# Conservation of momentum: linear and angular

1. Mar 15, 2015

### henry3369

1. The problem statement, all variables and given/known data
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?
2. Relevant equations
L = Iα = r x mv
p = mv
Moment of inertia of a door = (1/3)MR2
Conservation of angular momentumγ

3. 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.

Last edited: Mar 15, 2015
2. Mar 15, 2015

### Suraj M

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.

3. Mar 15, 2015

### Suraj M

I hope you didn't forget to include the -ve sign for $v'$

4. Mar 15, 2015

### henry3369

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?

5. Mar 15, 2015

### Suraj M

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?

6. Mar 15, 2015

### henry3369

The second equation is conservation of angular momentum = Iα = r x mv.

7. Mar 15, 2015

### Suraj M

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.

8. Mar 15, 2015

### henry3369

Sorry, I mean't Iω

9. Mar 17, 2015

### Suraj M

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. Mar 17, 2015

### haruspex

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?