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smmunday smmunday
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11 years ago
A bicycle has wheels of radius 0.23 m. Each wheel has a rotational inertia of 0.093 kg · m2 about its axle. The total mass of the bicycle including the wheels and the rider is 72 kg. When coasting at constant speed, what fraction of the total kinetic energy of the bicycle (including rider) is the rotational kinetic energy of the wheels?
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#1
11 years ago
The rotational kinetic energy is given by E(rot.) = (1/2)Iw^2 and the objects total kinetic energy is given by E(tot.) = (1/2)mv^2.

[where m is mass; v is velocity; I is moment of inertia; w is angular frequency = v/r]

So fraction of energy belonging to two wheels is:

I(v/r)^2 / (1/2)mv^2
(I.v^2/r^2) / (1/2)mv^2  [v^2 cancels]
(I/r^2) / (1/2)m            [plugging in data]
(0.093/(0.23)^2) / (72/2)

= ~0.05 or 5%
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#2
11 years ago
The TOTAL kinetic energy is the sum of translational AND rotational kinetic energy.

So (after much substitution, rearrangment and cancellation)

%rot = (2*I/r**2) / ( 2*I/r**2 + m)

where:
I = mass moment of inertia for the wheels = .093 kg*m**2
r = radius of wheels = .23 m
m = total mass of cycle and rider = 72 kg

%rot = [2*.093/.23**2] / { [2*.093/.23**2] + 72 }*100
        = (3.52)/(3.52 + 72) *100
       = 4.66 %
wrote...
#3
10 years ago
Thank you for posting this!
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