Question

Calculate the moment of inertia of a skater given the following information. The skater with arms extended is approximately 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:

The moment of inertia of a skater is : $I= 2.343 kgm^{2}$

Explanation:

we can use parallel theorem to find the moment of inertia of a skater.

• Arm extended cylinder mass: $M= 52.5 kg$

• Radius: $R= 0.110 m$

• Length: $L= 0.900 m$

• and $m=3.75 kg$

$I=\frac{1}{2} MR^{2} +[\frac{1}{2} mL^{2} +m(R+\frac{L}{2} )^{2} ]$

$I=\frac{1}{2} (52.5)(0.110)^{2} +[\frac{1}{12} (3.75)(0.900)^{2} +3.75(0.110+\frac{0.900}{2} )^{2} ]$

$I=2.343 kgm^{2}$