The triceps muscle in the back of the upper arm extends the forearm. this muscle in a professional boxer exerts a force of 2.00×103 n with an effective perpendicular lever arm of 3.10 cm, producing an angular acceleration of the forearm of 121 rad/s2. what is the moment of inertia of the boxer's forearm?

0.512 kgm²

Explanation:

Given

Force, F = 2*10^3 N

Angular acceleration, α = 121 rad/s²

Lever arm, r(⊥) = 3.1 cm = 3.1*10^-2 m

τ = r(⊥) * F

Also,

τ = Iα

Using the first equation, we have

τ = r(⊥) * F

τ = 0.031 * 2*10^3

τ = 62 Nm

Now we calculate for the inertia using the second equation

τ = Iα, making I subject of formula, we have

I = τ / α, on substituting, we have

I = 62 / 121

I = 0.512 kgm²

Thus, the moment of inertia of the boxers forearm is 0.512 kgm²

pretty sure it's false

i really dont know but i will try okay.

so i think u would just add all them up and get your answer..