# Angular velocity of a bullet and two masses

## Homework Statement:

Consider the picture attached. Suppose that the bullet has a mass ##m_0## and initial velocity ##v_0##, and suppose that the masses has mass ##m##. The length of the string is ##2b## and ##\theta=90°##
What's the final velocity and angular velocity

## Relevant Equations:

##P=ccte##
As ##P=ccte## we can find final velocity considering a plastic collision
##m_0 . v_0 =(2m+m_0).V##

But what about the angular velocity? Because, as the bullet hits the centre of mass of the string, it won't have angular velocity

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haruspex
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Relevant Equations:: ##P=ccte##

as the bullet hits the centre of mass of the string, it won't have angular velocity
I assume "ccte" is an abbreviation for constant in some language.

Angular velocity is, in general, in respect of a chosen axis. Maybe it would help to choose a different one.

Edit... on second thoughts, not sure that helps. But it is something you ought to try.

Last edited:
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Respect to the middle of the line, it's right that the total angular momentum of the system is zero(cause the angular velocity is zero), please verify that the angular momentum is still zero after the collision respect to the same point.

haruspex
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plastic collision
Did you mean inelastic? What makes you think so?

jbriggs444
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cause the angular velocity is zero
Angular velocity is a characteristic of rigid systems. You do not have a rigid system here. Pieces of the system may still rotate rigidly and have angular velocities about carefully chosen axes.

What is the expected final state here? Are the two masses expected to bump into each other? Or are we expecting a stable orbit of the two masses tethered to the still-moving bullet? What question is being asked?

kuruman
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My interpretation is that one is asked to describe the motion of the two masses and the bullet after the collision with the connecting chord and while the two masses are moving towards each other as suggested in the three figures. The angular velocity mentioned by OP would be that of either mass about the moving bullet. The middle figure (b) suggests that the connecting chord is elastic because it shows that the masses have not moved while the chord is stretched. Upon closer inspection, one must conclude that the middle figure is poorly drawn because the length of the connecting chord is explicitly labeled b + b in all three figures.

Did you mean inelastic? What makes you think so?
Because I considered that the two balls and the bullet remain as "one body" since they have the same velocity

What is the expected final state here? Are the two masses expected to bump into each other? Or are we expecting a stable orbit of the two masses tethered to the still-moving bullet? What question is being asked?
It's not said in the question. I think we can consider that the balls may collide inelastically or elastically.
I considered the first option so that the balls and the bullet move with same velocity

haruspex
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Because I considered that the two balls and the bullet remain as "one body" since they have the same velocity
Certainly there is a loss of KE associated with the forward velocity of the whole system, but this is not a simple coalescence. There is an obvious place for that energy to go.

My interpretation is that one is asked to describe the motion of the two masses and the bullet after the collision with the connecting chord and while the two masses are moving towards each other as suggested in the three figures. The angular velocity mentioned by OP would be that of either mass about the moving bullet. The middle figure (b) suggests that the connecting chord is elastic because it shows that the masses have not moved while the chord is stretched. Upon closer inspection, one must conclude that the middle figure is poorly drawn because the length of the connecting chord is explicitly labeled b + b in all three figures.
Maybe I am asked to find the angular velocity of the balls when they sweep 90°

Certainly there is a loss of KE associated with the forward velocity of the whole system, but this is not a simple coalescence. There is an obvious place for that energy to go.
And what if the string is not elastic?

haruspex
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And what if the string is not elastic?
It is not necessary for the string to be elastic. There is no abrupt change in velocity; it is a smooth transition.
You could solve this problem in terms of forces and accelerations, but on performing the first integration I believe you will simply get an energy conservation equation.

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kuruman