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\times 10^32.00×10 ^3 N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s^2 .
What is the moment of inertia of the boxer's forearm?

Answer 1

Answer:

Explanation:

From the given information:

The torque produced due to the force can be expressed as:

$\tau = F \times r$

where;

$\tau$ = torque

F = force exerted

r = lever's arm radius

$\tau$ = $2.00 \times 10^3 \times 0.03 m$

$\tau$ = 60 N.m

However, equating the torque with the moment of inertia & angular acceleration, we use the equation:

$\tau$ = I∝

60 Nm = I × 120 rad/s²

I = 60 Nm/120 rad/s²

I = 0.5 kg.m²