Final answer:
The moment of inertia of the boxer's forearm can be found using the formula: Moment of Inertia = (Torque)/(Angular Acceleration). Given that the force exerted by the triceps muscle is 2.00 × 10³ N and the lever arm is 3.00 cm (or 0.03 m), we can calculate the moment of inertia to be 0.5 kgm².
Step-by-step explanation:
The moment of inertia of the boxer's forearm can be found using the formula:
Moment of Inertia = (Torque)/(Angular Acceleration)
Given that the force exerted by the triceps muscle is 2.00 × 10³ N and the lever arm is 3.00 cm (or 0.03 m), we can calculate the torque by multiplying the force by the lever arm:
Torque = Force × Lever Arm = (2.00 × 10³ N) × (0.03 m) = 60 Nm
Plugging in the torque and the given angular acceleration of 120 rad/s² into the moment of inertia formula:
Moment of Inertia = (60 Nm)/(120 rad/s²) = 0.5 kgm²
Therefore, the moment of inertia of the boxer's forearm is 0.5 kgm².