While exercising in a fitness center, a man lies face down on a bench and lifts

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 13.5 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 1450 N, and its effective perpendicular lever arm is 3.60 cm. How much work is done if the leg rotates through an angle of 30.0° with a constant force exerted by the muscle?

Explanation / Answer

part a:

total torque applied=torque due to muscle force-torque due to weight on the lower leg

=1450*0.036-13.5*9.8*0.28=15.156 N.m

total momentum of inertia=moment of inertia due to the 13.5 kg mass+moment of inertia of the lower leg

=13.5*0.28^2+0.9=1.9584 kg.m^2

then angular acceleration=total torque/total moment of inertia=7.74 rad/s^2

part b:

work done=total torque*angle covered

=15.156*((30*pi/180) radians)

=7.9356 J

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