Angular momentum- rotating rod

[palaphys]

Homework Statement

A light rod is rotating on a frictionless horizontal table about one of its ends such that a mass is attached to its center and to its other end. What is the angular momentum of the mass at the other end with respect to the center?

Relevant Equations

L=Iw L=mvr

So my first approach was to use L=Iw.
Computing I, moment of inertia,I=$$I_{center} = \frac{md^2}{4}$$
$$L_{center} = I_{center} \omega = \frac{md^2\omega}{4}$$
And hence I obtain an answer.
However, when I further thought about the question, I realised that L is a vector and I can probably do this:

$$L_{ab} =L _{ao} - L_{bo} $$
Where b is the particle at the center and a is the particle at the end.
But that's when:
$$\begin{align*}
L_{B/A} &= md^2\omega - \frac{md^2\omega}{4} \\
L_{B/A} &= \frac{4md^2\omega}{4} - \frac{md^2\omega}{4} \\
L_{B/A} &= \frac{3md^2\omega}{4}
\end{align*}$$

Why are the results different?

 

[jbriggs444]

Homework Statement: A light rod is rotating on a frictionless horizontal table about one of its ends such that a mass is attached to its center and to its other end. What is the angular momentum of the mass at the other end with respect to the center?

So one end of the rod is fixed to an axis. The rest of the rod and attached weights are rotating on the frictionless surface. [That is just me talking to myself, making sure that I understand the set up].

Relevant Equations: L=Iw L=mvr

So my first approach was to use L=Iw.
Computing I, moment of inertia,I=$$I_{center} = \frac{md^2}{4}$$
$$L_{center} = I_{center} \omega = \frac{md^2\omega}{4}$$

That seems correct.

However, when I further thought about the question, I realised that L is a vector and I can probably do this:

$$L_{ab} =L _{ao} - L_{bo} $$

What are you trying to say here? What is $L_{ab}$? What is $L_{ao}$? What is $L_{bo}$? Why subtract them? What relationship do any of them have to $L$, the angular momentum of the assembly about its center?

Guessing...

$L_{ab}$ is the angular momentum of the end mass about the axis at the end of the rod.
$L_{bo}$ is the angular momentum of the middle mass about the axis at the end of the rod.
$L_{ab}$ is the angular momentum of the the end mass about a reference point at the center of the rod.

If this is correct then there is no motivation for the subtraction here. One could perform a "parallel axis" calculation to shift from angular momentum about the end to angular momentum about the center. But you've not provided a correct calculation for that.

Where b is the particle at the center and a is the particle at the end.
But that's when:
$$\begin{align*}
L_{B/A} &= md^2\omega - \frac{md^2\omega}{4} \\
L_{B/A} &= \frac{4md^2\omega}{4} - \frac{md^2\omega}{4} \\
L_{B/A} &= \frac{3md^2\omega}{4}
\end{align*}$$

Now you've switched notation again. What is $L_{B/A}$?

Why are the results different?

I do not understand the second approach at all.

 

[palaphys]

Let me explain. Lba is the angular momentum of b with respect to a. And Lao is the angular momentum of a wrt the fixed axis and b the same way. Sorry for switching th notation, B/a is the same thing what I mean by just ba.

 

[palaphys]

It's like how you add vectros- take relative velocity for instance- $$V_ba$$ = Vb- Va with respect to some other frame

 

[kuruman]

However, when I further thought about the question, I realised that L is a vector and I can probably do this:
$$L_{ab} =L _{ao} - L_{bo} $$Why are the results different?

Maybe you cannot "probably" do this. You need to derive the appropriate relation for this case starting from the definition of angular momentum of a point mass $m$ at position $\mathbf r$ moving with velocity $\mathbf v$ relative to the reference point, $~\mathbf L=m \mathbf r\times \mathbf v.$ Hint: In the definition of angular momentum you must use the velocity and position of the tip relative to the midpoint.

 

[palaphys]

Maybe you cannot "probably" do this. You need to derive the appropriate relation for this case starting from the definition of angular momentum of a point mass $m$ at position $\mathbf r$ moving with velocity $\mathbf v$ relative to the reference point, $~\mathbf L=m \mathbf r\times \mathbf v.$ Hint: In the definition of angular momentum you must use the velocity and position of the tip relative to the midpoint.

Yeah I noticed that finding the relative velocity and distance got me the right answer, but why can't I do that?

 

[kuruman]

Yeah I noticed that finding the relative velocity and distance got me the right answer, but why can't I do that?

Because it's wrong unless derived correctly?

 

[jbriggs444]

It's like how you add vectros- take relative velocity for instance- $$V_ba$$ = Vb- Va with respect to some other frame

The angular momentum of the assembly about its center is not the difference of the contributions from the two masses. It is the sum.

If you want to perform a proper parallel axis computation to shift the reference axis, you could start with the angular momentum about the one axis: The sum of the angular momenta of the two masses about the fixed axis at the end of the rod.

Then compute the linear momentum of the assembly.

Then multiply (vector cross product) the linear momentum of the assembly times the displacement by which you have moved the selected axis of rotation ($\frac{d}{2}$ in this case). This will provide the proper adjustment to convert from angular momentum about the one axis to angular momentum about the other.

 

[palaphys]

The angular momentum of the assembly about its center is not the difference of the contributions from the two masses. It is the sum.

If you want to perform a proper parallel axis computation to shift the reference axis, you could start with the angular momentum about the one axis: The sum of the angular momenta of the two masses about the fixed axis at the end of the rod.

Then compute the linear momentum of the assembly.

Then multiply (vector cross product) the linear momentum of the assembly times the displacement by which you have moved the selected axis of rotation ($\frac{d}{2}$ in this case). This will provide the proper adjustment to convert from angular momentum about the one axis to angular momentum about the other.

Yeah I know that, but how do you compute the angular momentum of A with respect to B?

 

[kuruman]

Yeah I know that, but how do you compute the angular momentum of A with respect to B?

You have
$\mathbf L_A^{\text{(O)}}=m\mathbf r_A\times \mathbf v_A$
$\mathbf L_B^{\text{(O)}}=m\mathbf r_B\times \mathbf v_B$
Then the relative angular momentum is
$\mathbf L_B^{\text{(A)}}=m(\mathbf r_B-\mathbf r_A)\times (\mathbf v_B-\mathbf v_A)$
Can you finish it now?
Hint: In general $\mathbf r_A$ is a fraction of $\mathbf r_B$, so you can write $\mathbf r_A=\alpha~\mathbf r_B~~~(0\leq\alpha\leq 1).$ Substitute and check that the expression makes sense in the limits $\alpha =0$ and $\alpha=1.$ Then find what it becomes when $\alpha =\frac{1}{2}$ which is what you have here. That's how a proper derivation is constructed.

 

[jbriggs444]

Yeah I know that, but how do you compute the angular momentum of A with respect to B?

The same way you would compute angular momentum about any arbitrary reference point.

The mass, if any, at B does not enter in. That mass is at distance zero from the reference point. So its angular momentum about that point is zero.

The velocity of the reference point does enter in. One judges the velocity of all other masses relative to the velocity of the reference point. That is to say that you use a frame of reference anchored on your selected reference point.

Then you add up the (relative) linear momentum of each mass times (vector cross product) the displacement of that mass from your reference at point B.

Alternately, for assemblies rotating rigidly about an unmoving reference point, one can multiply the moment of inertia of the assembly about the fixed point by the angular rotation rate of the assembly.

 

[palaphys]

You have
$\mathbf L_A^{\text{(O)}}=m\mathbf r_A\times \mathbf v_A$
$\mathbf L_B^{\text{(O)}}=m\mathbf r_B\times \mathbf v_B$
Then the relative angular momentum is
$\mathbf L_B^{\text{(A)}}=m(\mathbf r_B-\mathbf r_A)\times (\mathbf v_B-\mathbf v_A)$
Can you finish it now?
Hint: In general $\mathbf r_A$ is a fraction of $\mathbf r_B$, so you can write $\mathbf r_A=\alpha~\mathbf r_B~~~(0\leq\alpha\leq 1).$ Substitute and check that the expression makes sense in the limits $\alpha =0$ and $\alpha=1.$ Then find what it becomes when $\alpha =\frac{1}{2}$ which is what you have here. That's how a proper derivation is constructed.

The same way you would compute angular momentum about any arbitrary reference point.

The mass, if any, at B does not enter in. That mass is at distance zero from the reference point. So its angular momentum about that point is zero.

The velocity of the reference point does enter in. One judges the velocity of all other masses relative to the velocity of the reference point. That is to say that you use a frame of reference anchored on your selected reference point.

Then you add up the (relative) linear momentum of each mass times (vector cross product) the displacement of that mass from your reference at point B.

Alternately, for assemblies rotating rigidly about an unmoving reference point, one can multiply the moment of inertia of the assembly about the fixed point by the angular rotation rate of the assembly.

That was clear, thanks. But is there a general rule of thumb for when and when not to use Vba= Vb -Va for other quantities? Or is the only way actually understanding the new reference frame and applying the definition of the quantity?

 

[kuruman]

That was clear, thanks. But is there a general rule of thumb for when and when not to use Vba= Vb -Va for other quantities?

I don't understand what "other quantities" means. When one says that the angular momentum of mass $m$ relative to point O is $\mathbf L^{\text{(O)}}=m\mathbf r\times \mathbf v$, the understanding is that
$\mathbf r =~$ the position of the mass relative to O.
$\mathbf v =~$ the velocity of the mass relative to O.

There is no rule of thumb just consistent use of the definition. In this problem you have O at one end of the rod and B at the other end. Point A is at the midpoint. If you want to find the angular momentum of B relative to O, you put in the expression the velocity of B relative to O and the position of B relative to O. If you want to find the angular momentum of B relative to A, you put in the expression the velocity of B relative to A and the position of B relative to A.

 

[jbriggs444]

That was clear, thanks. But is there a general rule of thumb for when and when not to use Vba= Vb -Va for other quantities? Or is the only way actually understanding the new reference frame and applying the definition of the quantity?

Position and velocity work the way you appear to expect.

If you shift the origin, you shift all the position coordinates by the same amount.
If you shift the standard of rest, you shift all measured velocities by the same amount.

The same arguably holds for temperature and pressure.

If you shift the zero point on your temperature scale, you shift all measured temperatures by the same amount. [But don't try to use Fahrenheit or Celsius with the ideal gas law]

If you shift your standard for ambient pressure, all gauge pressures change by the same amount. [But don't try to use gauge pressures with the ideal gas law].

Angular momentum is different. Moving the reference point or changing your standard of rest does not change all angular momenta by the same fixed amount. It is more complicated than that. You have to do the parallel axis thing. If the axes are not parallel, then things are even worse.

As @kuruman advises:

There is no rule of thumb just consistent use of the definition.

Relative <something> is not always a synonym for measured <something> minus measured reference.

 

[palaphys]

Got it. Thanks.

 

[kickaxe]

If something is spinning, the weight at the center seems to be irrelevant in this case, is it not?

 

[haruspex]

If something is spinning, the weight at the center seems to be irrelevant in this case, is it not?

I assume you mean mass, not weight.
A point mass at the axis would be irrelevant. There is no mass at the axis in this question.

 

[kuruman]

A point mass at the axis would be irrelevant. There is no mass at the axis in this question.

I think @kickaxe's query stems from the question posed by the OP's problem,

What is the angular momentum of the mass at the other end with respect to the center?

in which a mass is attached at the midpoint (misnamed center) of the rod. It seems that @kickaxe conflates "angular momentum with respect to the center" with "spinning about the center".