# Conservation of Angular Momentum

• jisbon
In summary, @jisbon found an error in his calculation of the final moment of inertia, and is seeking guidance on when it is valid to add the masses.
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?

Delta2
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.

Delta2 said:
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?

jisbon said:
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.

Delta2
jisbon said:
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?

haruspex said:
You summed the masses, which only works if the radii of inertia are the same.
How will I know when I will need to compute them separately instead of together (since they stick together)
Not sure on this

Remember that I1winit = I 1w final + I2wfinal =I12wfinal 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?

jisbon said:
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.

jisbon said:
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.

PeroK said:
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.

Delta2

## What is conservation of angular momentum?

Conservation of angular momentum is a fundamental law of physics that states that the total angular momentum of a system remains constant over time, unless acted upon by an external torque. Angular momentum is a measure of rotational motion, and can be calculated by multiplying an object's moment of inertia by its angular velocity.

## How does conservation of angular momentum apply in real life?

Conservation of angular momentum can be observed in many everyday phenomena, such as the motion of spinning tops, the rotation of planets around the sun, and the movement of dancers and ice skaters. It also plays a crucial role in the stability of satellites and spacecraft in orbit.

## What is the relationship between torque and angular momentum?

Torque and angular momentum are closely related, as torque is the force that causes changes in angular momentum. When an external torque is applied to a rotating object, its angular momentum will either increase or decrease depending on the direction of the torque.

## Can angular momentum be conserved in a closed system?

Yes, angular momentum is conserved in a closed system, meaning that there is no external torque acting on the system. This is known as the law of conservation of angular momentum.

## How is the conservation of angular momentum used in engineering and technology?

The principle of conservation of angular momentum is used in various engineering and technological applications, such as designing gyroscopes, propelling rockets, and creating stable and efficient structures for buildings and bridges. It is also essential in understanding and controlling the motion of machines and vehicles, such as cars and airplanes.

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