Question

A person is spinning with his hands outstretched at the rate of $4$ rads $s^{-1}$. When he brings his hands close to the body, he spins at the rate of $16$ rad $s^{-1}$ . The ratio of $M.1$ . in the two cases successively is

• A. $4:1$
• B. $1:14$
• C. $16:1$
• D. $1:2$

Answer 1

Answer: (A)

$4:1$

$\begin{array}{c}\text{Let, the initial moment of inertia of person}\left(incase\right(1\left)\right)=I_0.\\ \text{and, given initial rate of rotation}\left({\omega }_1\right)=4\text{rad/s.}\\ \text{Let, the final moment of inertia}\left(case\right(2\left)\right)=I_f\\ \text{and, given final angular velocity}=16\text{rad/s.}\\ \left({\omega }_2\right)\end{array}$

$\begin{array}{c}\text{By angular momentum conservation Principal,}\\ \begin{array}{cc}& I_0{\omega }_1=I_f{\omega }_2\\ \Rightarrow & I_0\times 4=I_F\times 16.\\ & \text{Thus,}\frac{I_0}{I_f}=\frac{4}{1}\end{array}\end{array}$