Question

conservation of angular momentum about centre of masss​

Answer 1

general in the form it is expressed] that the angular momentum of and rigid body with respect to any point P can always be expressed as

LP=rcm×Mvcm+(∑imiR2i)ω

where rcm is the position of the centre of mass with respect to P, M the mass of the body, Ri the distance of the i-th point, having mass mi, composing the body, and ∑imiR2i=I its moment of inertia with respect to the instantaneous axis of the rotation around the centre of mass of angular velocity ω.

I know that the velocity vi of each point Pi, having mass mi, of a rigid body of mass M can be see as the sum of a translation velocity of one of its points C plus a rotation velocity around that point: vi=vC+ω×CPi−→−. If we chose C as the centre of mass I see that

Lcm=∑iCPi−→−×mivi=∑iCPi−→−×mivcm+∑iCPi−→−×mi(ω×CPi−→−)

=∑iCPi−→−×mi(ω×CPi−→−)

because, if I am not wrong, ∑iCPi−→−×mivC=(∑imiCPi−→−)×vC=0 since ∑imiCPi−→− is the position of the centre of mass with respect to itself, which is 0.

How can it be proved that ∑iCPi−→−×mi(ω×CPi−→−)=(∑imiR2i)ω? I have searched a lot on the Internet and on books, but I find nothing. To give some background of mine, I have studied nothing of analytical mechanics. I find the formula very, very interesting both in itself and because, if the moment of inertia does not depend upon time, ∀tI(t)=I(t0), the above expression can be differentiated to get the formula of the resultant torque with respect to the centre of mass ∑τcm=dLcmdt=Iαcm where α is the angular acceleration around the centre of mass. I heartily thank you for any answer!!

alternative answer or in simplified

words

By bringing part of the mass of her body closer to the axis, she decreases her body's moment of inertia. Because angular momentum is the product of moment of inertia and angular velocity, if the angular momentum remains constant (is conserved), then the angular velocity (rotational speed) of the skater must increase.

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Answer 2

Hey mate☺☺

This is your answer ⬇️⬇️⬇️

The spin angular momentum of an object is defined as the angular momentum about its centre of masscoordinate. ... Therefore, for a closed system (where there is no net external torque), the total torque on the system must be 0, which means that the totalangular momentum of the system is constant.

Hope it will help you ☺☺

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