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?

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

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.