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?(adsbygoogle = window.adsbygoogle || []).push({});

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)

**Physics Forums - The Fusion of Science and Community**

# Rotational Kinetic Energy of bicycle wheels

Have something to add?

- Similar discussions for: Rotational Kinetic Energy of bicycle wheels

Loading...

**Physics Forums - The Fusion of Science and Community**