Question

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2551 N2551 N with an effective perpendicular lever arm of 3.15 cm3.15 cm , producing an angular acceleration of the forearm of 115.0 rad / s2115.0 rad / s2 . What is the moment of inertia of the boxer's forearm?

Answer 1

Answer:

0.7 kg m²

Explanation:

F = force exerted applied by muscle in a professional boxer = 2551 N

r = length of lever arm = 3.15 cm = 0.0315 m

α = angular acceleration of the forearm = 115 rad/s²

I = moment of inertia of the boxer's forearm

τ = Torque applied by muscle in a professional boxer

Torque is given as

τ = I α = r F

Inserting the values

I (115) = (0.0315) (2551)

I = 0.7 kg m²