Question

While exercising in a fitness center, a man lies face down on a bench and lifts a weight with one lower leg by contacting the muscles in the back of the upper leg.

(a) Find the angular acceleration produced given the mass lifted is 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is 0.900 kg ⋅ m² , the muscle force is 1500 N, and its effective perpendicular lever arm is 3.00 cm.
(b) How much work is done if the leg rotates through an angle of 20.0º with a constant force exerted by the muscle?

Answer 1

The angular acceleration produced is 50 rad/s², and the work done by the muscle when the leg rotates through an angle of 20.0º is 15.7 Joules.

Step-by-step explanation:

To find the angular acceleration (α), we can use the formula α = τ / I, where τ is the torque and I is the moment of inertia.

The torque (τ) produced by the muscle force is calculated by the product of the force (F) and the effective perpendicular lever arm (r), so τ = F × r. Given that the force (F) is 1500 N and the effective lever arm (r) is 3.00 cm (0.03 m), we have τ = 1500 N × 0.03 m = 45 N·m.

With the moment of inertia (I) of the lower leg as 0.900 kg·m², we get α = 45 N·m / 0.900 kg·m² = 50 rad/s².

For part (b), the work (W) done by the muscle is the product of the torque and the angular displacement in radians (θ), W = τ × θ. The angle in degrees must be converted to radians by multiplying by π/180, so θ = 20.0º × (π/180) = 0.349 radians. Now we can calculate the work as W = 45 N·m × 0.349 rad = 15.7 Joules.