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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.

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Answer:

The moment of inertia of a skater is : [tex]I= 2.343 kgm^{2}[/tex]

Explanation:

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

  • Arm extended cylinder mass: [tex]M= 52.5 kg[/tex]
  • Radius: [tex]R= 0.110 m[/tex]
  • Length: [tex]L= 0.900 m[/tex]
  • and [tex]m=3.75 kg[/tex]

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

[tex]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} ][/tex]

[tex]I=2.343 kgm^{2}[/tex]

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