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3 years ago

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

Physics

1 answer:

Degger [83]3 years ago

6 0

Answer:

(a) I = 0.363 kgm^2

(b) I = 1.95 kgm^2

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:

$I_s=\frac{1}{2}MR^2$ (1)

M: mass of the skater = 60.0 kg

R: radius of the cylinder = 0.110m

$I_s=\frac{1}{2}(60.0kg)(0.110m)^2=0.363kg.m^2$

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:

$I=I_c+I_r\\\\I=\frac{1}{2}M_1R^2+\frac{1}{12}M_2L^2$ (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

$I=\frac{1}{2}(74.0kg)(0.150m)^2+\frac{1}{12}(6.00kg)(1.50m)^2\\\\I=1.95kg.m^2$

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

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