Physics KE and Rotational Energy Problem!!!!?

Question

A bicycle has wheels of radius 0.26 m. Each wheel has a rotational inertia of 0.095 kg·m2 about its axle. The total mass of the bicycle including the wheels and the rider is 77 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?

Answer 1

KE(trans) = mv^2/2
KE(rot) = Iω^2/2
ω = v/r
KE(rot) = Iv^2/(2r^2)
KE(tot) = KE(trans) + KE(rot) = v^2*(m/2 + I/(2r^2))
KE(rot) / KE(tot) = I/(2r^2) / (m/2 + I/(2r^2)) = I/r^2 / (m + I/r^2)
(Remember I = 2*0.095)

Answer 2

The other answer is on the right track. If you know the SUVAT (kinematic) equations for linear motion, there is an angular motion equivalent for each one. For example V = U - AT becomes Wf = Wo - alpha T in angular terms; where the two W's are angular speed in rad/sec and alpha is angular (-) deceleration in rad/sec^2. T is T for both angular and linear. Solve for T = (Wo - Wf)/alpha = ? sec, the time you're looking for. We need alpha. In SUVAT, we have V^2 = U^2 - 2aS; so the angular equivalent is Wf^2 = Wo^2 - 2 alpha Omega. Omega is the angular distance, the equivalent to S in linear terms. So alpha = Wo^2/(2 Omega) is the angular acceleration. Plug that back into the T equation. T = (Wo - Wf)//Wo^2/(2 Omega) = (Wo - Wf)(2 Omega)/Wo^2 = Wo(2 Omega)/Wo^2 = 2 Omega/Wo = 2*2*pi()*13.2/6.15 = 26.97 sec. ANS. Note that 13.2 revs is 2pi 13.2 radians = Omega. On the other hand, Wo = 6.15 rad/sec so the speed is already in radians and no conversion is needed.