Question

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.2×10^2N with an effective perpendicular lever arm of 2.2cm, producing an angular acceleration of the forearm of 10^2 radians/s^2. What is the moment of inertia of the boxer’s forearm?

a. 0.02kg⋅m^2
b. 0.04kg⋅m^2
c. 0.06kg⋅m ^2
d. 0.08kg⋅m^2

Answer 1

To find the moment of inertia of a boxer's forearm, we calculate the torque using the force and lever arm, then use the equation torque equals moment of inertia times angular acceleration. The computed moment of inertia is approximately 0.05kg·m2, though this value doesn't match any of the provided options.

Step-by-step explanation:

You are asking about calculating the moment of inertia of a boxer's forearm when a force is applied. To solve this problem, we use the relationship between torque (τ), moment of inertia (I), and angular acceleration (α), which is given by τ = Iα. The torque is also the product of the force (F) and the lever arm's perpendicular distance (r), so τ = Fr.

Given:

• Force, F = 2.2×102N
• Lever arm, r = 2.2cm = 0.022m (since 1cm = 0.01m)
• Angular acceleration, α = 102rad/s2

We calculate:

Torque, τ = Fr = 2.2×102N × 0.022m = 4.84Nm

Now, using the relationship:

τ = Iα → I = τ / α = 4.84Nm / 102rad/s2 = 0.0484kg·m2

Hence, the moment of inertia of the boxer's forearm is approximately 0.0484kg·m2, which we can round to 0.05kg·m2 considering the options provided. There is no option that exactly matches this result, which suggests there may have been an error in the question's values or options provided.