Question

The triceps muscle in the back of the upper arm is primarily used to extend the forearm. Suppose this muscle in a professional boxer exerts a force of 1.75 × 103 N with an effective perpendicular lever arm of 2.8 cm, producing an angular acceleration of the forearm of 150 rad/s2. What is the moment of inertia of the boxer's forearm?

Answer 1

To solve this problem it is necessary to address the concepts related to Torque as a function of the force and distance where it is applied and the moment of inertia from which the torque, moment of inertia and angular acceleration are related.

By definition the torque is defined as

$\tau = F*r$

Where,

$\tau = Torque$

F = Force

r = Radius

For our values we have:

$\tau = F*r$

$\tau = (1.75*10^3)(2.8*10^{-2})$

$\tau = 49Nm$

Consequently the calculation of the moment of inertia would then be given by the relationship

$\tau = I\alpha$

$I=\frac{\tau}{\alpha}$

Replacing with our values

$I = \frac{49}{150}$

$I = 0.322Kg.m^2$

The moment of inertia of the boxer's forearm $0.322Kg.m^2$