Conservation of momentum: linear and angular

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Homework Help Overview

The problem involves a uniform square wooden gate and a raven colliding with it, focusing on the conservation of momentum and angular momentum to determine the angular speed of the gate post-collision.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using conservation of linear momentum and angular momentum, questioning the validity of combining these approaches. There is also exploration of using conservation of energy as an alternative method.

Discussion Status

Participants are actively questioning the dimensional correctness of the equations presented and discussing the implications of different lengths of the door on the equations used. Some guidance has been offered regarding the application of angular momentum and the potential issues with energy conservation in this context.

Contextual Notes

There are discussions around the assumptions made regarding the point of impact and the moment of inertia, with some participants seeking clarification on the definitions and setups used in the equations.

henry3369
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Homework Statement


A uniform 4.00kg square solid wooden gate 2.00m on each side hangs vertically from a frictionless pivot at its upper edge. A 1.30kg raven flying horizontally at 4.50m/s flies into this door at its center and bounces back at 2.50m/s in the opposite direction.

What is the angular speed of the gate just after it is struck by the unfortunate raven?

Homework Equations


L = Iα = r x mv
p = mv
Moment of inertia of a door = (1/3)MR2
Conservation of angular momentumγ

The Attempt at a Solution


So I know how to get the answer to this using conservation of momentum, but I was wondering if you can combine linear momentum and angular momentum.

For example, initially only the raven is moving which has linear momentum and finally, the raven and the door are moving:
mv = Iα + mv'

Or using only conservation of angular momentum:
(rxmv) = Iα + (rxmv')

In this situation the answers are the same because the raven hits the center of the door which is 1 meters from the axis of rotation (r = 1), but if the door had a different length, would I have to use the second conservation equation? Such as if the length of the door were 4 meters instead, r = 2.
 
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I was thinking if it would be easier if you used conservation of energy.
loss of KE (bird's) = rotational kinetic energy, doesn't matter where the bird hits, in this case.
 
henry3369 said:
(rxmv) = Iα + (rxmv')
I hope you didn't forget to include the -ve sign for ## v'##
 
Suraj M said:
I hope you didn't forget to include the -ve sign for ## v'##
I didn't forget to negate the final velocity. If I wanted to use conservation of momentum, for doors with different lengths, would I have to use the second conservation of momentum equation?
 
Wait, I didn't notice
Your second equation looks to be dimensionally wrong! Do you think all the quantities in your equation have the same units?
 
Suraj M said:
Wait, I didn't notice
Your second equation looks to be dimensionally wrong! Do you think all the quantities in your equation have the same units?
The second equation is conservation of angular momentum = Iα = r x mv.
 
henry3369 said:
The second equation is conservation of angular momentum = Iα = r x mv.
It is dimensionally wrong...the LHS is the formula for torque which is nothing but the RATE of change of angular momentum. But your RHS has the units of just angular momentum.do you still think your equation is correct.
 
Suraj M said:
It is dimensionally wrong...the LHS is the formula for torque which is nothing but the RATE of change of angular momentum. But your RHS has the units of just angular momentum.do you still think your equation is correct.
Sorry, I mean't Iω
 
henry3369 said:
Sorry, I mean't Iω
Yeah, all good.
I'm starting to doubt if the energy method i mentioned in post #2, would it work? what do you think?
 
  • #10
Suraj M said:
I was thinking if it would be easier if you used conservation of energy.
loss of KE (bird's) = rotational kinetic energy, doesn't matter where the bird hits, in this case.
It would not be valid to assume energy is conserved. Since the answer can be obtained by angular momentum considerations, you can calculate whether energy is conserved.
Henry, wrt the moment of inertia, you quote (1/3)mr2. What are you taking r to be there (looks like you are taking it as half the door height) and what axis are you using?
 

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