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. Find the magnitude of the angular acceleration produced given the mass lifted is 10.5 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 1410 N, and its effective perpendicular lever arm is 3.50 cm.  How much work is done if the leg rotates through an angle of 10.0° with a constant force exerted by the muscle?

Answer 1

here,

m = 10.5 kg

d = 0.28 m

moment of inertia , I = 0.9 kg.m^2

F = 1410 N

r = 0.035 m

theta = 10 degree

net torque, t = 1410 N * 0.035 m - 10.5kg * 9.8 m/s² * 0.28m

t = 20.538 N·m

total I = 0.9 kg·m² + 10.5kg*(0.28m)² = 1.72 kg·m²

torque = I * alpha

20.538 N·m = 1.72 kg·m² * alpha

aplha = 11.9 rad/s²

the angular accelration is 11.9 rad/s^2

the work done, using the rotational equivalent of F*d

work = t * theta

W = 20.538 N·m * 0.174

W = 3.57 J

the work is done if the leg rotates through an angle of 10.0 degree is 3.57 J