However, in addition to possessing a forward motion, various parts of his body (

Question

However, in addition to possessing a forward motion, various parts of his body (such as the arms and legs) undergo rotation. Therefore, his total kinetic energy is the sum of the energy from his forward motion plus the rotational kinetic energy of his arms and legs. The purpose of this problem is to see how much this rotational motion contributes to the person's kinetic energy. Biomedical measurements show that the arms and hands together typically make up 14.0 of a person's mass, while the legs and feet together account for 37.0 . For a rough (but reasonable) calculation, we can model the arms and legs as thin uniform bars pivoting about the shoulder and hip, respectively. In a brisk walk, the arms and legs each move through an angle of about (a total of ) from the vertical in approximately 1 second. We shall assume that they are held straight, rather than being bent, which is not quite true. Let us consider a 70.0 person walking at 5.50 having arms 66.0 long and legs 95.0 long.

A:What is the average angular velocity of his arms and legs? W=
B:Using the average angular velocity from part A, calculate the amount of rotational kinetic energy in this person's arms and legs as he walks. Krot=
C:What is the total kinetic energy due to both his forward motion and his rotation? Ktot=
D:What percentage of his kinetic energy is due to the rotation of his legs and arms?

Explanation / Answer

I think they're saying each limb moves 30 degrees in one second. I'm going to ignore that phrase about "total of 60 degrees," which seems ambiguous. (a) The average angular speed of any limb is therefore (30*pi/180) per second; I could say "radians per second," but the radian is a dimensionless unit. In short, omega = pi/6 s^(-1). To make this into an "angular velocity," we need a directional vector, which will be parallel to the ground and perpendicular to the direction of the person's forward motion. (b) K = (0.5)*I*omega^2, where I (capital i) is the moment of inertia for a given shape. In this case, I guess you are supposed to assume each limb is just a rod, even though most people probably have more mass in their thighs than in their calves. The moment of inertia for a rod is (mL^2)/3. OK, the rotational kinetic energy of each leg is (0.37/4) (74 kg) (0.93 m)^2 (pi/6)^2 s^(-2) all divided by 3. Similarly the rotational kinetic energy of each arm is (0.13/4) (74 kg) (0.69 m)^2 (pi/6)^2 s^(-2) all divided by 3 and note that kg m^2 s^(-2) is Joules. Add up the K's for the four limbs and that'll be the answer for (b). (c) Probably what's wanted is just to add the answer from (b) to the K of the forward motion, which is (1/2) (74 kg) (6000m/3600s)^2

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