Angular momentum of a rod about hinge

In summary, the conversation discusses the calculation of angular momentum for a uniform rod rotated about a point L/3 from its left end. Equations for calculating angular momentum for a purely rotating body, orbital motion, and spin are mentioned, and it is noted that equation 1 is applicable even if the axis of rotation does not pass through the center of mass. The derivation of the equation L = L(orbit) + L(spin) is also briefly mentioned.
  • #1

Krushnaraj Pandya

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Homework Statement


A uniform rod (M, L) is rotated about a point L/3 from its left end. Angular momentum about O

Homework Equations


1) L=I(cm)w for purely rotating body
2) L(orbital)= M*v(cm)*perpendicular distance(r)
3) L(spin)= I*w

The Attempt at a Solution


I got the correct answer in two ways- in the first one I wrote L= I about O*w. But isn't equation 1 applicable only when axis of rotation passes through COM? or is it applicable here because the body is not translating and thus there are no non-inertial forces on any point?
In the second method I wrote L=L(orbital)+L(spin), L(orbital) is m*v(cm)*distance between O and COM while L(spin) has to be taken as I about com*w...Why is L(spin) taken about an axis through COM even though the body isn't even rotating about that point? please reveal my conceptual flaws
 
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  • #2
Krushnaraj Pandya said:
But isn't equation 1 applicable only when axis of rotation passes through COM?
Nope, equation 1 is general. If the body is purely rotating about the same point that the moment of inertia is taken about, then it will work.

Krushnaraj Pandya said:
or is it applicable here because the body is not translating and thus there are no non-inertial forces on any point?
Translation has nothing to do with forces if it’s uniform. Anyway no translation is involved in pure rotations, other than circular motions.

Krushnaraj Pandya said:
Why is L(spin) taken about an axis through COM even though the body isn't even rotating about that point?
You can derive the result “L = L_(orbit) + L_(spin)” from integrating (or summing for discrete distributions) the definition of the angular momentum L of a point particle.

It’s not fun to type latex on a phone, but maybe you can work it out yourself. Just consider the arbitrary position/velocity to be a composition of the CoM position/velocity plus the position/velocity in the CoM frame. Maybe I will type it up tomorrow if you still don’t know.
 
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  • #3
Nathanael said:
Nope, equation 1 is general. If the body is purely rotating about the same point that the moment of inertia is taken about, then it will work.Translation has nothing to do with forces if it’s uniform. Anyway no translation is involved in pure rotations, other than circular motions.You can derive the result “L = L_(orbit) + L_(spin)” from integrating (or summing for discrete distributions) the definition of the angular momentum L of a point particle.

It’s not fun to type latex on a phone, but maybe you can work it out yourself. Just consider the arbitrary position/velocity to be a composition of the CoM position/velocity plus the position/velocity in the CoM frame. Maybe I will type it up tomorrow if you still don’t know.
I'll try to derive it myself...thanks. I'll take a look at things again and see if I have any more confusions and then let you know
 
  • #4
Krushnaraj Pandya said:
I'll try to derive it myself...thanks. I'll take a look at things again and see if I have any more confusions and then let you know
seems like all my doubts are resolved...thank you very much!
 

1. What is the definition of angular momentum?

Angular momentum is the measure of an object's rotational motion, determined by its mass, velocity, and distance from the axis of rotation.

2. How is angular momentum calculated?

Angular momentum is calculated by multiplying the moment of inertia (a measure of an object's resistance to change in rotational motion) by the angular velocity (the rate at which the object rotates).

3. What is the relationship between angular momentum and angular velocity?

The greater an object's angular velocity, the greater its angular momentum will be. This means that the faster an object rotates, the more difficult it will be to stop its rotational motion.

4. Can the angular momentum of a rod about a hinge change?

Yes, the angular momentum of a rod about a hinge can change. This can happen if an external torque (a force that causes rotational motion) is applied to the rod or if the rod's mass distribution changes.

5. What is the significance of the hinge in the angular momentum of a rod?

The hinge acts as the axis of rotation for the rod, and its location can affect the angular momentum. If the hinge is closer to the center of mass of the rod, the angular momentum will be lower compared to if the hinge is farther from the center of mass.

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