# Homework Help: Angular Momentum Help

1. Jan 10, 2010

### jdboucher

1. The problem statement, all variables and given/known data
A projectile of mass m = 1.49 kg moves to
the right with speed v0 = 19.6 m/s. The pro-
jectile strikes and sticks to the end of a sta-
tionary rod of mass M = 6.55 kg and length
d = 2.29 m that is pivoted about a frictionless
axle through its center.
Find the angular speed of the system right
after the collision.

2. Relevant equations

Icom rod = (1/12)ML^2

3. The attempt at a solution[/b

I figured I would use conservation of angular momentum to determine the final angular speed. I'm unsure how to find the initial angular momentum. I can find the final moment of inertia.

(1/12)(1.49 + 6.55)(2.29)^2 = 3.51

HELP!

2. Jan 10, 2010

### Winzer

I think viewing it from an energy perspective will be easier.

3. Jan 10, 2010

### jdboucher

Does translational (i think thats what its called) kinetic energy relate to rotational kinetic energy?

4. Jan 10, 2010

### Winzer

What happens in the problem? The energy gets converted from ____ to _____.
It comes from the conversion of motion. From Linear to Rotational right?

5. Jan 10, 2010

### jdboucher

Is it kinetic to potential?

6. Jan 10, 2010

### jdboucher

so can I do this:

0.5mv^2 = Iw

7. Jan 10, 2010

### Winzer

I was looking for the energy gets converted from linear kinetic to rotational energy, right?
Use conservation of energy:
$$KE_{Linear}=KE_{Rotational}$$

8. Jan 10, 2010

### Winzer

It should be :
$$\frac{1}{2}mv^2 = \frac{1}{2}I \omega^2$$

9. Jan 10, 2010

### jdboucher

0.5(1.49)(19.6)^2 = 0.5(1/12)(1.49+6.55)(2.29)^2w^2

I solved this and got w = 12.76

The answer is wrong though. I checked my work. Did I do something wrong?

10. Jan 10, 2010

### vela

Staff Emeritus
The two bodies stick together, so this is an inelastic collision. Kinetic energy isn't conserved. You have to use angular momentum.

Remember $\vec{L}=\vec{r}\times\vec{p}$.

11. Jan 10, 2010

### jdboucher

So then I need to determine r. Is that d/2?

12. Jan 10, 2010

### danielatha4

Yes, r would be d/2. It's the distance to the pivot point.

13. Jan 11, 2010

### jdboucher

I assume the angle is 90. So L = mvrsin90. If I do that, set it equal to Iw, I get that w= 9.52. According to the online site I'm using thats wrong. Is it my angle?

14. Jan 11, 2010

### vela

Staff Emeritus
Your $I_{final}$ is wrong.

15. Jan 11, 2010

### jdboucher

Ifinal = (1/12)ML^2 = (1/12)(1.49 + 6.55)(2.29)^2

Ifinal = 3.51

Is that wrong?

16. Jan 11, 2010

### vela

Staff Emeritus
Yes, it's wrong. That would be the moment of inertia for a uniform rod of mass 8.04 kg. You have a total mass of 8.04 kg, but it's not distributed as a uniform rod.

17. Jan 11, 2010

### jdboucher

(1/12)(6.55+1.49)(2.29)^2 + (1.49)(2.29/2)^2 = 6.12

Its wrong again. Am I not supposed to use the parallel axis theorm?

18. Jan 11, 2010

### jdboucher

I mean that I assumed the bullet was a point and used mr^2 to solve for its moment of inertia. Thats wrong obviously

19. Jan 11, 2010

### vela

Staff Emeritus
No, the parallel axis theorem lets you calculate the moment of inertia of an object when it's rotating about an axis that doesn't go through its center of mass, like if you were rotating the rod by its end rather than through its center. That's not the case here, so the theorem doesn't apply.

In this problem, the total rotational mass is simply that of the rod (by itself) plus that of the projectile. You have the formula for the rod; just make sure you use the mass of the rod alone since the projectile isn't part of the rod. The projectile is a point mass at a distance $r$ from the rotation axis. Its rotational mass is given by $mr^2$, where $m$ is the mass.

20. Jan 11, 2010

### vela

Staff Emeritus
No, that's right. You just plugged the wrong mass in for the rod.

21. Jan 11, 2010

### jdboucher

Yes you're right. Silly me. I got it now. Thank you for your help.