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 2.00 cm, producing an angular acceleration of the forearm of 105 rad/s². what is the moment of inertia (in kg·m²) of the boxers forearm? kg·m²

Explanation:

Step1: Calculate the torque

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

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

The relation is $\tau = I\alpha$.
We know $\tau = 40\ N\cdot m$ and $\alpha=105\ rad/s^{2}$.
To find $I$, we can rewrite the formula as $I=\frac{\tau}{\alpha}$.
Substitute the values: $I=\frac{40\ N\cdot m}{105\ rad/s^{2}}$.
Calculate: $I=\frac{40}{105}\ kg\cdot m^{2}\approx0.381\ kg\cdot m^{2}$.

Answer:

$0.381\ kg\cdot m^{2}$