Question

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

a) Find the angular acceleration produced given the mass lifted is 10.0kg at a distance of 28.0cm from the knee joint, the moment of inertia of the lower leg is 0.900kgm{eq}^2 {/eq}, the muscle force is 1500N, and its effective perpendicular lever arm is 3.00cm.

b) How much work is done if the leg rotates through an angle of 20.0 degrees with a constant force exerted by the muscle?

Answer 1

The angular acceleration produced is 50 rad/s², and the work done is 10,000 rad.

Step-by-step explanation:

(a) To find the angular acceleration, we can use the equation:

\tau = I\alpha

Where \tau is the torque, I is the moment of inertia, and \alpha is the angular acceleration. Rearranging the formula to solve for \alpha, we have:

\alpha = \frac{\tau}{I}

Substituting the given values, we have:

\alpha = \frac{1500 \, \text{N} \times 0.03 \, \text{m}}{0.9 \, \text{kg} \, \text{m}2} = 50 \, \text{rad/s}^2

(b) The work done can be calculated using the formula:

W = \tau \theta

Where W is the work done, \theta is the angle, and \tau is the torque. Rearranging the formula to solve for \theta, we have:

\theta = \frac{W}{\tau}

Substituting the given values, we have:

\theta = \frac{1500 \, \text{N} \times 0.2 \, \text{rad}}{0.03 \, \text{m}} = 10,000 \, \text{rad}

Answer 2

The angular acceleration produced by the man exercising is 50 rad/s², and the work done during the rotational motion through an angle of 20.0 degrees is 15.7 Joules.

To determine the angular acceleration (α) produced by the man exercising, we can use the formula α = τ/I, where τ represents torque and I represents moment of inertia.

The torque (τ) caused by the muscle force (F) can be calculated by the product of the force and the effective perpendicular lever arm (r), so τ = F * r.

Converting units to be consistent, we get F = 1500 N and r = 3.00 cm = 0.03 m. Therefore, τ = 1500 N * 0.03 m = 45 N·m. Given the moment of inertia (I) of 0.900 kg·m², the angular acceleration is α = 45 N·m / 0.900 kg·m² = 50 rad/s².

To find out how much work is done when the leg rotates through an angle of 20.0 degrees (θ), we first convert the angle to radians: θ = 20.0 degrees * (π/180) = 0.349 radians.

Work (W) is the product of torque and the angular displacement in radians, so W = τ * θ. Substituting our values gives W = 45 N·m * 0.349 rad = 15.7 J. Hence, 15.7 Joules of work is done.