# Excellent problem on rotation

1. Oct 4, 2004

### Ganesh

Two masses of 'm/2' each are connected by a massless rigid rod of length 'l' and placed on a horizontal frictionless table. A third mass 'm' moving with velocity 'v' on the table strikes one of the masses in a head on elastic collision. Line of action is perpendicular to the length of the rod. Conservation of linear momentum and equation of elastic collision yields the velocity of the two colliding masses (as v/3 and 4v/3). Now, calculate initial and final angular momentum relative to the centre of mass of the masses attached at the ends of the rod. They will be different (mvl and mvl/6). (Here consider that the COM aquires a velocity of 2v/3). Where is the flaw since no external torque acts on the system in the frame of the COM of the two masses also?
Try !!!

2. Oct 4, 2004

### Ganesh

Two masses of 'm/2' each are connected by a massless rigid rod of length 'l' and placed on a horizontal frictionless table. A third mass 'm' moving with velocity 'v' on the table strikes one of the masses in a head on elastic collision. Line of action is perpendicular to the length of the rod. Conservation of linear momentum and equation of elastic collision yields the velocity of the two colliding masses (as v/3 and 4v/3). Now, calculate initial and final angular momentum relative to the centre of mass of the masses attached at the ends of the rod. They will be different (mvl and mvl/6). (Here consider that the COM aquires a velocity of 2v/3). Where is the flaw since no external torque acts on the system in the frame of the COM of the two masses also?
Try !!!

3. Oct 4, 2004

### Gokul43201

Staff Emeritus
I don't see the problem here. There are three equations, linear momentum; energy; and angular momentum conservation. Also, there are three unknowns (not two, as seems to be suggested), namely the final velocities of the three masses (or the final velocity of the point mass; the final linear velocity of the CoM of the double mass; and the final angular velocity of the double mass about the CoM) .

4. Oct 4, 2004

### Tide

If your concern is that angular momentum is not conserved then you're worried over nothing! Angular momentum is conserved! You must take into account the angular momentum of the incident ball - both before and after the collision.

5. Oct 4, 2004

### Gokul43201

Staff Emeritus
3 equations, 3 unknowns, no problem !

6. Oct 5, 2004

### Ganesh

Hi Tide and Gokul!

What I am saying is :
There are two equations for linear momentum and elastic collision.
(The third mass won't move as the rod can exert forces only along its length.)
So solving eqns.
mv = mv'/2 + mv" and
v - 0 = v' - v"
we get
v" = 4v/3 and v' = v/3.

NOW find angular momentum rel. to COM whose vel. is 2v/3.
It is m/2*(4v/3-2v/3)*l/2 + m/2*(2v/3-0)*l/2 + m*(v/3-2v/3)*l/2 = mvl/6

However initial angular momentum relative to COM which was at rest is mvl/2.

So, where is the problem?

7. Oct 5, 2004

### Gokul43201

Staff Emeritus
Now I'm confused. Could you label the masses 1, 2 and 3, and tell which one is which. I'm not sure which mass you are refering to by "third mass". I think you mean the mass (on the rod) that does not feel the collision from the free mass (which is not on the rod), but your original post refers to the mass not on a rod as the third mass.

Anyway, if you look at the statement v" = 4v/3, that should already tell you there's a problem, without having to look at angular momentum. How can v" > v ? How can the heavy free mass accelerate after collision ?

The error in your method is in the following assumption : "The third mass won't move as the rod can exert forces only along its length"

The flaw in this line of reasoning is that a rigidly attached rigid rod CAN exert a force which is not necessarily along its length.

Your argument is only valid for a non-rigid attachment, such as a rope.

8. Oct 5, 2004

### arildno

Ganesh:
Aside from the resulting linear velocities of the C.M's, you must also find the resulting angular velocity of the rod.
EDIT:
As Gokul already pointed out to you

Last edited: Oct 5, 2004
9. Oct 5, 2004

### chroot

Staff Emeritus
Ganesh,

Please do not cross-post. I have merged the two threads.

- Warren

10. Oct 5, 2004

### reilly

Here's a simple way to deal with this prolem.

First note that when particle A of mass m and speed v collides with B, also of mass m, and speed 0, A stops and B goes off with speed v in the same direction A was going.

Thus, the two connected masses move off, with the center of mass having velocity v.
The incident mass has angular momentum (L/2)*mv, just before collision. Each connected mass has speed v +(-) vR, an L/2 moment arm, and mass m/2, with vR the rotational velocity, defined in the cM frame. So each has angular momentum

mvRL/4 -(+) mvL/4.

. And, combined the system has angular momentum mvRL/2, so angular momentum is conserved if vR=v. Everything works

(

11. Oct 6, 2004

### Ganesh

I am sorry that a lot of confusion has arisen due to some improper wording.

Reilly, please note that masses connected at the ends are m/2 and m/2 while that of colliding mass is m.

Gokul, to avoid confusion, consider masses attached to the ends as A and B and that of the colliding mass as C.

As some of you have said, I can apply the three eqns of linear momentum, elastic collision and conservation of angular momentum in the GROUND frame.
Using that I get velocity of C as v/3 , A as 4v/3 and B as 0.
The same that I got before. So that is not exactly the point I am hinting at.

In the frame of the COM of A and B (whose vel is v/3), you will find that my statement above still holds, i.e, angular momentum is
m/2*(4v/3-2v/3)*l/2 + m/2*(2v/3-0)*l/2 + m*(v/3-2v/3)*l/2 = mvl/6

But initially ang. momentum is mvl/2.

So do try the problem again.

12. Oct 6, 2004

### reilly

In post 11, I made a serious error neglecting, as I did, the rotational motion's contribution to energy. Here's my take:

For simplicity, I'll set m=1, so the initial momentum is W, where W is the speed of the incident mass.

Let the incident mass have speed A after the collision, and the dumbell have speed V, and each rotating mass has linear rotational speed R.

Momentum Conservation W=V + A (M)
(The rotational momentum components cancel out)

Angular Momentum Conservation LW/2 = 2(1/2)(LR/2) + LA/2 (A)

or W/2 = R/2 + A/2

(This equation holds at the time of impact only. The moment arm, of course, increases with time.)

Conservation of energy W*W/2 = 2(V*V + R*R)/2 + A*A/2 (E)

From ((A), R=W - A, and, together, the momentum equations give R = V -- the linear and roational motion have the same speed -- this means that before and infitesimally just after the collision, the unhit mass has speed 0.

All this gives a quadratic equation for A, which gives A=1, or A=3/5. All the motion continues in the original direction.

Regards,
Reilly Atkinson

13. Oct 7, 2004

### reilly

Well, how about A=1 or A=1/3. The conservation of energy should be

W*W/2 = 1/2*(1/2)(V*V + R*R) + 1/2*(1/2)(V*V + R*R) + A*A/2.

Physics is easy, algebra is hard. RA

14. Oct 7, 2004

### arildno

Reilly&Ganesh:

Neither of you have been able to calculate initial angular momentum correctly!
Angular momentum is conserved for the system consisting of particle+rod.
It is preferable to calculate angular momentum with respect to that system's C.M.
The C.M of this system lies half-way between the end point where the impact happens, and the midpoint of the rod. Hence, the magnitude of initial angular momentum with respect to the C.M is mlv/4 and nothing else.

Initial quantities&configuration:
We put the origin at the midpoint of the rod.
We let the incident particle strike the end-point at $$\frac{l}{2}\vec{j}$$ with velocity $$v\vec{i}$$

Final quantities (subscript "p" for particle, "r" for rod system):
Final particle velocity: $$\vec{v}_{p}=\frac{1}{3}v\vec{i}$$
Final velocity of rod's C.M: $$\vec{v}_{r}=\frac{2}{3}v\vec{i}$$
Final angular velocity of rod:$$\vec{\omega}_{r}=-\frac{4}{3}\frac{v}{l}\vec{k}$$
There you have it.

Ganesh:
Please note that this DOES give your values for the velocities; your mistake was an incorrect set-up of conservation of angular momentum.
(Unless, of course, this has been intended from your side?)

To end it, lets look at the final angular momentum about the common center of mass:
Particle:
$$\frac{1}{4}\frac{mvl}{3}$$
Upper attached mass:
$$\frac{1}{4}\frac{1}{2}\frac{4mvl}{3}$$
Total:$$\frac{mvl}{4}$$
as it should be..

Last edited: Oct 7, 2004
15. Oct 7, 2004

### Gokul43201

Staff Emeritus
phew...that's a relief. I was getting pretty itchy too.

16. Oct 7, 2004

### arildno

I have muttered, growled and finally had to let out a primal scream..:grumpy:

17. Oct 7, 2004

### Gokul43201

Staff Emeritus
Say arildno, do you get even 5 hours of sleep at night ?

It's about midnight now, and I thought you were online really early this morning...or are you not in Oslo ?

You know your health is more important than PF !!

18. Oct 7, 2004

### arildno

Occasionally..

19. Oct 7, 2004

### Pyrrhus

Yes, arildno seems to take his job as science advisor verys erious :rofl:

20. Oct 8, 2004

Hi Arildno