Question

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

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

Answer 1

Answer: (B)

Initial moment of inertia,
$I_1=[6+2\times 5]\times (1{)}^2=16kg{m}^2$
Initial angular velocity, ${\omega }_1=1.2rev{s}^{-1}$
Initial angular momentum, $L_1=I_1{\omega }_1$
Final moment of inertia,
$I_2=[6+2\times 5]\times (0.2{)}^2=6.4kg{m}^2$
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 $I_1{\omega }_1-I_2{\omega }_2$
${\omega }_2=\frac{I_1{\omega }_1}{I_2}$
$=\frac{\left(16kg{m}^2\right)\left(1.2rev{s}^{-1}\right)}{\left(6.4kg{m}^2\right)}$
$=3rev{s}^{-1}$