Final answer:
The moment of inertia of the boxer's forearm is calculated using the formula for torque and the given force, lever arm, and angular acceleration to be 0.5 kg·m².
Step-by-step explanation:
The question is asking to determine the moment of inertia of a boxer's forearm when the triceps muscle, extending the forearm, exerts a known force with a specified effective perpendicular lever arm, causing an angular acceleration.
To find the moment of inertia, we use the formula τ=Iα, where τ is the torque, I is the moment of inertia, and α is the angular acceleration. The torque (τ) exerted by the muscle can be calculated by multiplying the force (F) by the effective perpendicular lever arm distance (r), so τ=F * r.
Given the force exerted by the triceps is 2.00 × 10³ N and the effective perpendicular lever arm is 3.00 cm (converted to meters as 0.03 m), we have τ = (2.00 × 10³ N) * (0.03 m). To find moment of inertia (I), we rearrange the formula to I = τ / α, substituting the given angular acceleration of 120 rad/s².
Therefore:
τ = (2.00 × 10³ N) * (0.03 m) = 60 N·m
I = τ / α = (60 N·m) / (120 rad/s²) = 0.5 kg·m²
The moment of inertia of the boxer's forearm is 0.5 kg·m².