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 10³ n with an effective perpendicular lever arm of 2.00 cm, producing an angular acceleration of the forearm of 1.05 rad/s². what is the moment of inertia (in kg m²) of the boxers forearm? kg-m²
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Explanation:

Step1: Calculate the torque

Torque formula is $\tau = rF$. Given $F = 2.00\times10^{3}\ N$ and $r=2.00\ cm=0.02\ m$.
$$\tau=(0.02\ m)\times(2.00\times 10^{3}\ N) = 40\ N\cdot m$$

Step2: Use the relation between torque, moment of inertia and angular acceleration

The formula is $\tau = I\alpha$. Given $\alpha = 1.05\ rad/s^{2}$.
$$I=\frac{\tau}{\alpha}=\frac{40\ N\cdot m}{1.05\ rad/s^{2}}\approx38.1\ kg\cdot m^{2}$$

Answer:

$38.1\ kg\cdot m^{2}$