Conservation of Angular Momentum

[jisbon]

Homework Statement

Thin rod of mass 5kg and length 2m rotates with an angular velocity of 1.5rad/s and translates with 2m/s towards another uniform rod with mass 20kg and length 5m. They collide and stick together such that the centres and axes align. Find angular velocity after collision.

Relevant Equations

NIL

Hi,

Since this is a question about COAM (Conservation of Angular Momentum), I will assume I can leave out the part on translation and just use the formula below:

$Initial Angular Momentum= Final Angular Momentum$
whereby $I = \frac {1}{12}ML^2$ (of rod)
So,
$\frac {1}{12}ML^2(1.5)=\frac {1}{12}(m+M)L^2(x)$
$\frac {1}{12}(5)2^2(1.5)=\frac {1}{12}(25)5^2(x)$
I find x to be 0.048rad/s, which appears to be wrong. Correct is 0.0577rad/s

Any ideas on this?

 

[gleem]

Are you sure your determination of the final moment of inertial is correct?

 

[Delta2]

I also think there is something wrong with taking the moment of inertia of the final rod to be $\frac{1}{12}(m+M)L^2$. That would be the moment of inertia of a uniform rod of length L that has mass m+M.

BUT the resulting rod after the collision, is not uniform. It has bigger mass density near the center, and less mass density in the outer regions.

 

[jisbon]

I also think there is something wrong with taking the moment of inertia of the final rod to be $\frac{1}{12}(m+M)L^2$. That would be the moment of inertia of a uniform rod of length L that has mass m+M.

BUT the resulting rod after the collision, is not uniform. It has bigger mass density near the center, and less mass density in the outer regions.

Hm, so to solve this mass density issue, I will need to integrate across the rod? Or are there any alternative methods to this?

 

[haruspex]

Hm, so to solve this mass density issue, I will need to integrate across the rod? Or are there any alternative methods to this?

You summed the masses, which only works if the radii of inertia are the same.
Instead, compute the moments of inertia separately and add those.

 

[PeroK]

Hm, so to solve this mass density issue, I will need to integrate across the rod? Or are there any alternative methods to this?

In terms of total AM does it matter whether the two rods are treated separately or together as a single body?

 

[jisbon]

You summed the masses, which only works if the radii of inertia are the same.
Instead, compute the moments of inertia separately and add those.

I got it :) Answer is 0.0577rad/s
How will I know when I will need to compute them separately instead of together (since they stick together)
Not sure on this

 

[gleem]

Remember that I₁w_init = I ₁w _final + I₂w_final =I₁₂w_final for this problem where I₁₂ is the moment of inertia of the combination of the joined rods which you incorrectly calculated by not applying the definition of moment of inertia which takes into account the distribution of the mass which is very very important .

How would you have handled the problem if the incoming rod collided with a 20 kg disk of diameter of 5 m? Would it have been more obvious that you could not just add the masses?

 

[haruspex]

I got it :) Answer is 0.0577rad/s
How will I know when I will need to compute them separately instead of together (since they stick together)
Not sure on this

How do you work out the total area of two rectangles? You find the area of each and sum them, right? You can only take a shortcut if one pair of sides of one rectangle happen to be the same length as one pair of the other.
More generally still, the two objects might not be rotating at the same rate, so you would not be able to factor out that either, and would have to compute the angular momenta separately and add those.

 

[PeroK]

I got it :) Answer is 0.0577rad/s
How will I know when I will need to compute them separately instead of together (since they stick together)
Not sure on this

You used the formula for the MoI of a uniform rod. Two rods of different lengths do not form a uniform rod.

 

[haruspex]

You used the formula for the MoI of a uniform rod. Two rods of different lengths do not form a uniform rod.

I think @jisbon understands that was the error, but seeks guidance on when it is valid to add the masses and when not.