Question

Angular Momentum question

Answer 1

An ice skater has a total mass 61 kg.

The ice skater initially has her arms extended. Assuming that each of her arms can be treated as rods 0.80 m long and with 5.2% of the total mass, and that the rest of the body can be treated as a solid cylinder with radius 0.25 m. Calculate the moment of inertia of the ice skater.

Answer 2

I can check this later.

Answer 3

I want to re-denote my things.

Answer 4

The total moment of inertia: \(I_{tot}\)
The moment of inertia of each arm: \(I_{arm}\)
The moment of inertia of her body: \(I_{bod}\)

WE ARE GIVEN
The "radius" of the arm: \(R_{arm}=0.8~{\rm m}\)
The "radius" of the body: \(R_{bod}=0.25~{\rm m}\)
The mass of the arm: \(M_{arm}=61\times 0.052~{\rm kg}\)
The mass of the body: \(M_{bod}=61\times (1-0.052)~{\rm kg}\)

Answer 5

So, the moment of inertia, since each arm is treated like a rod is:
\(I_{arm}=\frac{1}{12}M_{arm}(R_{arm})^2\)

and thus, the moment of inertia of both arms combined is:
\(2 I_{arm}=2\times\frac{1}{12}M_{arm}(R_{arm})^2\)
\(2I_{arm}=\frac{1}{6}M_{arm}(R_{arm})^2\)

ALSO,
Knowing that the moment of inertia for a solid cylinder is (1/2)MR^2,
therefore the moment of inertia for the body is,
\(I_{ bod}=\frac{1}{2}M_{bod}(R_{bod})^2\)
\(I_{bod}=\frac{1}{2}M_{bod}(R_{bod})^2\)

And, thus,
\(I_{tot}=2I_{arm}+I_{bod}\)
\(I_{tot}=\frac{1}{6}M_{arm}(R_{arm})^2+\frac{1}{2}M_{bod}(R_{bod})^2\)

Answer 6

And now, all I need to do is to plug everything in ...
\(I_{tot}=\frac{1}{6}(0.052\times 61~\color{gray}{\rm kg})(0.8~\color{gray}{\rm m})^2+\frac{1}{2}((1-0.052)\times 61~\color{gray}{\rm kg})(0.25~\color{gray}{\rm m})^2\)
\(I_{tot}=\color{gray}{\left[\color{black}{\frac{1}{6}(0.052\times 61)(0.8)^2+\frac{1}{2}((1-0.052)\times 61)(0.25)^2}\right]}~\color{gray}{\rm kg\cdot m^2}\)

Answer 7

\(I_{tot}=\color{gray}{\left[\color{black}{\frac{1}{6}(0.052\times 61)(0.8)^2+\frac{1}{2}((1-0.052)\times 61)(0.25)^2}\right]}~\color{gray}{\rm kg\cdot m^2}\)
\(I_{tot}=2.1~\color{gray}{\rm kg\cdot m^2}\)

Answer 8

Moment of inertia of the arms, is more like a rod rotated about one end. It's 1/3 ML^2 not 1/12.

Answer 9

tnx

Answer 10

I already got that fixed , but thanks for the feedback.
Takes me forever to do these problems:)