# A uniform rod allowed to rotate about an axis and then it breaks

#### Kaushik

Homework Statement
A uniform rod AB of length L is free to rotate about a horizontal axis passing through A. The rod is released from rest from the horizontal position. If the rod gets broken at midpoint C when it becomes vertical, then just after breaking of the rod. Multiple answers are correct
Homework Equations
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A uniform rod AB of length ℓ is free to rotate about a horizontal axis passing through A. The rod is released from rest from the horizontal position. If the rod gets broken at midpoint C when it becomes vertical, then just after breaking of the rod. Choose multiple answeres from the below options.
• Angular velocity of upper part starts to decrease while that of lower part remains constant
• Angular velocity of upper part starts to decrease while that of lower part starts to increase
• Angular velocity of both the parts is identical
• Angular velocity of lower part becomes equal to zero
I tried using conservation of angular momentum once it breaks. I found the angular velocity of the rod when it is vertical using conservation of energy.

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#### jbriggs444

Homework Helper
• Angular velocity of upper part starts to decrease while that of lower part remains constant
• Angular velocity of upper part starts to decrease while that of lower part starts to increase
• Angular velocity of both the parts is identical
• Angular velocity of lower part becomes equal to zero
Why not start by telling us what you think about the various possibilities being offered? For instance, does the angular velocity of the upper part start to decrease after the break takes place?

You should not need to calculate anything.

#### Kaushik

Why not start by telling us what you think about the various possibilities being offered? For instance, does the angular velocity of the upper part start to decrease after the break takes place?

You should not need to calculate anything.
When it breaks the angular velocity of the upper part decreases is my intuition. I dont have the exact reasoning for it. Is the reasoning related to the decrease in moment of inertia?

#### jbriggs444

Homework Helper
When it breaks the angular velocity of the upper part decreases is my intuition. I dont have the exact reasoning for it.
Let us attack that particular question then. What external influences is the upper part subject to after the break? Do any of these amount to an external torque?

#### Kaushik

Let us attack that particular question then. What external influences is the upper part subject to after the break? Do any of these amount to an external torque?
The upper part is subjected to hinge force and weight which does not provide external torque.( when it is perfectly vertical)

#### jbriggs444

Homework Helper
The upper part is subjected to hinge force and weight which does not provide external torque.( when it is perfectly vertical)
Given that parenthetical -- "when it is still perfectly vertical", I agree with you. Zero net external torque. And the object is rigid, so the moment of inertia is unchanging. The conclusion is that the angular acceleration is zero. At least momentarily.

However, my inclination is to read the question differently. I think we agree that the upper part is still swinging and that as it continues swinging it will no longer be vertical. As it goes past vertical there will be an external torque. So one could reasonably say that "the angular velocity of the upper part starts to decrease". That is, if you were to plot the angular velocity of the upper part over time you would see a peak in the graph where angular velocity reaches a maximum at the moment of the break and starts to decrease.

If we are in agreement so far, we can proceed to a next clause. Does the angular velocity of the lower part remain constant?

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#### Kaushik

Given that parenthetical -- "when it is still perfectly vertical", I agree with you. Zero net external torque. And the object is rigid, so the moment of inertial is unchanging. The conclusion is that the angular acceleration is zero. At least momentarily.

However, my inclination is to read the question differently. I think we agree that the upper part is still swinging and that as it continues swinging it will no longer be vertical. As it goes past vertical there will be an external torque. So one could reasonably say that "the angular velocity of the upper part starts to decrease". That is, if you were to plot the angular velocity of the upper part over time you would see a peak in the graph where angular velocity reaches a maximum at the moment of the break and starts to decrease.

If we are in agreement so far, we can proceed to a next clause. Does the angular velocity of the lower part remain constant?
Agreed!

The only force that acts on the lower part is weight but Angular velocity of the lower part has no external torque about the center of mass because $r =0$.

#### jbriggs444

Homework Helper
Agreed!

The only force that acts on the lower part is weight but Angular velocity of the lower part has no external torque about the center of mass because $r =0$.
Agreed.

So if we adopt my reading of the question, the first possibility below would be correct and the second possibility below would be incorrect.

Shall we proceed to the question of whether the angular velocity of both parts is identical?
• Angular velocity of upper part starts to decrease while that of lower part remains constant [correct]
• Angular velocity of upper part starts to decrease while that of lower part starts to increase [incorrect]
• Angular velocity of both the parts is identical
• Angular velocity of lower part becomes equal to zero

#### Kaushik

Agreed.

So if we adopt my reading of the question, the first possibility below would be correct and the second possibility below would be incorrect.

Shall we proceed to the question of whether the angular velocity of both parts is identical?
Yes!

Just when it breaks the angular velocity remains identical?

#### jbriggs444

Homework Helper
Just when it breaks the angular velocity remains identical?
It would be good to flesh that out with a little reasoning. There is no force (and no momentary angular impulse) associated with the break, so angular momentum conservation applies with full force, both for the upper part and the lower part. Both were rotating around their respective centers with the same speed prior to the break, so they continue rotating about their respective centers with the same speed after the break.

#### Kaushik

There is no force (and no momentary angular impulse) associated with the break, so angular momentum conservation applies with full force, both for the upper part and the lower part
Sorry for not responding to your message on time.
I’m not sure I understand the quoted line. Could you please elaborate?

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#### jbriggs444

Homework Helper
Sorry for not responding to your message on time.
I’m not sure I understand the quoted line. Could you please elaborate?
Let us consider the situation an instant prior to the break. The rod as a whole is rotating about the pivot with some angular velocity. Call that velocity $\omega$.

If we draw an imaginary dividing line at the point of the break, we can consider the top and bottom part of the rod separately. The top part is rotating about the pivot point with angular velocity $\omega$. The bottom part is rotating about its center point with angular velocity $\omega$ while that center point is itself translating horizontally with some velocity.

Are we in agreement so far?

#### Kaushik

Let us consider the situation an instant prior to the break. The rod as a whole is rotating about the pivot with some angular velocity. Call that velocity $\omega$.

If we draw an imaginary dividing line at the point of the break, we can consider the top and bottom part of the rod separately. The top part is rotating about the pivot point with angular velocity $\omega$. The bottom part is rotating about its center point with angular velocity $\omega$ while that center point is itself translating horizontally with some velocity.

Are we in agreement so far?
Yes!

#### jbriggs444

Homework Helper
Now the break takes place. Our imaginary dividing line becomes a real dividing line. That is all that happens. The motion of the top piece and the bottom piece are unchanged.

Are we still in agreement?

#### Kaushik

Now the break takes place. Our imaginary dividing line becomes a real dividing line. That is all that happens. The motion of the top piece and the bottom piece are unchanged.

Are we still in agreement?
Momentarily they are unchanged. Isn’t?
If yes, Agreed!

#### jbriggs444

Homework Helper
Momentarily they are unchanged. Isn’t?
If yes, Agreed!
Yes. Recall that this was an argument in favor of the correctness of:
Angular velocity of both the parts is identical
which I read to be about the angular velocity of both parts immediately after the moment the break.

#### Kaushik

Yes. Recall that this was an argument in favor of the correctness of:

which I read to be about the angular velocity of both parts immediately after the moment the break.
So as the angular velocities of both the parts are identical the last option cannot be true. Isn’t?

#### jbriggs444

Homework Helper
So as the angular velocities of both the parts are identical the last option cannot be true. Isn’t?
Yes, just so.

We have agreed that it is rotating just before and just after the break. Which means that we've agreed that it does not suddenly stop rotating.

#### Kaushik

Yes, just so.

We have agreed that it is rotating just before and just after the break. Which means that we've agreed that it does not suddenly stop rotating.
Thanks a lot for your help!

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