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**L= r x mv. R is the distance from the centre of mass. So can I say that angular momentum is 12 for a fixed position 4 metres away from centre of mass having mass of 1 kg and velocity of 3 metres per second north?**

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- Thread starter avito009
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- #2

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You mean the center of mass has a velocity of 3m/s?

If not, what has this velocity?

If not, what has this velocity?

- #3

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No not the centre of mass but the point 4 metres away from centre of mass.

- #4

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Maybe you can describe your system. Is it a rigid body, a set of several bodies (and what kind of bodies), etc.

- #5

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fixedposition 4 m away from the center of mass" is actually moving with 3m/s?

Maybe you can describe your system. Is it a rigid body, a set of several bodies (and what kind of bodies), etc.

It is a rigid body.

- #6

wrobel

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##\boldsymbol L_O= m\boldsymbol{OS}\times \boldsymbol v_S+J_S\boldsymbol\omega##, here ##S## is the center of mass, ##m## is the mass of the rigid body, ##\boldsymbol v_S## is the velocity of the center of mass, ##J_S## is the operator of inertia about the center of mass, ##\boldsymbol\omega## is the angular velocity

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Then what is moving with 3m/s?

- #8

Nugatory

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Angular momentum is calculated around a given point, and the value you calculate may be different for different points.L= r x mv. R is the distance from the centre of mass. So can I say that angular momentum is 12 for a fixed position 4 metres away from centre of mass having mass of 1 kg and velocity of 3 metres per second north?

- #9

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Angular momentum is calculated around a given point, and the value you calculate may be different for different points.

That answers one part of the question. But nasu confused me. So can you tell me that the velocity is the velocity of centre of mass or of the particular point? I mean what is v in the formula L=r x mv?

- #10

Khashishi

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But there is a similar formula for a rigid body

##\mathbf{L} = \mathbf{R}_{cm} \times M\mathbf{V}_{cm} + \overleftrightarrow{I}\mathbf{\omega}##

The first term is the orbital angular momentum, measured using the velocity of the center of mass, and the second term is the spin angular momentum (nothing to do with the quantum mechanics term). Basically, if it's not spinning, you can just treat the center of mass as a point particle.

- #11

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L= R x MV Now what is R in this formula. At some places it is mentioned as radius and at others the distance from centre of mass. So what is R? Is it radius or distance from centre of mass?

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- #12

CWatters

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https://www.physicsforums.com/threa...onentum-mean-faster-spin.874282/#post-5490162

https://www.physicsforums.com/threads/conservation-of-angular-momentum.874297/

Perhaps it helps to understand that an objects moment of inertia can depend on which axis it is being rotated about. For example a long thin rod will have a low moment of inertia when rotated about an axis that goes down the centre of the rod and a high moment of inertial when rotated about an axis through the middle of the rod. This is because mass further from the axis increases the moment of inertia more than mass close to the axis (think leverage).

I suggest reading up on how the moment if inertia is calculated for objects of different shapes. Basically you have to break the object down into small parts and then sum (integrate) the moment of inertia of all the component parts, each of which might be at a different radius from the axis of rotation.

- #13

CWatters

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a) Calculate the angular momentum of it's component parts and add them up

or

b) Calculate the moment of inertia of it's component parts, add them up, and then multiply by the angular velocity.

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