Question

Calculate the moment of inertia of a skater given the following information.

(a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius.
(b) The skater with arms extended is approximately a cylinder that is 74.0 kg, has a 0.150 m radius, and has two 0.750 m long arms which are 3.00 kg each and extend straight out from the cylinder like rods rotated about their ends.

Answer 1

(a) I = 0.363 kgm^2

(b) I = 1.95 kgm^2

Step-by-step explanation:

(a) If you consider the shape of the skater as approximately a cylinder, you use the following formula to calculate the moment of inertia of the skater:

`I_s=(1)/(2)MR^2` (1)

M: mass of the skater = 60.0 kg

R: radius of the cylinder = 0.110m

`I_s=(1)/(2)(60.0kg)(0.110m)^2=0.363kg.m^2`

The moment of inertia of the skater is 0.363 kgm^2

(b) In the case of the skater with his arms extended, you calculate the moment of inertia of a combine object, given by cylinder and a rod (the arms) that cross the cylinder. You use the following formula for the total moment of inertia:

`I=I_c+I_r\\\\I=(1)/(2)M_1R^2+(1)/(12)M_2L^2` (2)

M1: mass of the cylinder = 74.0 kg

M2: mass of the rod = 3.00kg +3.00kg = 6.00kg

L: length of the rod = 0.750m + 0.750m = 1.50m

R: radius of the cylinder = 0.150

`I=(1)/(2)(74.0kg)(0.150m)^2+(1)/(12)(6.00kg)(1.50m)^2\\\\I=1.95kg.m^2`

The moment of inertia of the skater with his arms extended is 1.95 kg.m^2