Question

If a person of mass *M* simply moved forward with speed
*V*, his kinetic energy would be 12*M**V*2.
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}
L^{2}_{a} /3 = (9.88) (0.66)^{2}/3 = 1.43
kgm^{2}

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}
L^{2}_{l} /3 = (28.12) (0.95)^{2}/3 = 8.5
kgm^{2}

Rotational kinetic energy is given as

KE_{r} = (0.5) (I_{a} + I_{l})
w^{2}

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 v^{2} = (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

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. Let us consider
a 70.0 kgkg person having arms 66.0 cmcm long and legs 95.0 cmcm
long. The person is running at 12.0 km/hkm/h, with his arms and...

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. Let us consider
a 74.0 kg person having arms 70.0 cm long and legs 90.0 cm long.
The person is running at 12.0 km/h , with his...

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Figure 9.31.
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