Question

A person stands at the center of a turntable, holding his arms extended horizontally with a 1.5 kg dumbbell in each hand. He is set rotation about a vertical axis, making one revolution in 2 seconds. Find the new angular velocity if he pulls the dumbbells in to his middle. His moment of inertia (without dumbbells) is 3 kgm2 when his arms are out stretched, dropping to 2.2 kgm2 when his hands are at his middle. The dumbbells are 1 m from the axis initially and 0.2 m from it when moved to his middle.

Answer #1

**Conservation of Angular momentum**

**L1 = L2**

**L1 is initial angular momentum of the system and
L2 is final angular momentum of the system**

**angular momentum is L = I*W**

**I is moment of inertial of the system**

**L1 = L2**

**I1*W1 = I2*W2 ==> W2 = I1*W1/I2**

**I1 = I1p+I1d , I2 = I2p+I2d**

**here I1 = 3+2*1.5*1^2 = 6 kgm^2
and I2 = 2.2+2*1.5*0.2^2 = 2.32 kg m^2
initial angular velocity is W1 = 1/2 = 0.5 rev/s**

**final angular velocity is W2 = ?**

**from conservation of angular momentum**

** L1 = L2
I1*W1 = I2*W2 ==> W2 = I1*W1/I2**

** W2 = 6*0.5/(2.32) rev/s**

** W2 = 1.24 rev/s
the angular velocity increases**

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