If a person of mass MM simply moved forward with speed VV, hiskinetic energy would be 12MV212MV2. However, in addition topossessing a forward motion, various parts of his body (such as thearms and legs) undergo rotation. Therefore, his total kineticenergy is the sum of the energy from his forward motion plus therotational kinetic energy of his arms and legs. The purpose of thisproblem is to see how much this rotational motion contributes tothe person's kinetic energy. Biomedical measurements show that thearms and hands together typically make up 14.0 %% of a person'smass, while the legs and feet together account for 38.0 %% . For arough (but reasonable) calculation, we can model the arms and legsas thin uniform bars pivoting about the shoulder and hip,respectively. In a brisk walk, the arms and legs each move throughan angle of about ±30∘±30∘ (a total of 60∘60∘) from the vertical inapproximately 1 second. We shall assume that they are heldstraight, rather than being bent, which is not quite true. Let usconsider a 70.0 kgkg person walking at 5.00 km/hkm/h having arms70.0 cmcm long and legs 90.0 cmcm long.

1.What is the average angular velocity of his arms and legs?

2.Using the average angular velocity from part A, calculate theamount of rotational kinetic energy in this person's arms and legsas he walks.

3.What is the total kinetic energy due to both his forwardmotion and his rotation?

4.What percentage of his kinetic energy is due to the rotationof his legs and arms?