Question

If a person of mass M simply moved forward with speed V, his kinetic energy would be 12MV2. 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 13.0 % 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 ±30∘ (a total of 60∘) 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 76.0 kg person walking at 6.00 km/h having arms 66.0 cm long and legs 95.0 cm long.

a) What is the average angular velocity of his arms and legs?

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.

c) What is the total kinetic energy due to both his forward motion and his rotation?

d) What percentage of his kinetic energy is due to the rotation of his legs and arms?

Answer 1

a)

= total angular displacement = 60 degree = /3 = 1.05 rad

t = time taken = 1 sec

average angular velocity = w = /t = 1.05/1 = 1.05 rad/s

b)

for arms :

M_a = mass of arms = 0.13 x 76 = 9.88 kg

L_a = length of arm = 66 cm = 0.66 m

I_a = moment of inertia of arms = M_a L²_a /3 = (9.88) (0.66)²/3 = 1.43 kgm²

for legs :

M_l = mass of arms = 0.37 x 76 = 28.12 kg

L_l = length of arm = 95 cm = 0.95 m

I_l = moment of inertia of arms = M_l L²_l /3 = (28.12) (0.95)²/3 = 8.5 kgm²

Rotational kinetic energy is given as

KE_r = (0.5) (I_a + I_l) w²

KE_r = (0.5) (1.43 + 8.5) (1.05)²

KE_r = 5.5 J

c)

M = mass of person = 76 kg

v = speed = 6 km/h = 1.67 m/s

translational kinetic energy is given as

KE_t = (0.5) M v² = (0.5) (76) (1.67)² = 106 J

Total kinetic energy is given as

KE = KE_r + KE_t = 5.5 + 106 = 111.5 J

d)

%age = KE_r x 100/KE = 5.5 x 100/111.5 = 4.93