Question

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 5.00 cm, producing an angular acceleration of the forearm of 125 rad/s2. What is the moment of inertia (in kg·m2) of the boxer's forearm?

Answer 1

To solve the problem it is necessary to apply all the concepts related to the definition of Torque, both linear and angular.

From the linear definition the torque is defined as

$\tau = F*r$

Where,

F = Force

r = radius

On the other hand we have that,

$\tau = I \alpha$

Where,

I = Moment of inertia

$\alpha =$ Angular Acceleration

Using the first equation we can find the Torque, there,

$\tau = F*r$

$\tau = (2*10^3)(0.05)$

$\tau = 100Nm$

Therefore the Inertia moment can be calculated from the second equation,

$\tau = I \alpha$

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

$I = \frac{100}{125}$

$I = 0.8 kg.m^2$

Therefore the value of moment of inertia is $0.8 Kg.m^2$