# Can someone check my answer to a rotational problem?

• toesockshoe
In summary, the problem involves a stick with a point mass colliding and adhering to it. The resulting motion can be found by setting the initial and final angular momentums equal to each other and solving for the final angular velocity. It is important to take a non-accelerating reference point and account for the post-impact momentum of the stick's center and the point mass.
toesockshoe

## Homework Statement

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.

L=IW
L=rxp

## 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 wouldn't really be affected by the rotations right? [/B]

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##.

haruspex said:
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##.
ok so the origin would be the center of the stick and stay at that position (so it doesn't move along with the stick right)? would the post impact of the mass be: r x P + mbw?

toesockshoe said:
ok so the origin would be the center of the stick and stay at that position (so it doesn't move along with the stick right)? would the post impact of the mass be: r x P + mbw?
It depends how you are defining r and P.

haruspex said:
It depends how you are defining r and P.
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?

toesockshoe said:
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?
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##

haruspex said:
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##
wait do you mean the total momentum of the stick would be: $mbv + mb^2\omega$ ? isn't that what I said when I said you simply add mvb?

toesockshoe said:
wait do you mean the total momentum of the stick would be: $mbv + mb^2\omega$ ? isn't that what I said when I said you simply add mvb?
Are you mixing up m with M?
Your post #5 is somewhat confusing because you wrote
toesockshoe said:
m would be the velocity of the mass (which is the same as the velocity of the stick)
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##.

haruspex said:
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##.
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?

haruspex said:
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##.
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?

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

toesockshoe said:
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?
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.

haruspex said:
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.
yes yes, idk why I am 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?

toesockshoe said:
yes yes, idk why I am 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?
Yes.

haruspex said:
Yes.
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?

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?

toesockshoe said:
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?
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.

toesockshoe said:
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?
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.

haruspex said:
Yes, but rather than work out where the mas centre 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.
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?

haruspex said:
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.
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 isn't the final angular momentum of the mass r x p instead of Iw^2?

toesockshoe said:
would the equation be:##v_{fp} = \frac{m_pv_i-m_sv_{fs}}{m_p}##
No, I mean express ##v_{fp}## in terms of ##v_{fs}##, ##\omega## and b.

haruspex said:
No, I mean express ##v_{fp}## in terms of ##v_{fs}##, ##\omega## and b.
i think there is a problem with my understand of rotations first. i want to see where I am 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 isn't the final angular momentum of the mass r x p instead of Iw^2?

toesockshoe said:
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 isn't the final angular momentum of the mass r x p instead of Iw^2?
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.

haruspex said:
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.
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?

toesockshoe said:
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?
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.

haruspex said:
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.
oh ok, so how would it be possible to use L=Iw for the point mass's momentum here? It would not be possible right... becuase it is a single point that is not rotating around its center.

haruspex said:
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.
also, we don't know $v_{fs}$ or $v_{fp}$. to find the final linear velocity of the stick, do we do $p_i = p_f$?

in which case: $m_pv_i = (m_p+m_s)v_f$ and thus: $v_f = \frac{m_pv_i}{m_p+m_s}$ ?

toesockshoe said:
how would it be possible to use L=Iw for the point mass's momentum here?
If I here is the MoI about the particle's mass centre, that will be zero. The particle's angular momentum about the reference point, post collision, will be ##m_pv_{fp}b##.
toesockshoe said:
we don't know ##v_{fs}## or ##v_{fp}##.
No, but as I keep pointing out there is a relationship between them. It involves w and b, but does not involve any masses.
toesockshoe said:
##v_f = \frac{m_pv_i}{m_p+m_s}##
No. That would be true if ##v_{fs}=v_{fp}##.

We seem to be going around in circles here.
First, find the relationship between ##v_{fs}##, ##v_{fp}##, w and b. It's simply a relative velocity relationship. Just after collision, what is the particle's relative velocity compared with the mass centre of the stick?
When you have that, write out the equation for conservation of linear momentum.

haruspex said:
If I here is the MoI about the particle's mass centre, that will be zero. The particle's angular momentum about the reference point, post collision, will be ##m_pv_{fp}b##.

No, but as I keep pointing out there is a relationship between them. It involves w and b, but does not involve any masses.

No. That would be true if ##v_{fs}=v_{fp}##.

We seem to be going around in circles here.
First, find the relationship between ##v_{fs}##, ##v_{fp}##, w and b. It's simply a relative velocity relationship. Just after collision, what is the particle's relative velocity compared with the mass centre of the stick?
When you have that, write out the equation for conservation of linear momentum.
the relationship is $v_{fp} = v_{fs} + \omega b$

this is the conservation of linear momentum:
$p_i = p_f$
$m_pv_i^2 = m_pv_p + m_sv_s$
$m_pv_{pi}^2 = m_p(v_{fs}+\omega b) + m_sv_{fs}$

toesockshoe said:
the relationship is $v_{fp} = v_{fs} + \omega b$

this is the conservation of linear momentum:
$p_i = p_f$
$m_pv_i^2 = m_pv_p + m_sv_s$
$m_pv_{pi}^2 = m_p(v_{fs}+\omega b) + m_sv_{fs}$
Good, except for the v2 on the left.
Fix that, then use the above to determine $v_{fs}$.

haruspex said:
Good, except for the v2 on the left.
Fix that, then use the above to determine $v_{fs}$.
uhhh, that gets really messy cause there's a quadratic doesn't it?
anyway, after I find $v_{fs}$ ... I find $v_{fp}$ using the vfs + wb forumla right?

also, my friend did the priblem this way:

$L_i = L_f$
$m_pv_{pi}b = i\omega$
$m_pv_{pi}b = (\frac{m_sL^2}{12} + m_pb^2) w_f$ because the angular velocity are the same for both masses right?

isnt this method much easier?

toesockshoe said:
that gets really messy cause there's a quadratic doesn't it?
But it shouldn't be quadratic. I said that was an error in your equation:
haruspex said:
Good, except for the v2 on the left.
Correct that equation and try again.
toesockshoe said:
isnt this method much easier?
Oh yes, that's much easier. Unfortunately it's wrong. It's the same error you made in your initial post on this thread.

haruspex said:
Oh yes, that's much easier. Unfortunately it's wrong. It's the same error you made in your initial post on this thread.

you can't do it because the point is not rotating on its around its own mass center right?

haruspex said:
it shouldn't be quadratic. I said that was an error in your equation.

I did try solving it after I fixed it... I got $m_p v_{pi}^2 = m_p(v_{fs}+\omega b)^2 + m_sv_{fs}^2$... and that gets messy because you need to square the $v_{fs} + wb$ and that makes it difficult to factor out $v_{fs}$ becasue there is two v_fs terms to the 2nd power and one v_fs term to the first power. where am i going wrong?

toesockshoe said:
you can't do it because the point is not rotating on its around its own mass center right?
Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just ##m_pb^2\omega_f##. There is also the ##m_pv_{fs}b## term. Look at your relative velocity equation.
toesockshoe said:
I did try solving it after I fixed it... I got $m_p v_{pi}^2 = m_p(v_{fs}+\omega b)^2 + m_sv_{fs}^2$
I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.

haruspex said:
Not exactly. It's wrong because the particle's angular momentum about the point in space where the mass centre of the stick had been is not just ##m_pb^2\omega_f##. There is also the ##m_pv_{fs}b## term. Look at your relative velocity equation.

I'll try once more... IT SHOULD NOT BE QUADRATIC! You appear to be adding up energies, not angular momenta.
OHH STUPID ME. yes sorry. i added up energies.

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