# Calculate Internal Angular Momentum and Energy

• Nicolas Gallardo
In summary: And how would you calculate VCoM?To calculate VCoM, you would take the average of the velocities of all four particles. So, VCoM = (5+3+3+2)/4 = 3.25 m/s.

## Homework Statement

Four particles of mass 1 Kg each, are moving on a plane with the velocities given in the figure.

## The Attempt at a Solution

First I calculated the position of the CoM:
Xcm=7/4(i + j)
Then I calculated the velocity of the CoM:
Vcm= ½i + ¼j

For the internal energy I did the following:

Kint= (1/2)(1.52) + (1/2)(1.32) + (1/2)(1.32) + (1/2)(1.22) = 23,5J

But the answer is 22,9 J. And for the internal angular momentum I do not actually know with respect to which point I should calculate it. Any hint would be appreciated. Thank you.

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Nicolas Gallardo said:
Xcm=7/4(i + j)
Then I calculated the velocity of the CoM:
Vcm= ½i + ¼j

For the internal energy I did the following:

Kint= (1/2)(1.52) + (1/2)(1.32) + (1/2)(1.32) + (1/2)(1.22) = 23,5J
The numbers in the figure are hard to read. If I'm reading them correctly, it appears that your results for the position and velocity of the center of mass are correct. But, it doesn't appear to me that your values for the "internal" speeds of the particles are correct. For example, how did you get 1.5 m/s for the speed that you used in the first term of Kint? To which particle does this refer?

TSny said:
The numbers in the figure are hard to read. If I'm reading them correctly, it appears that your results for the position and velocity of the center of mass are correct. But, it doesn't appear to me that your values for the "internal" speeds of the particles are correct. For example, how did you get 1.5 m/s for the speed that you used in the first term of Kint? To which particle does this refer?

That is (1/2)(mv^2)⇒ (1/2)(1(5^2)). One of the velocities is 5 m/s.
The other 3 velocities are: 3 m/s, 3 m/s, 2m/s

Nicolas Gallardo said:
That is (1/2)(mv^2)⇒ (1/2)(1(5^2)). One of the velocities is 5 m/s.
5 m/s is the speed of the particle relative to the reference frame of the Cartesian axes shown in the picture. You would use 5 m/s if you wanted the kinetic energy of the particle relative to this frame. But, you are looking for the "internal kinetic energy". This would be the kinetic energy relative to what frame of reference?

TSny said:
5 m/s is the speed of the particle relative to the reference frame of the Cartesian axes shown in the picture. You would use 5 m/s if you wanted the kinetic energy of the particle relative to this frame. But, you are looking for the "internal kinetic energy". This would be the kinetic energy relative to what frame of reference?
Relative to the CoM? But how would that change in the equation?

Nicolas Gallardo said:
Relative to the CoM?
Yes
But how would that change in the equation?
Can you find the velocity of each particle relative to the CoM frame?

TSny said:
Yes
Can you find the velocity of each particle relative to the CoM frame?
If we denote V' to the velocity of the particle related to the CoM frame, V to the velocity of the particle related to the cartesian axe and VCoM to the velocity of the CoM, then the velocity of the particle would be:
V'= Vcartesian - VCoM
Am I correct?

Nicolas Gallardo said:
If we denote V' to the velocity of the particle related to the CoM frame, V to the velocity of the particle related to the cartesian axe and VCoM to the velocity of the CoM, then the velocity of the particle would be:
V'= Vcartesian - VCoM
Am I correct?
Yes.

## 1. What is internal angular momentum and energy?

Internal angular momentum and energy refer to the rotational motion and energy within a system of particles. This includes the individual angular momentum and energy of each particle, as well as the overall angular momentum and energy of the system.

## 2. How is internal angular momentum and energy calculated?

Internal angular momentum can be calculated by taking the cross product of the position vector and the momentum vector for each particle in the system, and then summing these values. Internal energy can be calculated by taking the sum of the kinetic and potential energy of each particle in the system.

## 3. Why is internal angular momentum and energy important in scientific research?

Internal angular momentum and energy are important because they play a crucial role in understanding the dynamics and behavior of systems at the atomic and molecular level. They also have practical applications in fields such as astrophysics, quantum mechanics, and thermodynamics.

## 4. How does the conservation of angular momentum and energy apply to internal motion?

Just like in macroscopic systems, the law of conservation of angular momentum and energy also applies to internal motion. This means that the total internal angular momentum and energy of a system will remain constant unless acted upon by an external force.

## 5. Can internal angular momentum and energy be manipulated or controlled?

Yes, internal angular momentum and energy can be manipulated or controlled through external forces such as torque or energy transfer. This is often used in experiments and technologies that involve manipulating the rotation or energy of individual particles in a system.