Question

A ballet dancer is rotating about his own vertical axis on smooth horizontal floor. $I$,$\omega$, $L$ and $K$ are moment of inertia, angular velocity, angular momentum and rotational kinetic energy of ballet dancer respectively. If ballet dancer folds himself close to his axis of rotation then,(assuming that there is no role of friction)

• A. $I$, $K$ both increase and $\omega$ decreases but $L$ is constant
• B. $I$, $K$ both increase and $\omega$ increases but $L$ is constant
• C. $\omega$, $K$ both decrease and $I$ increases but $L$ is constant
• D. $\omega$, $K$ both increase and $I$ decreases but $L$ is constant

Answer 1

Answer: (D)

$\omega$, $K$ both increase and $I$ decreases but $L$ is constant

I $\to$ decrease
because mass will come closer to the center
$\omega \to$ will increase because Angular momentum
is conserved.
$\therefore I\omega =constant$
$\therefore I\propto \frac{1}{\omega .}$
$\because I$ decreases $\therefore \omega$ increases.
L $\to$ Angular momentum will remain conserved as no external torque is applied.
$\therefore L=$ Constant.
Rotational $KE=\frac{1}{2}\left(I\omega \right).\omega .$
$=\frac{1}{2}k\omega (\because I\omega$ is constant)
$\therefore KE\propto \omega .$
$\therefore$ $KE$ will increase.
$\because \omega$ is increased.