Final answer:
The moment of inertia of the boxer's forearm is calculated using the given force, lever arm, and angular acceleration. By calculating the torque and dividing by the angular acceleration, the moment of inertia is determined to be 0.50 kg·m².
Step-by-step explanation:
The question is asking for the moment of inertia of the boxer's forearm, which can be calculated using the relationship between torque, angular acceleration, and moment of inertia. Torque (τ) can be determined by multiplying the force (F) exerted by the triceps with the effective perpendicular lever arm distance (r). The angular acceleration (α) is given, and the formula for torque in relation to moment of inertia (I) and angular acceleration is τ = Iα.
Given values are:
Force, F = 2.00 × 10³ N
Lever arm, r = 3.00 cm = 0.03 m (converted to meters)
Angular acceleration, α = 120 rad/s²
First, we calculate the torque:
τ = Fr = (2.00 × 10³ N)(0.03 m)
Using the formula for torque, we then rearrange to find the moment of inertia:
I = τ/α
Therefore, the moment of inertia of the boxer's forearm is:
I = (2.00 × 10³ N × 0.03 m) / 120 rad/s²
I = 0.50 kg·m²
The moment of inertia of the boxer's forearm is thus 0.50 kg·m².