Question

A man stands on a rotating platform with his arms stretched holding a $5\, kg$ weight in each hand. The angular speed of the platform is $1.2 \,rev \,s^{-1}$. The moment of inertia of the man together with the platform may be taken to be constant and equal to $6 \,kg \,m^2$ . If the man brings his arms close to his chest with the distance of each weight from the axis changing from $100\, cm$ to $20 \,cm$. The new angular speed of the platform is

• A. $2 \,rev\,s^{-1}$
• B. $3 \,rev\,s^{-1}$
• C. $5 \,rev\,s^{-1}$
• D. $6 \,rev\,s^{-1}$

Answer 1

The Correct Option is B

Initial moment of inertia, $I_1 = [6 + 2 \times 5] \times (1)^2 = 16 \,kg \,m^2$ Initial angular velocity, $\omega_1 = 1.2 \,rev\, s ^{-1}$ Initial angular momentum, $ L_1= I_1 \omega_1$ Final moment of inertia, $I_2 = [6 + 2 \times 5] \times (0.2)^2 = 6.4\, kg\, m^2$ Final angular speed $= \omega_2$ Final angular momentum, $L_ 2 = I_2\omega_2$ According to law of conservation of angular momentum, $L_1 = L_2$ or $I_1 \omega_1 - I_2 \omega_2$ $\omega_{2} = \frac{I_{1}\omega_{1}}{I_{2}} $ $ = \frac{\left(16 \,kg\, m^{2}\right)\left(1.2 \,rev \,s^{-1}\right)}{\left(6.4\, kg \,m^{2}\right)} $ $ = 3 \,rev\, s^{-1}$