Question

A physics teacher does a classroom demonstration on a rotating table illustrating angular momentum. His body has a moment of inertia of 6.21 kg-m^2. He puts a one kilogram mass in each hand and with his arms outstretched, he begins spinning at 27 rpm. The masses are initially at 54 cm from the axis of rotation. He then pulls his arms into his body and the masses end up at a radius of 8 cm from the axis of rotation. Ignore the effect of the mass of his arms on the moment of inertia. Calculate his final rotational speed in revolutions per minute. Round your answer to the nearest 0.1 rpm. (Do not round your numbers until the very end)

Answer #1

moment of inertia of body I_{b} = 6.21 kg-m2;

moment of inertia of two point mass Ip = 2mr^{2} =
2*1*(0.54)^{2};

Ip = 0.5832 kg-m2;

total moment of inertia I_{1} = 6.21 + 0.5832;

**I _{1} = 6.7932 kg-m2; w_{1} 27 rpm (when
arms are out streched);**

**---------------------------------------------------------------------**

when, he then pulls his arms into his body and the masses end up at a radius of 8 cm from the axis of rotation

total moment of inertia I_{2} = 6.21 + 2mr^{2} =
6.21 + 2*1*(0.08)^{2}

**I _{2} = 6.2228 kg-m2, w_{2}
?;**

apply conservation of angular momentum;

I_{1}w_{1} = I_{2}w_{2};

**6.7932* 27 = 6.2228* w _{2;}**

**w _{2} = 29.45 rpm. (revolution per
minute);**

**w _{2} = 29.45/60 rps. = 0.491 rps (revolution
per second);**

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