A dancer is standing on a rotating platform h | vedclass

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(C) Concept $1$: The moment of inertia $I$ of a mass $m$ at a distance $r$ from the axis of rotation is given by $I = mr^2$. Thus,$I \propto r^2$. Whe

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angular momentum will remain unchanged and angular velocity will increase

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(C) Concept $1$: The moment of inertia $I$ of a mass $m$ at a distance $r$ from the axis of rotation is given by $I = mr^2$. Thus,$I \propto r^2$. When the dancer drops the spheres,the mass $m$ is removed from the system at a distance $r$,which decreases the total moment of inertia $I$ of the dancer.Concept $2$: The angular momentum $L$ of the system is conserved if the net external torque $	au$ acting on the system is zero. Since $\frac{d\vec{L}}{dt} = \vec{	au}$,if $\vec{	au} = 0$,then $\vec{L}$ remains constant.Concept $3$: We know that $L = I\omega$,which implies $\omega = \frac{L}{I}$. Since $L$ is constant,$\omega \propto \frac{1}{I}$.Conclusion: When the dancer drops the spheres,the moment of inertia $I$ decreases. Because the angular momentum $L$ remains constant,a decrease in $I$ leads to an increase in the angular velocity $\omega$.

$A$ dancer is standing on a rotating platform holding two spheres in her hands. If she drops the spheres to the ground,the dancer's...

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