The triceps muscle in the back of the upper arm extends the forearm.
this muscle in a professional boxer exerts a force of 2.00 x 103n with an effective perpendicular arm of 3.0 cm, producing an angular acceleration of the forearm of 130 rad/s2

what is the moment of inertia of the boxer's forearm?

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

Where,

F = Force

r = Distance

Replacing,

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

I = Moment of inertia

= Angular acceleration

Replacing with our values we have that

Therefore the value of moment of inertia is

15.3 cos 65 degrees trig

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