Question

If a person of mass MM simply moved forward with speed VV, his kinetic energy would be 12MV212MV2. 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 38.0 %% . 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∘±30∘ (a total of 60∘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 70.0 kgkg person walking at 5.00 km/hkm/h having arms 70.0 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 the amount of rotational kinetic energy in this person's arms and legs as he walks.

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

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

Answer 1

Part (1)

average angular velocity of arms and legs, = angular displacement / time = /3 rad/s

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Part-(2)

Rotational kinetic energy of legs = (1/2) I_l²

where I_l is moment of inertia of legs , I_l = (1/12) M_l L_l²

I_l = (1/12) { 0.38 70 0.9 0.9 ) = 1.796 kg m²

Rotational kinetic energy of legs = (1/2) 1.796 (/3)² = 0.985 J

Rotational kinetic energy of hands = (1/2) I_h²

where I_l is moment of inertia of hands , I_h = (1/12) M_h L_h²

I_l = (1/12) { 0.14 70 0.7 0.7 ) = 0.4 kg m²

Rotational kinetic energy of hands = (1/2) 0.4 (/3)² = 0.219 J

Rotational kinetic energy of hands + legs, E_R = 0.985 + 0.219 = 1.204 J

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Part-(3)

Kinetic energy due to forward motion , E_F = (1/2) m v² = (1/2) 70 [ 5 (5/18) ]² = 67.515 J

Kinetic energy due to forward motion and rotational energy of arms and legs = E_F + E_R

E_F + E_R = 67.515 + 1.204 = 68.719 J

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Part-(4)

Percentage of rotational kinetic energy due to rotational motion of legs and hands

in total kinetic energy is given as, ( 1.204 / 68.719 ) 100 = 1.752 %