Question

A rigid body rotates with an angular momentum L. If it's kinetic energy is halved, the angular momentum becomes

Answer 1

Angular momentum is given as

$L=I \omega$

and rotational kinetic energy is given as

$K.E=\frac{1}{2}I (\omega)^2 \\ where\\ I=moment\: of\: inertia\\ \omega=angular \: speed$

We will first create a relationship between the angular momentum and rotational kinetic energy.

let's express the angular speed in terms of Kinetic energy

$K.E=\frac{1}{2}I (\omega)^2$

multiplying with 2 we get

$2K.E=I (\omega)^2$

dividing by "I" we get

$\frac{2K.E}{I}= (\omega)^2$

taking square root we get

$\sqrt{\frac{2K.E}{I}}= \omega$

so we got the value of angular speed is terms of kinetic energy so now putting this in the equation of angular momentum

$L=I \omega=I\sqrt{\frac{2K.E}{I}}$

simplifying it first we get

$L=\sqrt{I^2\times \frac{2K.E}{I}}$

hence we get the relation of angular momentum and kinetic energy as

$L=\sqrt{2IK.E}$

now if we halved the kinetic energy

$L_{new}=\sqrt{2I\times \frac{K.E}{2}}$

we get

$L_{new}=\sqrt{\frac{2IK.E}{2}}$

from this

$L_{new}=\frac{\sqrt{2IK.E}}{\sqrt{2}}$

so we get the relationship of old momentum and new momentum as

${\bf L_{new}=\frac{1}{\sqrt{2}}\times L}$

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