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The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2615 N with an effective perpendicular lever arm of 2.85 cm , producing an angular acceleration of the forearm of 110.0 rad / s2 . What is the moment of inertia of the boxer's forearm?

Respuesta :

Answer:

0.68 kg-m²

Explanation:

F = Force applied by the muscle = 2615 N

r = effective perpendicular lever arm = 2.85 cm = 0.0285 m

α = Angular acceleration of the forearm = 110.0 rad/s²

I  = moment of inertia of the boxer's forearm = ?

Torque is given as

τ = I α                                   eq-1

Torque is also given as

τ = r F                                   eq-2

using eq-1 and eq-2

r F =  I α

(0.0285)(2615) = (110.0) I

I = 0.68 kg-m²

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