Angular velocity after collision?

[naianator]

Homework Statement

A small circular block of mass M traveling with a speed v on a frictionless table collides and sticks to the end of a thin rod of with length D and mass M. The picture shows a top down view of the block and rod on the frictionless table. What is the rod's angular velocity after the collision? Express your answer in some combination of M, v, and D.

Homework Equations

conservation of angular momentum
I_cm = M*D^2/12
I_point mass = Mr^2

The Attempt at a Solution

So I guess the part that I'm struggling with is where to put the reference point. I tried the center of mass first:

L_i = M*v*D/2 + I*omega
and since omega_i = 0 its just
L_i = M*v*D/2

L_f = omega(I_cm + I_point mass) = omega(M*D^2/12+M*D^2/4)

and using conservation of angular momentum:
M*v*D/2 = omega(M*D^2/12+M*D^2/4)

M*v*D/2 = omega*M*D^2/3

omega = 3*v/(2*D)

I tried the same thing with the upper end of the rod and got 3*v/(4*D) so I'm obviously doing something wrong. Do I have to include the translation of the center of mass? If I do I'm not quite sure how.

 

[TSny]

The Attempt at a Solution

So I guess the part that I'm struggling with is where to put the reference point. I tried the center of mass first:

OK, just to be clear, it appears that you are choosing your origin to be at the initial position of the center of mass of the rod, not the center of mass of the whole system.

L_i = M*v*D/2 + I*omega
and since omega_i = 0 its just
L_i = M*v*D/2

OK

L_f = omega(I_cm + I_point mass) = omega(M*D^2/12+M*D^2/4)

No, the equation above isn't correct.

Do I have to include the translation of the center of mass?

Yes. Just after the collision, the center of the rod will have a horizontal "recoil" velocity V_c,f. The velocity of the point mass just after the collision will have a contribution from V_c,f and a contribution from ω.

Both V_c,f and ω are unknowns. So, you're going to need another equation. Can you think of another conservation principle that is applicable?

Note: there is a tool bar for formatting. The Σ icon will bring up some mathematical symbols.

 

[naianator]

OK, just to be clear, it appears that you are choosing your origin to be at the initial position of the center of mass of the rod, not the center of mass of the whole system.OK
No, the equation above isn't correct.
Yes. Just after the collision, the center of the rod will have a horizontal "recoil" velocity V_c,f. The velocity of the point mass just after the collision will have a contribution from V_c,f and a contribution from ω.

Both V_c,f and ω are unknowns. So, you're going to need another equation. Can you think of another conservation principle that is applicable?

Note: there is a tool bar for formatting. The Σ icon will bring up some mathematical symbols.

Conservation of energy right? Or can you use conservation of linear momentum to find the velocity of the center of mass?

 

[TSny]

Conservation of energy right? Or can you use conservation of linear momentum to find the velocity of the center of mass?

Is the collision elastic or inelastic?

 

[naianator]

Is the collision elastic or inelastic?

inelastic, so I should use conservation of linear momentum?

 

[naianator]

inelastic, so I should use conservation of linear momentum?

in which case v_f = v/2 but how do i incorporate that into the angular momentum equation?

 

[TSny]

in which case v_f = v/2 but how do i incorporate that into the angular momentum equation?

It's not that easy to get the final velocity of the center of the rod, V_c,f. You'll need to set up the linear momentum equation carefully.

 

[naianator]

It's not that easy to get the final velocity of the center of the rod, V_c,f. You'll need to set up the linear momentum equation carefully.

p_i = M*v right?
and p_f = M*v'+M*v' = 2*M*v'
What am i doing wrong?

 

[haruspex]

in which case v_f = v/2 but how do i incorporate that into the angular momentum equation?

In general, a reference point for conservation of angular momentum should be either a fixed point or the mass centre of the system. You appear to have taken the mass centre of the rod, but not as a fixed point. Take it it as fixed at the initial position of the rod's mass centre. Then, as TSny says, you get a post-collision contribution from the linear motion of the disc. You can treat that either as purely linear motion, or as the linear motion of the rod's centre plus a contribution from the rotation of the rod.

Although it's easy to find the post-collision velocity of the system's mass centre, it is, as you found, awkward to use. I would introduce an unknown for the post-collision velocity of the rod's mass centre. The velocity of the disc can easily be expressed in terms of that and the rotation rate. The two conservation equations allow you to solve for those two unknowns.
I believe this is the path TSny laid out.

 

[naianator]

In general, a reference point for conservation of angular momentum should be either a fixed point or the mass centre of the system. You appear to have taken the mass centre of the rod, but not as a fixed point. Take it it as fixed at the initial position of the rod's mass centre. Then, as TSny says, you get a post-collision contribution from the linear motion of the disc. You can treat that either as purely linear motion, or as the linear motion of the rod's centre plus a contribution from the rotation of the rod.

Although it's easy to find the post-collision velocity of the system's mass centre, it is, as you found, awkward to use. I would introduce an unknown for the post-collision velocity of the rod's mass centre. The velocity of the disc can easily be expressed in terms of that and the rotation rate. The two conservation equations allow you to solve for those two unknowns.
I believe this is the path TSny laid out.

I'm not sure I completely understand, if I take the center of mass of the system 3*D/4 as the reference point will the linear conservation be less awkward to use?

 

[TSny]

I believe this is the path TSny laid out.

Yes, I was considering the origin as a fixed point that coincides with the initial location of the center of the rod.

 

[haruspex]

I'm not sure I completely understand, if I take the center of mass of the system 3*D/4 as the reference point will the linear conservation be less awkward to use?

No, I'm saying that I regard it generally as simpler not to bother working out where the mass centre of the system is. I just assign a variable to the velocity of each component. But it's a matter of taste.

 

[naianator]

Yes, I was considering the origin as a fixed point that coincides with the initial location of the center of the rod.

so how do I find the final velocity?

 

[TSny]

Try to express the final linear momentum of the system in terms of M, V_c,f, ω, and L.

 

[naianator]

No, I'm saying that I regard it generally as simpler not to bother working out where the mass centre of the system is. I just assign a variable to the velocity of each component. But it's a matter of taste.

So would it be correct to say that L_f = m*D^2/3*ω+2*m*D/4*v_f

 

[haruspex]

So would it be correct to say that L_f = m*D^2/3*ω+2*m*D/4*v_f

It depends... what's v_f here?

 

[naianator]

It depends... what's v_f here?

the final velocity of the center of mass of the system

 

[haruspex]

the final velocity of the center of mass of the system

Then the equation is wrong. $\omega \frac D2$ is the velocity of the disc relative to what?

 

[naianator]

Then the equation is wrong. $\omega \frac D2$ is the velocity of the disc relative to what?

the center of the rod right?

 

[naianator]

Try to express the final linear momentum of the system in terms of M, V_c,f, ω, and L.

m*v_cf+m*ω*d/2?

 

[haruspex]

the center of the rod right?

Right. You have accounted for the angular momentum from that relative motion in the m*D^2/3*ω term, so what do you need still to add in for the disc's angular momentum?

 

[naianator]

Right. You have accounted for the angular momentum from that relative motion in the m*D^2/3*ω term, so what do you need still to add in for the disc's angular momentum?

md^2/16*ω?

 

[haruspex]

md^2/16*ω?

No.
You already have the part of the disc's angular momentum that comes from the rotation of the rod. But that is only relative to the linear velocity of the mass centre of the rod. So what part of the disc's motion does it not account for?

 

[naianator]

No.
You already have the part of the disc's angular momentum that comes from the rotation of the rod. But that is only relative to the linear velocity of the mass centre of the rod. So what part of the disc's motion does it not account for?

the translational? How do I account for that?

 

[naianator]

the translational? How do I account for that?

I thought the translation of the center of mass would account for the translation of the system

 

[haruspex]

I thought the translation of the center of mass would account for the translation of the system

The rod's mass centre is moving at velocity v_c,f. That has no angular momentum about the reference point because the path is through the reference point.
The rod is rotating about its mass centre. That has angular momentum $\frac 1{12}m*D^2\omega$ about the reference point.
The disc is rotating about the rod's mass centre. That has angular momentum $\frac 1{4}m*D^2\omega$ about the reference point.
The disc's total velocity is $\vec v_{c,f}+\vec{\omega} \times \frac 12 \vec D$. The $\vec v_{c,f}$ part is not through the reference point in this case. So what is the disc's angular momentum about the reference point?