Write the kinetic energy of the sphere

[ritwik06]

Homework Statement

I have a string(length= r) and negligible mass to which a uniform sphere of radius r is attached on one end. I fix upper end of the string to the ceiling and release the sphere when it made angle θ₀ with the vertical. I have to write the kinetic energy of the sphere when it makes an angle θ with the vertical.

The Attempt at a Solution

Say, the angular velocity of the center of mass about the point of suspension be ω when it makes an angle θ.
I have a doubt:
Way 1:
KE=0.5*Iω²
Using parallel axis theorem;
'I' about the point of suspension= (2/5)mr²+m(2r)²
=4.4mr²
KE=2.2mr²ω²

Way 2:
KE=0.5*mv_cm²
v=2rω
KE=2mr²ω²

Which of these is correct. My book gives the method 1 as a correct solution. But I cannot understand what's wrong with method 2?

What I think is that method 1 gives kinetic energy when the sphere itself is rotating about its own axis with the same angular velocity 'ω' as the center of mass has about the point of suspension. Am I right?
But in the case specified does the sphere rotate about its axis? If yes, from where does the torque come from? The tension and the mass both pass from the center of mass, so they will have zero torque.
Please help me. I am confused!

 

[tiny-tim]

Way 2:
KE=0.5*mv_cm²
v=2rω
KE=2mr²ω²

Which of these is correct. My book gives the method 1 as a correct solution. But I cannot understand what's wrong with method 2?

Hi Ritwik!

Method 2 is the one we usually use for a sphere, because we usually treat the sphere as a point mass (so that the whole mass moves with the same speed, rω).

When the string is a lot longer than the radius of the sphere, that's accurate enough.

But here, the string equals the radius, and the approximation isn't good enough …

different parts of the sphere are moving with different speeds, and the average speed isn't rω …

so we can't treat the sphere as a point mass, and we have to do it "properly".

 

[davieddy]

Further complication:
Since the sphere has angular acceleration, the string is not
in line with the centre of the sphere.

To make it agree with the expected answer, the string should
be replaced by a light stiff rod rigidly attached to the sphere.

The original problem is close to an example of chaos theory!
Although I would see it as an example of 2 "normal modes" of
oscillation. The 2 variables are the angles of rotation of the sphere,
and the angle of the string. In one normal mode these are in phase,
and in the other antiphase.




David

 

[ritwik06]

Hi Ritwik!

Method 2 is the one we usually use for a sphere, because we usually treat the sphere as a point mass (so that the whole mass moves with the same speed, rω).

When the string is a lot longer than the radius of the sphere, that's accurate enough.

But here, the string equals the radius, and the approximation isn't good enough …

different parts of the sphere are moving with different speeds, and the average speed isn't rω …

so we can't treat the sphere as a point mass, and we have to do it "properly".

I understood that exactly! the sphere will also rotate about its center, right? But from where did the torque to cause this roatation came from?

 

[davieddy]

I understood that exactly! the sphere will also rotate about its center, right? But from where did the torque to cause this roatation came from?

That was exactly the point I was making.
In fact the sphere's angle of rotation about its centre is slightly greater
than the string's angle. This enables the string to provide the torque required.


David

 

[ritwik06]

Further complication:
Since the sphere has angular acceleration, the string is not
in line with the centre of the sphere.

To make it agree with the expected answer, the string should
be replaced by a light stiff rod rigidly attached to the sphere.

The original problem is close to an example of chaos theory!
Although I would see it as an example of 2 "normal modes" of
oscillation. The 2 variables are the angles of rotation of the sphere,
and the angle of the string. In one normal mode these are in phase,
and in the other antiphase.



David

First please tell me, the angular velocity about the sphere about its axis should be equal to the angular velocity of the center of mass about the suspension point for method 1 to be correct, right?
That means the there is rotation in the sphere about its center of mass! Now to cause this rotation, there must be a torque on the body. From where did this torque come from?

 

[tiny-tim]

I understood that exactly! the sphere will also rotate about its center, right? But from where did the torque to cause this roatation came from?

No!

This has nothing to do with rotation!

(and there's no torque!)

The bits of the sphere at height h above the centre are moving at speed (r+h)ω,

so the moment of inertia of those bits has a factor of (r+h)²ω, and the average of that is not r²ω … the lower bits have more contribution than the upper bits …

to put it another way, the centre of mass is not at the same place as the centre of moment of inertia.

 

[ritwik06]

That was exactly the point I was making.
In fact the sphere's angle of rotation about its centre is slightly greater
than the string's angle. This enables the string to provide the torque required.


David

Oh thanks! I understood. And if I replace the string with a stiff rod, will there be any rotation about the center? Then I presume method two will work there. Right?

 

[ritwik06]

No!

This has nothing to do with rotation!

(and there's no torque!)

The bits of the sphere at height h above the centre are moving at speed (r+h)ω,

so the moment of inertia of those bits has a factor of (r+h)²ω, and the average of that is not r²ω … the lower bits have more contribution than the upper bits …

to put it another way, the centre of mass is not at the same place as the centre of moment of inertia.

I see through that, but the thing is that it can be possible only when the sphere rotates. I don't wish to be rude, I just want to clarify my concepts.
Please see through my viewpoint: suppose the sphere does not undergo rotation about its centre. Then can I say that all the particles on the sphere will have the same velocity as the centre of mass? Without rotation about its own axis, is there any other way that the velocities of the particles in/on the sphere are different? I don't think so.

 

[davieddy]

Oh thanks! I understood. And if I replace the string with a stiff rod, will there be any rotation about the center? Then I presume method two will work there. Right?

No. You still need to take into account the sphere's rotation about its
centre of mass. But now the "rigid" connection between rod and sphere
ensures that they have the same angular velocity, and the rigid connection
provides the necessary torque, which string can't do.

 

[tiny-tim]

… suppose the sphere does not undergo rotation about its centre. Then can I say that all the particles on the sphere will have the same velocity as the centre of mass? Without rotation about its own axis, is there any other way that the velocities of the particles in/on the sphere are different? I don't think so.

The sphere is rotating about the top of the string.

That is why the moment of inertia needs to be calculated relative to an axis through the top of the string.

Each point on the sphere is moving in an arc whose centre is the top of the string.

That is all you need for Way 1 in the book.

 

[ritwik06]

The sphere is rotating about the top of the string.

That is why the moment of inertia needs to be calculated relative to an axis through the top of the string.

Each point on the sphere is moving in an arc whose centre is the top of the string.

That is all you need for Way 1 in the book.

Thanks a lot. I get it now.

 

[ritwik06]

No. You still need to take into account the sphere's rotation about its
centre of mass. But now the "rigid" connection between rod and sphere
ensures that they have the same angular velocity, and the rigid connection
provides the necessary torque, which string can't do.

Thank you! I see through it! Can you find a practical example where all the points on the sphere do have the same velocity?
Will this do? -----> Say a rigid rod is attached to the center of the sphere and the attachment is such that the sphere does not rotate about it centre. Now will the way 1 be applicable?

 

[davieddy]

No!

This has nothing to do with rotation!

(and there's no torque!)

The bits of the sphere at height h above the centre are moving at speed (r+h)ω,

so the moment of inertia of those bits has a factor of (r+h)²ω, and the average of that is not r²ω … the lower bits have more contribution than the upper bits …

to put it another way, the centre of mass is not at the same place as the centre of moment of inertia.

?

 

[davieddy]

Thank you! I see through it! Can you find a practical example where all the points on the sphere do have the same velocity?

Yes. Cut the sphere almost in two so that it is like a yoyo,
and attach the string at its centre.

 

[ritwik06]

Yes. Cut the sphere almost in two so that it is like a yoyo,
and attach the string at its centre.

I get it completely now, i think a rod will be better than a string to avoid rotations about the centre?
A string might still allow rotation about the center. What do you say?

 

[ritwik06]

I get it completely now, i think a rod will be better than a string to avoid rotations about the centre?
A string might still allow rotation about the center. What do you say?

I am sorry. A string would work too if attached to the centre as even if there are forces there on centre of mass their net torque woul be zero. Am I right?
Thanks a lot!

 

[davieddy]

I get it completely now, i think a rod will be better than a string to avoid rotations about the centre?
A string might still allow rotation about the center. What do you say?

It might "allow" it, but with the only forces on the sphere now being
gravity and tension applied to the centre, there is now nothing to cause it.

The sphere could be spinning with constant angular velocity, (string loop
round the axle slips rather than winds up) but the sphere still moves as if
it were a point mass at its centre.

 

[davieddy]

I am sorry. A string would work too if attached to the centre as even if there are forces there on centre of mass their net torque woul be zero. Am I right?
Thanks a lot!

exactly

OTOH a rod welded to the axle would constrain the sphere to
rotate with it, and such a joint can transmit the required torque.

 

[ritwik06]

exactly

OTOH a rod welded to the axle would constrain the sphere to
rotate with it, and such a joint can transmit the required torque.

Thank you very very much!

 

[davieddy]

Thank you very very much!

Enjoyed. There's a simple principle I still want to explain (PM me or I will you)

Meantime enjoy this one.


Love,
David

 

[tiny-tim]

Since the sphere has angular acceleration, the string is not in line with the centre of the sphere.

i] there is no reason why the string should not stay in line with the centre of the sphere, so long as the sphere was released carefully enough
ii] the answer in the book makes it very clear that that is the correct interpretation of the question

… I would see it as an example of 2 "normal modes" of
oscillation. The 2 variables are the angles of rotation of the sphere, and the angle of the string. In one normal mode these are in phase, and in the other antiphase.

and iii] yes, that's absolutely correct …

if you don't release the sphere carefully at the start, it will have an "extra" rotation (in addition to the rotation about the top of the string) …

but this is a wholly different, more advanced, topic (probably involving Lagrangians) which I assume is not on this part of Ritwik's syllabus

Ritwik, to pass your exams, you must be able to understand why the way given in your book works! ​

 

[davieddy]

i] there is no reason why the string should not stay in line with the centre of the sphere, so long as the sphere was released carefully enough

Both Ritwik and I have explained that while we, you or the textbook author might
be capable of visualizing the string/sphere system as a rigid body, it ain't.
The sphere has rotational acceleration, and this requires some moment about its c of m.

ii] the answer in the book makes it very clear that that is the correct interpretation of the question

The important point to stress is that the kinetic energy and angular momentum
of any set of particles is that of the total mass located at the c of m PLUS that
relative to the c of m. The same applies to linear momentum, but more or less by
definition, the linear momentum relative to the c of m is zero.

and iii] yes, that's absolutely correct …

if you don't release the sphere carefully at the start, it will have an "extra" rotation (in addition to the rotation about the top of the string) …

but this is a wholly different, more advanced, topic (probably involving Lagrangians) which I assume is not on this part of Ritwik's syllabus

admittedly, but since the problem focuses attention on the rotation of the sphere
(as you pointed out in your first post) it is relevant.

Ritwik, to pass your exams, you must be able to understand why the way given in your book works! ​

I have every confidence in him/her!

David

 

[davieddy]

Homework Statement

I have a string(length= r) and negligible mass to which a uniform sphere of radius r is attached on one end. I fix upper end of the string to the ceiling and release the sphere when it made angle θ₀ with the vertical. I have to write the kinetic energy of the sphere when it makes an angle θ with the vertical.

The Attempt at a Solution

Say, the angular velocity of the center of mass about the point of suspension be ω when it makes an angle θ.
I have a doubt:
Way 1:
KE=0.5*Iω²
Using parallel axis theorem;
'I' about the point of suspension= (2/5)mr²+m(2r)²
=4.4mr²
KE=2.2mr²ω²

Way 2:
KE=0.5*mv_cm²
v=2rω
KE=2mr²ω²

Which of these is correct. My book gives the method 1 as a correct solution. But I cannot understand what's wrong with method 2?

What I think is that method 1 gives kinetic energy when the sphere itself is rotating about its own axis with the same angular velocity 'ω' as the center of mass has about the point of suspension. Am I right?
But in the case specified does the sphere rotate about its axis? If yes, from where does the torque come from? The tension and the mass both pass from the center of mass, so they will have zero torque.
Please help me. I am confused!

I think all of us are guilty of ignoring the first rule of passing exams:
READ THE QUESTION.

If we were told that a mass m was released from rest at height h0,
and were asked for its kinetic energy when it had reached height h, we wouldn't
expect many marks for the answer "1/2mv^2".
[Or even 1/2(mv^2 + Iω^2)]

Ritwik, could you give us the full question as posed?