# Homework Help: Can someone check my answer to a rotational problem?

1. Jun 6, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
A stick of length L and mass M is in free space and not rotating. A point mass m has an initial velocity v heading in a trajectory perpendicular to the stick. The mass collides and adheres to the stick a distance b from the center of the stick. Find the resulting motion of the two together in terms of their center of mass velocity and final angular velocity.

2. Relevant equations
L=IW
L=rxp
3. The attempt at a solution

I set the origin at the center of rotation (in the middle of the stick)
LET M STAND FOR THE SMALL MASS AND S STAND FOR THE STICK.
this is a momentum problem so i said $L_i=L_f$ at the time of contact.
$L_{mi} + L_{si} = L_{mf} + L_{sf}$
initial momentm of the stick is 0.
$mv_ib = I_{stick}w+I_{mass}w$
$m_pv_ib=\frac{m_sL^2}{12}w_f + mb^2w$
solve for $w_f$
$w_f = \frac{m_pv_ib}{\frac{mL^2}{12}+m_pb^2}$

i know i didnt solve for center of mass velocity but would we just do a regular momentum problem? (not rotational) becasue there are no external forces and the center of mass wouldnt really be affected by the rotations right?

2. Jun 6, 2015

### haruspex

It is important to take a non-accelerating reference point. The mass centre of the stick gets accelerated by the impact. You can use the position in space of the mass centre before impact as your reference point. If you do that, the mass m has a post-impact moment which is more than just $mb\omega$.

3. Jun 6, 2015

### toesockshoe

ok so the origin would be the center of the stick and stay at that position (so it doesnt move along with the stick right)? would the post impact of the mass be: r x P + mbw?

4. Jun 6, 2015

### haruspex

It depends how you are defining r and P.

5. Jun 6, 2015

### toesockshoe

well, P=mv right? and m would be the velocity of the mass (which is the same as the velocity of the stick) and we can take $rsin(\theta)$ (where r is the distance from the origin to the mass )as b (because rsin(theta) is always b). would that be correct?

6. Jun 6, 2015

### haruspex

If P is the momentum of the point mass in an inertial frame, you don't need to add $mb^2\omega$ as well. $b\omega$ is the velocity of the mass relative to the rod's centre. Alternatively, if you were to define v as the velocity after impact of the rod's centre then you would use $mbv+mb^2\omega$

7. Jun 6, 2015

### toesockshoe

wait do you mean the total momentum of the stick would be: $mbv + mb^2\omega$ ? isnt that what I said when I said you simply add mvb?

8. Jun 6, 2015

### haruspex

Are you mixing up m with M?
Your post #5 is somewhat confusing because you wrote
I assume you meant v, not m, and the velocity of the mass is not the same as the velocity of the stick. If v is the velocity of the stick's mass centre then the velocity of the mass is $v+b\omega$. Its angular momentum about the reference point is mb times that. The total angular momentum about the reference point is $ML^2\omega/12+mbv+mb^2\omega$.

9. Jun 6, 2015

### toesockshoe

ugh yes i messed up. in post 7, i mean the total angular momentum of the mass.... not the stick. that would be correct right?

10. Jun 6, 2015

### toesockshoe

also, you would just set that equal to the inital momentum, which I said was $m_pv_ib$ and solve for w, and that would be the correct answer right?

11. Jun 6, 2015

### haruspex

If you are defining v as the post impact velocity of the stick's centre, yes.

12. Jun 6, 2015

### haruspex

Yes, but you also need to use linear momentum to find the post-impact velocity of the stick. I fear there is confusion here because you have used v for that unknown, but v is the given initial velocity of the mass.

13. Jun 6, 2015

### toesockshoe

yes yes, idk why im making so many stupid mistakes. ok, so to do that would just just use $p_f=p_i$ and this would give you the linear velocity right?

14. Jun 6, 2015

### haruspex

Yes.

15. Jun 6, 2015

### toesockshoe

ok, so doing that I get : $v_f = \frac{m_pv_i}{m_p+m_s}$ , then its a simple plug into the rotational mometum problem. Also, this would be the velocity of the center of mass correct?

16. Jun 6, 2015

### toesockshoe

also, do you mind explain to me why only the momentum of the small mass is different than what I posted initially? How come the angular momentum of the stick has no linear velocity factored into it?

17. Jun 7, 2015

### haruspex

Yes, but rather than work out where the mass centre of the whole is, it's simpler to let vf be the velocity of the mass centre of the rod and write an expression for the linear velocity of the mass (just after impact) in terms of that.

18. Jun 7, 2015

### haruspex

You are taking moments about the point where the mass centre of the rod was just before impact. The velocity of the mass centre of the rod just after impact is along a line passing through that point, so has no angular momentum about it.

19. Jun 7, 2015

### toesockshoe

oh is it because the center of the rod is moving in a straight line whereas the mass is moving in a circle?
would the equation be: $v_{fp} = \frac{m_pv_i-m_sv_{fs}}{m_p}$.... now i have 2 unknowns in this equation. how would be able to solve that?

20. Jun 8, 2015

### toesockshoe

hey haruspex... i had another question on this problem. why are we using $L = I\omega^2$ for the momentum of the point mass after the collision? in another thread I created, some said that Iw^2 is only used for spinning and L= r x p is used for rotating around another object (which the bullet is doing). why isnt the final angular momentum of the mass r x p instead of Iw^2?

21. Jun 8, 2015

### haruspex

No, I mean express $v_{fp}$ in terms of $v_{fs}$, $\omega$ and b.

22. Jun 8, 2015

### toesockshoe

i think there is a problem with my understand of rotations first. i want to see where Im going wrong there. do you mind answering the following:

had another question on this problem. why are we using L=Iω2 for the momentum of the point mass after the collision? in another thread I created, some said that Iw^2 is only used for spinning and L= r x p is used for rotating around another object (which the bullet is doing). why isnt the final angular momentum of the mass r x p instead of Iw^2?

23. Jun 8, 2015

### haruspex

Are we using $L = I\omega^2$ for the point mass? I'm using r x p, where r = b, $p = m v_{fp}$, and $v_{fp}$ is ... some function of $v_{fs}$, b and ω.
Of course, they're the same really. Using the original location of the stick's mass centre as the reference point, you can write $I_p= mb^2$, but then you have to work out the instantaneous angular velocity of the particle about that point. It will not be just ω.

24. Jun 8, 2015

### toesockshoe

is $v_{fp} = v_{fs}b\omega$ ? the point mass is moving in a circle so I'm not sure how the linear velocity would look. also, you didnt answer my question on when we have to use L=Iw^2 and when we have to use L=p x r.... i seem to have some confusion here.... are both the exact same? Do we get the same answer if we use either one of them? is there a general rule other than use Iw^2 if the object is spinning and use p x r if the object is rotating around an axis?

25. Jun 8, 2015

### haruspex

I assume you mean $v_{fp} = v_{fs}+b\omega$ , in which case, yes.
L=Iw (not w^2) is generally used in regard to an object rotating about its mass centre. If we break the object into tiny mass elements, we can apply L=mr^2w to each and add them up. I is just the sum of the mr^2 terms, and all have the same w.
Alternatively, each of those mass elements has velocity v=rw orthogonal to its radius vector, so we can obtain its moment about the mass centre by L=r x p = mvr = mr^2w, same thing.
If we are considering the moment about some other point, we can either use the parallel axis theorem and consider rotation about the instantaneous centre of rotation, or treat the motion as the combination of a rotation about the mass centre plus a linear motion of the mass centre. Suppose the mass centre is distance r from the instantaneous centre of rotation, O. With the first approach, IO = I+mr^2, L=IOw. With the second, L=Iw+mrv=Iw+mr^2w. Same thing.