# Motion of rolling metallic shell filled with water

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## Homework Statement

A metallic spherical shell of mass M and radius R is fully filled with water of mass M is set into pure pure rolling motion on a smooth horizonal surface (w clockwise, v to the right). Now water is freezed suddenly, and the ice sticks with the shell. What is-
a) moment of Inertia of shell about point of contact
b) angular momentum of water about point of contact
c) MI of shell+ice system about point of contact
d) MI of shell+water system about point of contact
e) new velocity of center of mass just after water freezes

## Homework Equations

All equations pertaining to rotational mechanics

## The Attempt at a Solution

a) since it is asked just for shell, I can forget about the water and write I(com) as 2/3mr^2 then apply parallel axes theorem to get answer as 5/3 Mr^2 (correct according to my textbook)
b) My intuition is that the water won't roll and will just have a straight translational velocity (since this question hasn't called fluid mechanics into question- we can forget about the drag of the liquid, internal friction between sphere and water etc.) and just write is as MvR (the correct answer as per the book) am I right?
c) I considered them as 2 separate spheres- since ice is solid, it'll behave as a normal solid sphere; so adding 2/3 + 2/5 + 2(parallel axis theorem) I got the correct answer 46/15 MR^2
d) here is where I'm getting confused, MI of a fluid is more since it tries to get away from the axis of rotation- I don't know how that'll come into play here and how I can calculate total MI
e) Angular momentum about point of contact is conserved therefore I(shell+water about COM)*w + 2Mvr=I(shell+ice) about com*w1 + 2Mv1r where w1 and v1 are new angular and linear velocitites, and v=rw is followed everywhere

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## Answers and Replies

haruspex
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write I(com) as 2/3mr^2
I take it that it is a spherical shell.
the water won't roll and will just have a straight translational velocity
Yes.
MI of a fluid is more since it tries to get away from the axis of rotation
Not sure what you mean. The water is constrained.
Angular momentum about point of contact is conserved therefore I(shell+water about COM)*w + 2Mvr=I(shell+ice) about com*w1 + 2Mv1r where w1 and v1 are new angular and linear velocitites, and v=rw is followed everywhere
Could be right, but I'm not certain what you intend in parts. What answer do you get?

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I take it that it is a spherical shell.
yes it is

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Not sure what you mean. The water is constrained.
for example if we take a hard-boiled egg and a raw egg and spin them, the one with the fluid inside spins slower because MI of the fluid is more since it tries to get away from axis of rotation

Ray Vickson
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for example if we take a hard-boiled egg and a raw egg and spin them, the one with the fluid inside spins slower because MI of the fluid is more since it tries to get away from axis of rotation

I have not done the experiment you suggest, so I cannot judge the accuracy of your explanation.

Anyway, you are not dealing with an egg; you are dealing with a spherical mass of water constrained by a rigid external surface. Water is essentially "incompressible", so there can be no net movement away from the axis of rotation. For every molecule of water that tries to get away there will be another one that is forced back towards the axis of rotation, so as to keep the volume unchanged.

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Could be right, but I'm not certain what you intend in parts. What answer do you get?
about the point of contact, initially L about point of contact is L(orbital) = 2MvR + (L(spin) = 2/3mr^2*w=2/3mvr (since water doesnt spin))=8/3 MvR. Finally, L about the same point is L(orbital) 2Mv'R + (L(spin)=I(shell+ice)*w') now I(shell+ice)=2/3 mr^2 + 2/5 mr^2= 16/15 mr^2 and w' and v' are the new angular and linear velocities so total final L=46/15mv'r. equating both, I get v'=46/15 which is wrong

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I have not done the experiment you suggest, so I cannot judge the accuracy of your explanation.

Anyway, you are not dealing with an egg; you are dealing with a spherical mass of water constrained by a rigid external surface. Water is essentially "incompressible", so there can be no net movement away from the axis of rotation. For every molecule of water that tries to get away there will be another one that is forced back towards the axis of rotation, so as to keep the volume unchanged.
Ah! yes, I did not think of that. this means I can create another interesting question in which the water is filled only upto h/3 or something but i'll post that as another thread

Nathanael
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The problem sounded unsolvable to me... we’re given the mass of water as M, so how do we know the water fills the sphere entirely unless we’re told M = 4πR2ρwater/3?

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The problem sounded unsolvable to me... we’re given the mass of water as M, so how do we know the water fills the sphere entirely unless we’re told M = 4πR2ρwater/3?
sorry for the misunderstanding. There is a simple figure given, but I don't have a camera. Ill edit the question to add all information I might have overlooked (I think it was only this though)

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The problem sounded unsolvable to me... we’re given the mass of water as M, so how do we know the water fills the sphere entirely unless we’re told M = 4πR2ρwater/3?
corrected it! are there any mistakes in my work now?

Nathanael
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Ok. In that case either M or R is redundant (one can be replaced in terms of the density of water).

e) Angular momentum about point of contact is conserved therefore I(shell+water about COM)*w + 2Mvr=I(shell+ice) about com*w1 + 2Mv1r where w1 and v1 are new angular and linear velocitites, and v=rw is followed everywhere
The problem states that the ground is smooth, a.k.a. frictionless. The ball is rolling at first by virtue of how we set it into motion, but since there is no friction it isn’t required to always roll.

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Ok. In that case either M or R is redundant (one can be replaced in terms of the density of water).

The problem states that the ground is smooth, a.k.a. frictionless. The ball is rolling at first by virtue of how we set it into motion, but since there is no friction it isn’t required to always roll.
what force will change its rolling state then? also if v=rw doesn't apply everywhere, even angular momentum conservation won't give the right answer since we'll have one more variable. How do we solve this then. P.S. my textbook says Vcom remains same before and after water freezes so I suspect rolling continues...but not sure

Nathanael
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what force will change its rolling state then?
When it freezes, the rotation rate slows down. If there were friction it would mean the ball is now sliding forward, which would cause friction pointing backwards speeding the rotation back up until it’s rolling again (as in your problems from a couple days ago). Since there is no friction this doesn’t happen... it simply slides.

also if v=rw doesn't apply everywhere, even angular momentum conservation won't give the right answer since we'll have one more variable. How do we solve this then.
When the water freezes, does the linear momentum change?

P.S. my textbook says Vcom remains same before and after water freezes so I suspect rolling continues...but not sure
That shows that it doesn’t continue Gold Member
When it freezes, the rotation rate slows down. If there were friction it would mean the ball is now sliding forward, which would cause friction pointing backwards speeding the rotation back up until it’s rolling again (as in your problems from a couple days ago). Since there is no friction this doesn’t happen... it simply slides.
since MI increases, w decreases (AM conservation), right? I understood the rest of the part

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When the water freezes, does the linear momentum change?
oh right, linear momentum doesn't change...I got so much into angular momentum problems that I forgot thinking about this

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That shows that it doesn’t continue
all I could infer from this is that translational KE doesn't change.......oh, and therefore rotational KE doesn't change (from conservation of energy), but we know MI increases so w must have changed while v remained the same...so, v is not equal to rw anymore. Am I reasoning correctly?
duh! It seems so obvious after you give a hint...I'm hopefully correct now. Let me try the question again,

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hey, it seems like we have to do practically nothing...there's no net external force on the system so Vcom must remain the same. All changes are in I and w caused due to the freezing

Nathanael
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but we know MI increases so w must have changed while v remained the same...so, v is not equal to rw anymore. Am I reasoning correctly?
Yes that is good. We only assume rolling when there is (enough) friction.
all I could infer from this is that translational KE doesn't change.......oh, and therefore rotational KE doesn't change (from conservation of energy)
I would be hesitant to apply conservation of energy as a first solution. It is an interesting question, which I invite you to check, but it’s not obvious if it’s valid. There are internal energies we are ignoring and there is some kind of energy transfer we are ignoring (water doesn’t spontaneously freeze) but maybe all this becomes irrelevant.

My point is, conservation of energy is a nice shortcut in situations where you know it will work. In new situations it is a principle to be checked by other means.

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Yes that is good. We only assume rolling when there is (enough) friction.

I would be hesitant to apply conservation of energy as a first solution. It is an interesting question, which I invite you to check, but it’s not obvious if it’s valid. There are internal energies we are ignoring and there is some kind of energy transfer we are ignoring (water doesn’t spontaneously freeze) but maybe all this becomes irrelevant.

My point is, conservation of energy is a nice shortcut in situations where you know it will work. In new situations it is a principle to be checked by other means.
yes, but if we don't apply energy conservation can we infer that w decreases from AM conservation? since L(orbital) is same, L(spin) must be constant too but MI definitely increases therefore w decreases. correct?

Nathanael
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yes, but if we don't apply energy conservation can we infer that w decreases from AM conservation? since L(orbital) is same, L(spin) must be constant too but MI definitely increases therefore w decreases. correct?
Yes that is correct and yes we can infer it, quantitatively. It is a law of its own, independent of conservation of energy.

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Yes that is correct and yes we can infer it, quantitatively. It is a law of its own, independent of conservation of energy.
seems like this is solved then. One last question remaining for the day, thank you very much :)

haruspex
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for example if we take a hard-boiled egg and a raw egg and spin them, the one with the fluid inside spins slower because MI of the fluid is more since it tries to get away from axis of rotation
No, that's not how that works.
Although there is a small airspace in an egg, spinning a raw egg will not drive that to the centre. The egg contents are too viscous.
In the usual way this is done, the egg is set spinning with one motion. It is not kept spinning at a steady rate for a while before release. So in a raw egg the innards will not have reached the same spin rate as the shell and the angular momentum is less.
This is like the unfrozen water case in this thread, but with some viscosity.

• Nathanael
haruspex
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how do we know the water fills the sphere entirely
i'm sure I read that it was filled, but it has been edited since then.
I would be hesitant to apply conservation of energy
Indeed, it is a classic coalescence problem, but in a rotational mode instead of a linear one.