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 approximated by a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.

Answer 1

Answer:

a) I = 0.363 kg*$m^{2}$

b) $I_{T}$ = 0.82385 kg*$m^{2}$

Explanation:

a) If we approximate the skater how a cylinder his moment of inertia is:

I = $\frac{mr^{2} }{2}$

I = $\frac{(60)(0.110)^{2} }{2}$

I = 0.363 kg*$m^{2}$

b)  If the skater has his arms extended then:

$I_{T} = I_{B} + I_{A}$

       where   $I_{B}$: Body’s moment of inertia

                     $I_{A}$: Moment of inertia of the arms

$I_{B}$ = $\frac{mr^{2} }{2}$

$I_{B}$ = $\frac{(52.5)(0.110)^{2} }{2}$ = 0.3176 kg*$m^{2}$

$I_{A}$ = 2 * $\frac{mL^{2} }{12}$

$I_{A}$ = 2 * $\frac{(3.75)(0.9)^{2} }{12}$ = 0.50625 kg*$m^{2}$

$I_{T}$  = 0.3176 + 0.50625 = 0.82385 kg*$m^{2}$

Answer 2

(a) The moment of inertia of the skater at the given radius is 0.363 kgm².

(b) The moment of inertia of the skater when he externs his arm is 0.824 kgm².

Moment of inertia of the skater

The moment of inertia of the skater at the given radius is calculated as follows;

I = ¹/₂MR²

I = 0.5 x 60 x (0.11)²

I = 0.363 kgm²

Moment of inertia when the skater extends his arm

The moment of inertia of the skater when he externs his arm is calculated as follows;

I₁ = ¹/₂MR²

I₁ = 0.5 x 52.5 x (0.11)²

I₁ = 0.318 kgm²

Moment of inertia of the long arms about the end is calculated as follows;

I₂ = 2(¹/₁₂ML²)

I₂ = ¹/₆ML²

I₂ =  ¹/₆ x 3.75 x (0.9)²

I₂ = 0.506 kgm²

Total moment of inertia = I₁ + I₂ = 0.318 kgm² + 0.506 kgm² = 0.824 kgm².

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