Question

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2231 N with an effective perpendicular lever arm of 3.90 cm, producing an angular acceleration of the forearm of 130.0 rad / s2. What is the moment of inertia of the boxer\'s forearm?

Answer 1

Given that,

Force = F = 2231 N ; Length of the lever arm = L = 3.9 cm = 0.039 m

angular acceleration = alpha = 130 rad/s².

We need to find the moment of inertia of the boxer.

We know that, the torque is related to the angular acceleration alpha and moment of inertia I, as:

torque = I alpha

Also , Torque = F x perpendicular distance

torque = F x L

Equating the above two values of torque, we get:

I alpha = F L => I = F x L / alpha

I = 2231 x 0.039 / 130 = 0.6693 Kg -m²

Hence, the moment of inertia of the boxer's arm is = I = 0.6693 Kg-m²