Angular momentum of two particles connected by a rigid bar

[Jenny Physics]

Homework Statement:

Find the angular momentum of the system of two particles in the figure connected by a (massless) rigid bar.

Relevant Equations:

$$L=r_{left}\times mv_{left}+r_{right}\times mv_{right}$$

Lets do it for the left (the right will be similar): $r_{left}=[(L-a\sin\theta)\sin\phi,(L+a\cos\theta)\cos\phi]$ so $v_{left}=[-a\dot{\theta}\cos\theta\sin\phi+(L-a\sin\theta)\dot{\phi}\cos\phi,-a\dot{\theta}\sin\theta\cos\phi-(L+a\cos\theta)\dot{\phi}\sin\phi]$. Is this right?

 

[PeroK]

How is this system moving?

 

[Jenny Physics]

How is this system moving?

It is a pendulum swinging back and forth and so both $\phi,\theta$ change with time.

 

[Orodruin]

Are you trying to find just the angular momentum or go via the Lagrangian of the system?

 

[Jenny Physics]

Are you trying to find just the angular momentum or go via the Lagrangian of the system?

Just trying to find the angular momentum using the definition.

 

[haruspex]

I assume this is all in one plane.
You have omitted a from your expressions, and anyway your expressions for the r coordinates are wrong. Draw a diagram showing where $(L-a\sin(\theta))\sin(\phi)$ is.

But I think it will be much easier treating both particles together and using the parallel axis theorem.

 

[PeroK]

Homework Statement:: Find the angular momentum of the system of two particles in the figure connected by a (massless) rigid bar.
Homework Equations:: $$L=r_{left}\times mv_{left}+r_{right}\times mv_{right}$$

View attachment 254771
Lets do it for the left (the right will be similar): $r_{left}=[(L-a\sin\theta)\sin\phi,(L+a\cos\theta)\cos\phi]$ so $v_{left}=[-a\dot{\theta}\cos\theta\sin\phi+(L-a\sin\theta)\dot{\phi}\cos\phi,-a\dot{\theta}\sin\theta\cos\phi-(L+a\cos\theta)\dot{\phi}\sin\phi]$. Is this right?

Try $\theta = 0$ in your equation for $r_{left}$. Something is not right there.

 

[Orodruin]

Also, no, the expressions are not correct. You cannot have factors such as $L-\cos\theta$ as the physical dimensions of the terms are different.

 

[Jenny Physics]

It should then be $r_{left}=[(L-a\cos\theta)\sin\phi,(L-a\cos\theta)\cos\phi]$? That will not work when $\theta=\pi/2$

 

[haruspex]

It should then be $r_{left}=[(L-a\cos\theta)\sin\phi,(L-a\cos\theta)\cos\phi]$? That will not work when $\theta=\pi/2$

Wild guesses won't get you there.
Draw a horizontal line through the top pivot (O) and vertical lines up to that from the left mass and the lower pivot, meeting it at A, B. Can you figure out the lengths of OB, AB?

 

[Orodruin]

Wild guesses won't get you there.
Draw a horizontal line through the top pivot (O) and vertical lines up to that from the left mass and the lower pivot, meeting it at A, B. Can you figure out the lengths of OB, AB?

I still think the parallel axis theorem is a better way forward, as you mentioned in #6.

 

[Jenny Physics]

Wild guesses won't get you there.
Draw a horizontal line through the top pivot (O) and vertical lines up to that from the left mass and the lower pivot, meeting it at A, B. Can you figure out the lengths of OB, AB?

The length of $OB=L\sin\phi$. The length of AB requires some trigonometry involving $\theta$ which is not obvious to me.

 

[vela]

Can you figure out the angle $\alpha$ in this diagram in terms of $\phi$ and $\theta$?

 

[Jenny Physics]

Can you figure out the angle $\alpha$ in this diagram in terms of $\phi$ and $\theta$?

View attachment 254779

$\phi+\theta+\alpha=\pi/2$ that would mean that $\alpha=\pi/2-\phi-\theta$ and hence $AB=a\cos\alpha=a\sin(\phi+\theta)$?

 

[haruspex]

$\phi+\theta+\alpha=\pi/2$ that would mean that $\alpha=\pi/2-\phi-\theta$ and hence $AB=a\cos\alpha=a\sin(\phi+\theta)$?

Yes