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 x 103N with an effective perpendicular arm of 3.0 cm, producing an angular acceleration of the forearm of 130 rad/s2

What is the moment of inertia of the boxer's forearm?

Answer 1

To solve this problem it is necessary to apply the concepts related to Torque as a function of Force and distance. Basically the torque is located in the forearm and would be determined by the effective perpendicular lever arm and force, that is

$\tau = F \times r$

Where,

F = Force

r = Distance

Replacing,

$\tau = 2*10^3*0.03$

$\tau = 60N\cdot m$

The moment of inertia of the boxer's forearm can be calculated from the relation between torque and moment of inertia and angular acceleration

$\tau = I \alpha$

I = Moment of inertia

$\alpha$ = Angular acceleration

Replacing with our values we have that

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

$I = \frac{60}{120}$

$I = 0.5kg\cdot m^2$

Therefore the value of moment of inertia is $0.5kg\cdot m^2$