What fraction of the total kinetic energy is the rotational kinetic energy of the wheels on a bike?
A bicycle has wheels of radius 0.23 m. Each wheel has a rotational inertia of 0.093 kg ยท 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?
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 %
October 31st, 2009 at 12:53 pm
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%
References :
October 31st, 2009 at 1:35 pm
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 %
References :