# Tricky torque problem I have for you!

1. Nov 9, 2012

### NasuSama

1. The problem statement, all variables and given/known data

A uniform thin rod (M = 7.29 kg, L = 2.25 m) sits motionless on a horizontal, frictionless table (as in the previous problem). Now, however, it is pinned to the table by a frictionless hinge placed distance x = 0.68 m above the center of mass, and the rod is free to rotate about the hinge. A point particle of mass m = 0.761 kg) moves at speed vo = 77.7 m/s and strikes the rod on the end below the center of mass, and perpendicular to the rod.

Suppose the block is made of clay, and it sticks to the rod. Find ω, the angular speed of the clay-plus-rod after the collision.

2. Relevant equations

Uh...

→conservation of energy

3. The attempt at a solution

mv_0 * x = (I_rod + mx²)ω
ω = mv_0 * x/(I_rod + mx²)
ω = mv_0 * x/(1/3 * mL² + mx²)

I know that...

v_0 = 77.7
m = 7.29

But I don't know if...

L = 2.25
x = 1.805 [0.68 from the center of mass. Then, I believe that 2.25/2 + .68 = 1.805 from the hinge to the bottom of the rod]

I'm stuck. :(

2. Nov 9, 2012

### NasuSama

Not sure if I'm on the right track. I would assume that it's the wrong answer, but there is something I have for some right part. I wonder if someone can assist me for this problem.

3. Nov 9, 2012

### haruspex

Not sure I understand the set-up. Is the rod standing vertically on the table? How does the rod rotate about the hinge? In a vertical plane that contains the line of motion of the block?
Anyway, the impact is clearly inelastic, so energy will not be conserved. You can use conservation of both linear and angular momentum, as long as you take into account that there is an unknown impulse from the hinge at moment of impact of the block. You should be able to eliminate that from your equations.

4. Nov 9, 2012

### NasuSama

Here is the diagram I visualize..

File size:
7.1 KB
Views:
119
5. Nov 9, 2012

### NasuSama

See below the post.

Last edited: Nov 9, 2012
6. Nov 9, 2012

### NasuSama

The rod is vertical, but there is a hinge ON the rod. This is the thing you should consider.

7. Nov 9, 2012

### haruspex

OK, so I have the right model. What equations do you get for linear and angular momentum?

8. Nov 9, 2012

### NasuSama

Here are the equations I have...

mv_0 = (M + m)v_f
v_f = mv_0/(M + m)

That is the linear momentum before and after the collision.

Let's see if I can figure out the angular momentum. This is rather the complicated problem here!

I_0 * ω_0 = I_f * ω_f
(m + M)v_f * (.68 + 2.25/2) = (I_rod + mx²)ω_f
(m + M)v_f * (.68 + 2.25/2) = (1/3 * m * L² + m(.68 + 2.25/2)²)ω_f

Seems like I lack direction here. -___-

9. Nov 9, 2012

### haruspex

What is the linear speed of M after impact?
I did mention there will be an impulse from the hinge. You need to include that in the equation.
Let's take this one step at a time. What point are you taking moments about? What is the moment of inertia of the rod about that point? What are the angular velocities of the rod and particle about that point?
Please don't substitute numerical values at this stage. Just use letters for distances etc.

10. Nov 9, 2012

### NasuSama

Why does impulse take in account of this situation? I don't think I see the point here. The linear speed of M after impact would be just v, which is the same final velocity as the velocity m after impact. Isn't that true?

The hinge x away from the center of mass of the rod is the point. I guess that I need to apply some Parallel Axis Theorem to get the moment of inertia; am I right?

If so, then it's something like...

I_parallel = I_cm of rod + (M + m)d² [after impact]
= 1/12 * (m + M)L² + (M + m)d²

Last edited: Nov 9, 2012
11. Nov 9, 2012

### haruspex

Imagine the hinge consists of you holding the rod between finger and thumb. When the particle hits you will feel a kick. This is the impulse which the rod and hinge apply to each other. The complete linear momentum equation will be:
m v0 - I = m vf + M uf
That's how fast the rod will be moving at the tip touching the particle, but its centre of mass will be moving more slowly.
OK, in that case we can avoid worrying about I.
Where d is? (It can't be the distance from hinge to centre of mass for both.)

12. Nov 10, 2012

### NasuSama

Yes, I know that, and I believe that it's the end from the hinge.

...and how did you get this equation?

m v_0 - I = m v_f + M u_f

13. Nov 10, 2012

### haruspex

If you're defining d as the distance from the hinge to the lower tip of the rod then that is the right distance to use for m, but not for M. Your expression 1/12 * (m + M)L² + (M + m)d² is treating the two masses as though they're the same effective distance from the hinge. 1/12 * L² + d² is not the right factor for either. Think about each separately.

m v0 - I = m vf + M uf says that the impulse from m (m v0) less the opposing impulse from the hinge (I) = the net linear momentum immediately after impact, m vf + M uf. Note that uf and vf are not equal. Their ratio is determined by the relative position of the hinge. We won't need this equation because, for the angular momentum, you have taken moments about the hinge. If moments had been taken about any other point then I would have turned up in that equation and we would have needed the linear momentum equation to eliminate it.

14. Nov 10, 2012

### NasuSama

Then... As you just said, M doesn't take place at the end of the hinge, so...

I_parallel = 1/12 * ML² + md²

m takes in account for the lower tip as you just said. I believe that since both masses don't take in effect of the distance from the hinge, I must use M for the center of mass for the Parallel Axis Theorem I just applied. Both masses don't behave the same way for this situation.

Then, I have...

P_intial = P_final
mv_0 * d = (1/12 * ML² + md²)ω where d is the distance from the hinge of the rod to the end of the rod [where m mass hit]

Still a lot of thinking to do. Not sure if this is totally right. For the initial momentum, only the smaller mass hit the bottom tip of the rod. Then, for the final momentum, Parallel Axis Theorem takes place. I took the center of mass of the rod without the mass. Then, I combined that with the momentum of the smaller mass at the lower tip of the rod. That is how I got to that equation.

There are much considerations to deal with. You said that I don't need to use the linear momentum since I took the angular moments already.

Also, how would the momentum occur when a block doesn't stick on the rod for the same situation? Would it be this?

mv_0 * d = 1/12 * ML² * ω_f [Then, md² doesn't exist for the final momentum since the mass doesn't stick on the rod]

Last edited: Nov 10, 2012
15. Nov 10, 2012

### haruspex

Closer, but still not right for the rod. The above would be right for the rod's rotation about its mass centre, but you're taking moments about the hinge. Use the parallel axis theorem here, but think carefully about the distance to use.
More precisely, it's because you took moments about the hinge, which means the impulse from the hinge does not appear in the equation.
That would only be the case if the block came to rest. It would always be correct to have a term m*d*vf, where vf is the speed of the block after impact. The fact that the block sticks merely tells you the relationship between vf and how the rod moves after impact.

16. Nov 10, 2012

### NasuSama

Nvm. Let me figure out...

Last edited: Nov 10, 2012
17. Nov 10, 2012

### NasuSama

"Closer, but still not right for the rod. The above would be right for the rod's rotation about its mass centre, but you're taking moments about the hinge. Use the parallel axis theorem here, but think carefully about the distance to use."

Then, the inertia would change by being increased in the way the hinge is positioned.

Last edited: Nov 10, 2012
18. Nov 10, 2012

### NasuSama

I had the very hard time finding the right length of rotation of the rod. Wonder if it's possible to do integration here instead of algebra. Here is what I thought.

We know that the hinge is not on the center of rod nor the end. This is interesting, yes? Here is my work..

I = ∫^(x = (-(L_s/2 - f)/L_s) * L, (d)/L_s * L) x²λ dx
= 1/3 * x³λ | x = -((L_s/2) - f)/L_s * L, ((L_s/2) + f)/L_s * L

Here, I treat L_s [the length of the rod] as the dummy variable for the integration. f is the distance between the hinge and the center of mass. Well...

1/3 * λ * [((d)/L_s * L)³ + (((L_s/2) - f)/L_s * L)³]
= 1/3 * λ * L³ * [((d)³ + ((L_s/2) - f)³)/(L_s)³]
= 1/3 * mL² * ((d)³ + ((L_s/2) - f)³)/(L_s)³

So by the conservation of momentum, I should get...

mv_0 * d = (I_rod + md²)ω
mv_0 * d = (1/3 * ML² * ((d)³ + ((L_s/2) - f)³)/(L_s)³ + md²)ω where...

d = distance between the hinge and the end of the rod where it's hit by the block
f = distance between the hinge and the center of mass
L = length of the rod = L_s [This time, I substitute L_s with 2.25 m]
m = mass of the block
M = mass of the rod
v_0 = initial velocity of the block

Let...

d = 2.25/2 + 0.68
f = 0.68
L = L_s = 2.25
m = 0.761
M = 7.29
v_0 = 77.7

So I get...

ω = mv_0 * d/(1/3 * ML² * ((d)³ + ((L_s/2) - f)³)/(L_s)³ + md²)

So...

ω = 0.761 * 77.7 * (.68 + 2.25/2)/(1/3 * 7.29 * 2.25² * (((2.25/2) + 0.68)³ + ((2.25/2) - 0.68)³)/(2.25)³ + 0.761 * (.68 + 2.25/2)²)

Haha! Actually, I'm on the right track.

Last edited: Nov 10, 2012
19. Nov 10, 2012

### SammyS

Staff Emeritus
I doubt that the rod is vertical.

How could that be if it's pinned to the horizontal table in the manner described?

I would use conservation of angular momentum to solve this problem.

20. Nov 10, 2012

### haruspex

I agree the terminology is puzzling, but I don't know how else to make sense of "strikes the rod on the end below the center of mass".
That is what we are endeavouring to do.

NasuSama, the rod rotates about a point distance d-L/2 from its mass centre, right? So by the parallel axis theorem the moment of inertia to use is ML2/12 + M(d-L/2)2, right?