Question

A dancer spins about a vertical axis at $60$$rpm$ with her arms folded.If she streches her hands so that M.I. about the vertical axis increases by $25\%,$ the new rate of revolution is

• A. $48$ $rpm$
• B. $75$ $rpm$
• C. $15$ $rpm$
• D. $24$ $rpm$

Answer 1

The correct option is A $48$ $rpm$

$\begin{array}{c}\text{Let, the initial moment of inertia of dancer}=I_0.\\ \text{and, initial rate of revolution}=60rpm\text{.}\\ \text{Given, final moment of inertia}=I_0+I_0\times \frac{25}{100}=\frac{5}{4}\text{Io.}\end{array}$

$\begin{array}{c}\text{Here, there is no any external torque, so angular}\\ \text{momentum remains conserved.}\end{array}$

$\begin{array}{c}\text{Thus,}{I}_i{\omega }_i=I_f{\omega }_f\Rightarrow I_0\times 60=\frac{5}{4}{I}_0\times {\omega }_F\\ \text{Thus,}{\omega }_F=48rpm\end{array}$