Comments on: What fraction of the total kinetic energy is the rotational kinetic energy of the wheels on a bike? http://www.mathstudenthelp.info/fraction/what-fraction-of-the-total-kinetic-energy-is-the-rotational-kinetic-energy-of-the-wheels-on-a-bike Let us help you add it up! Mon, 21 May 2012 11:50:35 +0000 http://wordpress.org/?v=2.7 hourly 1 By: SCAR http://www.mathstudenthelp.info/fraction/what-fraction-of-the-total-kinetic-energy-is-the-rotational-kinetic-energy-of-the-wheels-on-a-bike/comment-page-1#comment-6894 SCAR Sat, 31 Oct 2009 18:35:59 +0000 http://www.mathstudenthelp.info/fraction/what-fraction-of-the-total-kinetic-energy-is-the-rotational-kinetic-energy-of-the-wheels-on-a-bike#comment-6894 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 %<br><b>References : </b><br> 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 :

]]> By: Alex http://www.mathstudenthelp.info/fraction/what-fraction-of-the-total-kinetic-energy-is-the-rotational-kinetic-energy-of-the-wheels-on-a-bike/comment-page-1#comment-6893 Alex Sat, 31 Oct 2009 17:53:59 +0000 http://www.mathstudenthelp.info/fraction/what-fraction-of-the-total-kinetic-energy-is-the-rotational-kinetic-energy-of-the-wheels-on-a-bike#comment-6893 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%<br><b>References : </b><br> 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 :

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