# Homework Help: Angular momentum

1. Oct 27, 2007

### Microzero

A uniforum stick AB of mass M and length L is moving on a smooth plane with speed v. The motion is normal to the stick. The stick collides with a fixed nail on the plane and impact is at the middle of AC, where C is the C.M. of the stick.

Q: What is the instantaneous speed of the C.M. of the stick?

The attempt at a solution:
Let F be the force between the stick and the nail
vf be the instantaneous speed of the C.M. of the stick
ω be the angular velocity of the stick after impact

△P = F△t
M(vf - v)= F△t

Take moment @ C
∴ I=1/12 ML^2
△L (angular momentum) = τ(torque)△t
∵τ= rXF
∴I(ω-0)= (L/4)F△t
(1/12 ML^2)ω = (L/4) M(vf - v)
...
(vf - v) = 1/3 Lω

How can I let ω interms of v or vf ?
Or..Is there anything I have missed?

2. Oct 28, 2007

### rl.bhat

Whether the motion of the stick is only tanslatory or both transalatory and rotaional? And again after the collision what is the nature of the motion? In both the cases use conservation of energy and angular momentum.

3. Oct 28, 2007

### Microzero

Thx~
I have not try to use conservation of energy becox I'm afraid that there will be energy loss during the collision

If there are energy loss, how can I solve this problem?

4. Oct 28, 2007

### Shooting Star

Suppose that instead of a rod, it was a small sphere given in the problem. Then could you have found the instantaneous speed of the sphere without assuming conservation of KE?

You have to assume that the KE of the rod is conserved.

5. Oct 28, 2007

### Microzero

By the use of conservation of energy and the motion of the stick is transalatory and rotational:

(1/2)Iω^2 + (1/2) mvf^2 = (1/2) mv^2

ω^2 = 12/L (v^2 - vf^2)
(vf - v) = (1/3) Lω
∴ (vf - v)^2 = (1/9) L^2 ω^2 = (1/9) L2 (12/L) (v^2 - vf^2)

(7/3) vf^2 + (-2v) vf - (1/3) v^2 = 0

vf = v or (-1/7) v

Then i take vf = (-1/7) v becox the final velocity must be smaller the initail one ( as there are rotation, RKE exist)
Am I right?

Last edited: Oct 28, 2007
6. Oct 28, 2007

### Shooting Star

(Initial KE= KE of CM + RKE about CM. That's right.)

Let’s call the mid pt of AC as D. Then, instantaneously, D is at rest. Calculate the RKE only taking D to be fixed. Both the segments AD and BD have the same omega, and vf of CM is given by vf=(length of CD)*omega.

So, mv^2/2=(1/2)I1w^2 +(1/2)I2w^2, where I1 and I2 is about D, and vf=(L/4)*w.

I get the new speed as sqrt(3/7)v. Cross check it.

7. Oct 28, 2007

### rl.bhat

The rod hits the nail normally. So it must return back in the same direction. The rod is not rotating about D. Hence Vf = r*w, where r is the radius of the rod.

8. Oct 28, 2007

### Microzero

I'm agree that the rod is not rotating about D, but why Vf = r*w?
r*w is the tangential velocity of the end pt, not the final velocity of the CM
Am I correct?

9. Oct 28, 2007

### Staff: Mentor

Maybe I haven't had enough coffee yet, but I don't get the point of this problem. Instantaneous speed when? Realize that by symmetry the interaction exerts no torque about the c.m. of the stick, so no rotation occurs.

10. Oct 28, 2007

### Shooting Star

The rod hits the nail off-centre. So, why will it not rotate?

The resulting motion will be translation and rotation.

The OP is not very clear about the kind of impact. We have to assume that the KE is conserved.

11. Oct 28, 2007

### Staff: Mentor

Ah... I misread the problem. Impact is at at the middle of AC, not AB. (D'oh!) ...<reaches for the coffee pot>...

12. Oct 28, 2007

### Staff: Mentor

The way I read the problem (reread it, actually), what they are looking for is the speed of the C.M. at the "instant" of the collision, which I take to be that instant where the midpoint of AC is momentarily at rest. At that instant the rod can be viewed as in pure rotation about the impact point.

13. Oct 28, 2007

### Microzero

Someone solve this Q without using conservation of energy:
Let P = linear impulse acted on the stick by the nail
Let I = MI of stick about C
Let V1 = speed of C after impact

By conservation of linear momentum,
MV - P = MV1...............................(1)

By conservation of angular monentum about C,
P(L/4) = wI................................(2)

By kinematics,
V1 = (L/4)w................................(3)

I don't know why (3) is correct

14. Oct 28, 2007

### Shooting Star

Have you read my earlier post? Looks as if we are thinking on the same lines?

15. Oct 28, 2007

### Microzero

vf=(L/4)*w <--- then we have the relationship between w and vf
we don't need to use covservaton of energy

16. Oct 28, 2007

### Shooting Star

How do we find the relationship between vf and v? We don't know omega.

17. Oct 28, 2007

### Microzero

Let P = linear impulse acted on the stick by the nail
Let I = MI of stick about C
Let Vf = speed of C after impact

Change in linear momentum,
MV - P = MVf...............................(1)

Change in angular monentum about C,
P(L/4) = wI................................(2)

By kinematics,
Vf = (L/4)w................................(3)

(1) and (2) give the relationship between w, v, vf
(3) give the relationship between w, vf
then vf can be interms of v
and the ans is vf=3/7 v

Last edited: Oct 28, 2007
18. Oct 28, 2007

### Staff: Mentor

Just to be clear, angular momentum of the rod about C is not conserved. (The impact imparts an angular impulse about C.) Instead, consider angular momentum about the point of impact.

It's correct if you presume that the rod is momentarily in pure rotation about the impact point.

19. Oct 28, 2007

### Shooting Star

What's going on? Why would the ang mom be "conserved" about C? P(L/4) is not ang mom, more like a change in ang mom, like impulse.

20. Oct 28, 2007

### Microzero

Actually..I think the word 'By conservation of angular monentum about C' is not correct
It should be 'change in ang mom abt C'

sorry abt that

21. Oct 28, 2007

### saket

Please Do Not Take "Doc Al" Easily!

"Doc Al" is the PF Mentor and a recognized homework helper. I would suggest you all, not to take him casually. I think this problem was explained best by him. (I hope you people will agree with me that arriving at a result is one thing, properly explaining every step of yours is a different thing!)
Let me explain myself in the context of this problem.
Everyone else, other than "Doc Al", had to make some assumption or the other to solve this problem. He alone, gave a different look to the problem, and I think he did so brilliantly!
The question was "What is the instantaneous speed of the C.M. of the stick?". And, he took that instant of the collision when the point of contact came at rest. He didn't demand anything extra-ordinary because, that is what most of us will define as the instant of collision. (Although, collision happens over a short interval of time.)
And, going by the method he suggested (conservation of angular momentum about the fixed nail) one can easily arrive at the result:
M*v*(l/4) = M*vf*(l/4) + M*w*(l^2)/12
and since the point of contact on the rod is at rest, vf - w*(l/4) = 0.
Solving the above two, we get vf = 3*v/7

But those, who do not agree with the proposition of "Doc Al", will run short of information and they have, knowingly or unknowingly, made some assumption or the other to derive at the result.
Let us try to do the problem without any assumption of physics. If the question meant to ask final velocity of the centre of the mass of the rod, the solution would have been slightly different.
I shall work with variables and substitute the values at last, in order to get deeper insight on the problem.
Let us denote the nail by N and the point on the rod which makes contact with the nail be denoted by D. (I am using the notations used by you people to avoid any confusion.)
Further, although CD = l/4 is given, I assume CD = l/k. Please note that, for this problem k = 4.
I, further, assume that coefficient of restitution = e.
Conservation of angular momentum of the rod about N (because the impulse on the rod during collision passes through N, and therefore torque-impulse about N is zero) gives:
(l/k)*M*v = (l/k)*M*vf + M*w*(l^2)/12
and, using "e" between N and D,
w*(l/k) - vf = e*v
Solving the above two equations, we get:
vf = v*{12 - (k^2)*e}/{12 + (k^2)}.

Assumption I : It is perfectly elastic collision.
Then, e = 1 and k = 4.
We get, vf = -v/7. (As was obtained by you people.)

Assumption II : It is perfectly inelastic colliion.
Then, e = 0 and k = 4.
We get, vf = 3*v/7 (As was obtained by you people.)

So, knowingly or unknowingly, you people (other than "Doc Al") were using either of the assumptions. Please, note that, if you don't use the assumption you will get stuck with one unknown variable, here e, as was "Microzero" when he posted this problem.

Thus, I think, "Doc Al" did his job brilliantly, didn't he?

Well, I also wanted to draw attention to few results.
Note that 0 <= e <= 1. Also, |k| >= 2, for the collision to take place.

Perfectly Elastic Collision (e = 1)
1. The rod can recoil back with same speed, v, only if D coincides with C and e = 1. vf = -v, w = 0.
2. When D coincides with one of the ends (i.e., the rod grazes against the nail), |k| = 2, e = 1, gives vf = v/2, w = 7*v/(4*l).
3. When CD = sqrt(3)*l/6, vf = 0, w = 2*sqrt(3)*v/l. Thus, the rod comes to (translationary) rest and starts pure rotation.

Perfectly Inelastic Collision (e = 0)
1. The rod comes to complete rest if D coincides with C. vf = 0, w = 0.
2. When D coincides with one of the ends (i.e., the rod grazes against the nail), |k| = 2, e = 0, gives vf = 3*v/4, w = 3*v/(2*l). Note that, this situation is also as if the rod is being rotated about the end with angular speed, w = 3*v/(2*l)

I hope, I haven't bored you people much!

Last edited: Oct 28, 2007