Where Is the Mistake in Calculating the New Angular Velocity?

[PhMichael]

http://img821.imageshack.us/img821/2338/angular.jpg

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Homework Statement

A massless rod connects two similar particles of mass M and is rotating on the plane of a frictioless horizontal table with an angular velocity $$\omega$$. At some instant, a particles of mass 2M is put on the table such that it sticks to one of the M particles and the system rotates about its new center of mass. Find the new angular velocity.

2. The attempt at a solution

Angular momentum is conserved about the center of the "circle", A.

The angular momntum before the coliision is:

$$\vec{L}_{A} = I_{A} \vec{\omega} = -2Ma^{2} \omega \hat{z}$$

The new center of mass is: (where the origin is taken at the center and the y-axis is along the rod pointing to the upper mass and the x-axis is pointing to the right relative to y)

$$\vec{r}_{cm}=\frac{M(0,a)+3M(-a,0)}{4M}=(-0.75a,0.25a)$$

The angular momentum after the collision will have the form:

$$\vec{L'}_{A}=I_{A}\vec{\omega '}=(I_{cm}+4MD^{2}) \vec{\omega '}$$

where,

$$I_{cm} = 3M[(0.25a)^{2}+(0.25a)^{2}]+M[(a+0.75a)^{2}+(0.25a)^{2}]=3.5Ma^{2}$$

$$D^{2}=(0.75a)^{2}+(0.25a)^{2}=0.625a^{2}$$

Therefore,

$$\vec{L'}_{A}=(3.5Ma^{2}+4M(0.625a^{2})) \vec{\omega '}=-6Ma^{2} \omega ' \hat{z}$$

and equating $$\vec{L}_{A}$$ with $$\vec{L'}_{A}$$ yields:

$$\omega ' = \frac{\omega}{3}$$

However, the correct answer is:

$$\omega ' = \frac{2 \omega}{3}$$

Where is my mistake? =/

 

[Doc Al]

The new center of mass is: (where the origin is taken at the center and the y-axis is along the rod pointing to the upper mass and the x-axis is pointing to the right relative to y)

$$\vec{r}_{cm}=\frac{M(0,a)+3M(-a,0)}{4M}=(-0.75a,0.25a)$$

I think you messed up the coordinates of the 3M mass.

 

[PhMichael]

Here's how I've calculated it:

http://img225.imageshack.us/img225/838/21313967.jpg

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so, according to this, the coordinates of the 3M mass (bottom) is $$(-a,0)$$.

No?

 

[Doc Al]

so, according to this, the coordinates of the 3M mass (bottom) is $$(-a,0)$$.

No?

No.

What's the x-coordinate of all the masses?

 

[hikaru1221]

Angular momentum is conserved about the center of the "circle", A.

Though this is true, it's not easy to see/understand it correctly.

The angular momentum after the collision will have the form:

$$\vec{L'}_{A}=I_{A}\vec{\omega '}=(I_{cm}+4MD^{2}) \vec{\omega '}$$

Besides from the mess with coordinates, this is the 2nd mistake.
Only when the rigid body rotates around an axis do we have its angular momentum L=Iw, where L is angular momentum about that axis, I is counted also about that axis and w is also about the same axis.
Because in this problem, after the collision, the system doesn't rotate around A, $$\vec{\omega}'$$ simply describes the motion of the system around the new CM, not A. Therefore, we cannot apply L=Iw to calculate angular momentum of the system about A after collision.
Hint: Go back to the original definition of angular momentum for a system: $$\vec{L}=\Sigma m_i\vec{r}_i\times\vec{v}_i$$.

 

[PhMichael]

Before I'll retry solving this according to hikaru1221's explanation, I don't get what x coordinate I need to account for in calculating the center of mass' new position as all of the masses are located along the y axis, just before the collision occurs.

 

[Doc Al]

as all of the masses are located along the y axis, just before the collision occurs.

Exactly. So what is their x-component? (Compare that to what you actually wrote earlier.)

 

[PhMichael]

oh shoot, I switched between x and y ... I'm such an idiot =/

 

[PhMichael]

Ok, great, I've obtained the correct answer but I have two more questions:

(1)

"Angular momentum is conserved about the center of the "circle", A"

Though this is true, it's not easy to see/understand it correctly.

What explanation, really, I should add after using this conservation about point A so that my answer will be complete?

(2)

Suppose the 2M mass isn't just put there, instead it's shot with some velocity $$\vec{v}$$ towards the system so that it also sticks to one of the M masses.
Is this following equation right for obtaining the new c.m. velocity?

$$\vec{P}_{before}=\vec{P}_{after}$$

$$0=2M \vec{v} + 4M \vec{V}_{cm}$$

 

[hikaru1221]

For (1), it's quite lengthy so I'll leave it back. But that's just to alert you that there must be caution when applying the angular momentum conservation law. That torque = 0 is just one condition to apply, and there are more. If you're interested in it, try this problem:

Consider a system of point mass m1, m2, ... which have velocity v1, v2,... in a reference frame (R). The angular momentum of the system about a point A in the reference frame (R) is: $$\vec{L}_A(R)=\Sigma m_i(\vec{r}_i-\vec{r}_A)\times\vec{v}_i$$ (does it sound logical to you?).
Take the time derivative of that formula in reference frame (R) and derive under what the conditions we have $$d\vec{L}_A(R)/dt=0$$. Notice that we also have: $$d\vec{r}_A/dt = \vec{v}_A \neq 0$$ in general.
Also notice that the role of reference frame (R) is important. If you switch to another different reference frame, velocities changes, which leads to that angular momentum and its time derivative change.​

For (2), almost correct, a little mistake in the 2nd equation. But what's the point of calculating CM velocity?

 

[PhMichael]

For (1), do you have any source where from I can read about this?

For (2),

the angular momentum about some point A is:

$$\vec{L}_{A}= \vec{R} \times M_{total} \vec{V}_{cm} + \vec{L}_{cm}$$

So, I would need the velocity of the center of mass to apply this equation.

Can you please tell what is the mistake?

 

[Doc Al]

The angular momentum after the collision will have the form:

$$\vec{L'}_{A}=I_{A}\vec{\omega '}=(I_{cm}+4MD^{2}) \vec{\omega '}$$

You want the angular momentum about point A. That's the angular momentum about the new center of mass plus the angular momentum of the center of mass about A. But what's the speed of the new center of mass?

 

[hikaru1221]

For (1): I haven't found any popular book that mentions this (by the way, I don't read much book). You can try the problem I wrote in my previous post. Then compare to other popular books on how they derive the conditions for angular momentum conservation, and see the difference.
For (2): Try my problem, think about it and you may figure out some subtleties behind the concept of angular momentum. Then you will see what is wrong

 

[Doc Al]

For (1), do you have any source where from I can read about this?

The point that hikaru1221 is making is covered in any classical mechanics textbook. The key point is that no external torque implies no change in angular momentum only under certain conditions--it's not true in general. But as long as you measure your angular momentum with respect to a fixed reference point (point A, in this problem), that statement holds true and angular momentum about that point is conserved.

 

[PhMichael]

I've worked on that problem, and my work is attached here. However, I don't see the "insight" from this...
I still can't figure out what is the mistake in obtaining the center of mass velocity if I shoot that 2M mass with a velocity $$\vec{v}$$ by:

$$0 = 4M \vec{V}_{cm} + 2M \vec{v}$$ Note: I see that docx files can't be uploaded in this forum so here's an external link-

http://www.fileflyer.com/view/OlTtBCK

 

[Doc Al]

For (2),

the angular momentum about some point A is:

$$\vec{L}_{A}= \vec{R} \times M_{total} \vec{V}_{cm} + \vec{L}_{cm}$$

So, I would need the velocity of the center of mass to apply this equation.

Can you please tell what is the mistake?

So, in your original problem, what's the velocity of the center of mass?

 

[PhMichael]

So, in your original problem, what's the velocity of the center of mass?

Zero! because the mass is just put there is isn't given any velocity so by momentum conservation we would have:

$$\vec{P}_{before} = 0$$

$$\vec{P}_{after} = (2M)(0)+ \mathbf{(4M)(\vec{V}_{cm})}$$

and $$\vec{V}_{cm}=0$$

Whatever the bold part is, I will get a zero velocity so in this case I can't verify whether this is the right expression or not.

 

[Doc Al]

Yes, the speed of the center of mass will be zero. (Assuming it starts out as zero. )

Before the 2M mass 'attaches' to the other mass the total momentum is zero and it remains zero after they merge. The center of mass remains fixed.

Revise the calculation in your first post accordingly.

 

[PhMichael]

Yes, the speed of the center of mass will be zero. (Assuming it starts out as zero. )

Before the 2M mass 'attaches' to the other mass the total momentum is zero and it remains zero after they merge. The center of mass remains fixed.

Revise the calculation in your first post accordingly.

eventually, I got the correct answer:

$$\vec{r}_{cm}=(0,-0.5a)$$

$$\vec{L}_{before} = -2Ma^{2} \omega \hat{z}$$

$$\vec{L}_{after} = \vec{R} \times M_{tot} \vec{V}_{cm} + \vec{L}_{cm}=\vec{L}_{cm} = [3M(0.5a)^{2}+M(a+0.5a)^{2}]\vec{\omega'} = -3Ma^{2} \omega ' \hat{z}$$

and equating both terms yield:

$$\omega ' = 2 \omega / 3$$


What I still don't get is how to obtain the center of mass velocity for the case in which the 2M mass is shot with some velocity $$\vec{v}$$. Is the follwing correct for obtaining it?

$$0 = 2M \vec{v} + 4M \vec{V}_{cm}$$

hikaru1221 said that it's wrong .. but what is the mistake?

 

[Doc Al]

eventually, I got the correct answer:

$$\vec{r}_{cm}=(0,-0.5a)$$

$$\vec{L}_{before} = -2Ma^{2} \omega \hat{z}$$

$$\vec{L}_{after} = \vec{R} \times M_{tot} \vec{V}_{cm} + \vec{L}_{cm}=\vec{L}_{cm} = [3M(0.5a)^{2}+M(a+0.5a)^{2}]\vec{\omega'} = -3Ma^{2} \omega ' \hat{z}$$

and equating both terms yield:

$$\omega ' = 2 \omega / 3$$

Good.

What I still don't get is how to obtain the center of mass velocity for the case in which the 2M mass is shot with some velocity $$\vec{v}$$. Is the follwing correct for obtaining it?

$$0 = 2M \vec{v} + 4M \vec{V}_{cm}$$

I'd calculate the center of mass velocity like this:

$$4M \vec{V}_{cm} = 2M \vec{v} + 2M(0)$$

 

[PhMichael]

ummm can you explain how you received this? and is this $$\vec{V}_{cm}$$ the one to substitute in the expression of the angular momentum, after the collision?

 

[PhMichael]

nevermind ... the momnetum afterwards is all "contained" in the cm ... got ya!

 

[Doc Al]

ummm can you explain how you received this?

It follows from the definition of center of mass. The first two masses have a combined momentum of zero, so they do not contribute to the total momentum. So the overall momentum must equal 2Mv.

and is this $$\vec{V}_{cm}$$ the one to substitute in the expression of the angular momentum, after the collision?

Yes. You'd solve this version in much the same way. Angular momentum about point A will be conserved, only now the center of mass has non-zero velocity. (The initial angular momentum will no longer be zero, of course.)

 

[PhMichael]

Just to make sure I understand this, suppose now that the 3 Masses were connected by a massless rod and the system is rotating with $$\omega$$. At some instant, one of the edge masses is released.

[PLAIN]http://img85.imageshack.us/img85/8148/70474369.jpg

Is the following correct for obtaining $$\vec{V}_{cm}$$ (of the new center of mass) ?

the velocity of the edge mass just after it's released:

$$\vec{v} = \vec{\omega} \times \vec{r}$$

where $$\vec{r}$$ is the position vector from the center to that mass.

SO,

$$0=m \vec{\omega} \times \vec{r} + 2m \vec{V}_{cm}$$

 

[Doc Al]

Is the following correct for obtaining $$\vec{V}_{cm}$$ (of the new center of mass) ?

the velocity of the edge mass just after it's released:

$$\vec{v} = \vec{\omega} \times \vec{r}$$

where $$\vec{r}$$ is the position vector from the center to that mass.

SO,

$$0=m \vec{\omega} \times \vec{r} + 2m \vec{V}_{cm}$$

Sure, if by Vcm you mean the velocity of the center of mass of the two remaining masses only.

 

[PhMichael]

Sure, if by Vcm you mean the velocity of the center of mass of the two remaining masses only.

Yeah, of course, as those two are what's left in the system (system=what's connected to the rod).

Thanks alot!