Rotational Kinetic Energy of bicycle wheels

AI Thread Summary
The discussion focuses on calculating the fraction of the total kinetic energy of a bicycle that is attributed to the rotational kinetic energy of its wheels. The bicycle has wheels with a radius of 0.33 m and a rotational inertia of 0.082 kg*m². The total mass of the bicycle and rider is 74 kg. The equations for rotational kinetic energy (KE) and linear KE are presented, with an emphasis on how to combine them to find the total KE. The user expresses confusion about the calculations, specifically regarding the number of wheels and the final result, indicating a need for clarification on the problem.
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A bicycle has wheels of radius 0.33 m. Each wheel has a rotational inertia of 0.082 kg* m2 about its axle. The total mass of the bicycle including the wheels and the rider is 74 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?

I thought this: Rotational KE = (1/2)Iw^2
=(1/2)(second bold number)w^2

Linear KE= (1/2)mv^2
= (1/2)(third bold number)(radius*w)^2 (i.e. plug in r*w for v)
Total KE is equal to Rotational KE + Linear KE
add the two eqns
(1/2)Iw^2/ (some # * w^2)
 
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i think you are correct
eventually one takes away the w^2 in both numerator and denominator, then gets a result independent of w
 
i keep getting .041/4.07 and that is not right
 
How many wheels does a bicycle have?

Please post in the HW forums.
 
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