that is what you meant.
For angular momentum of a system to be conserved about an axis, you must choose the axis so that the external forces have no net torque about it. What external forces are there? What choice of axis does that lead to?

Axis might be one which is vertical, passing through CM and along the plane of paper. Then normal and Mg will have zero torque.
 
  • #22
haruspex
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Axis might be one which is vertical, passing through CM and along the plane of paper. Then normal and Mg will have zero torque.
You can ignore normal and weight since they will cancel and not produce a net torque, regardless of axis.
What force does that leave?
 
  • #23
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You can ignore normal and weight since they will cancel and not produce a net torque, regardless of axis.
What force does that leave?

Friction|||
 
  • #24
haruspex
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Friction|||
Right. So what axis can you pick so that friction exerts no torque?
 
  • #25
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Right. So what axis can you pick so that friction exerts no torque?
One which passes through both CM and point of contact. ..
 

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