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 10.0 kg at a distance of 28.0 cm from the knee joint, the moment of inertia of the lower leg is 0.900 kg⋅m20.900 \textrm{ kg} \cdot \textrm{m}^2, 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 20.0∘20.0^\circ with a constant force exerted by the muscle?
Question by OpenStax is licensed under CC BY 4.0
Final Answer

a) 10.4 rad/s210.4 \textrm{ rad/s}^2

b) 6.12 J6.12 \textrm{ J}

Note: This video assumes there isn't any torque due to the weight of the lower leg. With the weight of the leg included in the torque calculation, the answer to part (a) would be 7.55rad/s27.55 rad/s^2. For all the questions in chapter 10, the torque due to the weight of a limb is ignored. Like ignoring friction, this is a simplification done to make real-world complexity more manageable. For instance, for part (b) of this question, it would be challenging to account for how the torque due to gravity on the lower leg would change as the angle of the lower leg changes. As the lower leg is lifted, the perpendicular component of the lever arm of gravity changes. Dealing with quantities that change gradually like this is the subject of calculus and is beyond the scope of this textbook.

Solution video

OpenStax College Physics, Chapter 10, Problem 29 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. This man who’s exercising is lying on a bench and his foot is here, and there is a weight behind this hill and here’s his knee and he’s going to curl his leg up this direction because of the force exerted by this muscle on the back of his thigh which is exerting a force on the lower leg at the attachment point it’s here. We’re told that the effective level arm of this force is three centimeters, and so I’ve labelled that distance here perpendicular distance between the force acting and the knee. And so we have a torque going this way due to the muscle and then there’s a torque going clockwise due to the weight of the weight, which has magnitude of mg and it’s located at a distance 28 centimeters from the knee. Now the moment of inertia of the lower leg we’re told is 0.9 kilograms-meters squared. Well, we have the net torque is going to equal the moment of inertia of the entire system, which by the way this is not, this is just the moment of inertia of the leg, and we’ll multiply that by the angular acceleration. And so we’ll solve for angular acceleration by dividing both sides by I, and so we have the net torque divided by moment of inertia is going to be the angular acceleration that we have to find. So the net torque is the counter-clockwise torque which is the force of the muscle multiplied by its effective level arm and then minus the torque due to the weight which is mg times level arm of the weight, and we divide that by the total moment of inertia of the system, which is the moment of inertia of the leg plus the moment of inertia of the weight, which we modelled as a point mass, and so it’s formula is mr squared. Then, we plug in some numbers so we have 1500 newtons force of the muscle times three centimeters written in meters, minus ten kilograms times 9.81 newtons per kilogram times 28 times ten to the minus two meters, divide that by 0.9 kilograms-meters squared moment of inertia of the lower leg plus ten kilograms times 28 times ten to the minus two meters squared, and this gives 10.4 radians per second squared will be the angular acceleration. Then the question is, what is the net work done in this case. Well that’s going to be the net torque multiplied by the angular displacement. The net torque is moment of inertia times angular acceleration and so we substitute IL plus mw rw squared just as we did over here, and multiply that by the angular acceleration we just found in part a, and then times by the angular displacement that were given, 20 degrees. So we have 0.9 kilograms-meters squared plus ten kilograms times 28 times ten to the minus two meters squared times 10.411 radians per second squared angular acceleration, multiply by 20 degrees converted into radians by multiplying by pi radian for every 180 degrees, and this gives 6.12 Joules of work done.

Comments

Submitted by djab on Fri, 11/12/2021 - 10:03

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In calculating the net torque why wasn’t the weight of the bottom part of the leg to its center of mass considered?

Hi djab, this is a great question. I think you're right to consider the weight of the leg, and perhaps this video deserves a re-do to take the torque due to the weight of the leg into account. Assuming the lower leg has mass uniformly distributed and is like a rod rotating about an axis through one end (the knee in this case), then its moment of inertia would be I=Ml23I = \dfrac{Ml^2}{3}, giving a mass for the lower leg of 34.44 kg. With the weight assumed to be acting at the center of the lower leg, at 1:42 you could introduce a term for the clockwise torque due to the weight of the leg which would subtract from the counterclockwise torque due to the muscle. The new answer for the angular acceleration in part (a) would be 7.55 rad/s, which would also change the answer for the work done in part (b). I'll put a note in the final answer to refer to this comment, and I've flagged this video for a re-do.
All the best,
Shaun

This video is not going to get a re-do since it makes a reasonable simplification that's common in this chapter.