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Physics Question on System of Particles & Rotational Motion

A man stands on a rotating platform with his arms stretched holding a 5 kg5\, kg weight in each hand. The angular speed of the platform is 1.2 rev s−11.2 \,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 m26 \,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 cm100\, cm to 20 cm20 \,cm. The new angular speed of the platform is

A

2 rev s−12 \,rev\,s^{-1}

B

3 rev s−13 \,rev\,s^{-1}

C

5 rev s−15 \,rev\,s^{-1}

D

6 rev s−16 \,rev\,s^{-1}

Answer

3 rev s−13 \,rev\,s^{-1}

Explanation

Solution

Initial moment of inertia, I1=[6+2×5]×(1)2=16 kg m2I_1 = [6 + 2 \times 5] \times (1)^2 = 16 \,kg \,m^2 Initial angular velocity, ω1=1.2 rev s−1\omega_1 = 1.2 \,rev\, s ^{-1} Initial angular momentum, L1=I1ω1 L_1= I_1 \omega_1 Final moment of inertia, I2=[6+2×5]×(0.2)2=6.4 kg m2I_2 = [6 + 2 \times 5] \times (0.2)^2 = 6.4\, kg\, m^2 Final angular speed =ω2= \omega_2 Final angular momentum, L2=I2ω2L_ 2 = I_2\omega_2 According to law of conservation of angular momentum, L1=L2L_1 = L_2 or I1ω1−I2ω2I_1 \omega_1 - I_2 \omega_2 ω2=I1ω1I2\omega_{2} = \frac{I_{1}\omega_{1}}{I_{2}} =(16 kg m2)(1.2 rev s−1)(6.4 kg m2)= \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 = 3 \,rev\, s^{-1}