Kinetic energy of bicycle wheels A 55 kg woman is riding a 2

Kinetic energy of bicycle wheels. A 55 kg woman is riding a 24 kg bike at a speed of 7 m/s. The wheels of the bike can be treated as thin rings, each of mass 3 kg and radius 0.33 m. What percentage of the total kinetic energy of the woman-bike system is carried in the rotational kinetic energy of the wheels?

Solution

Total kinetic energy= 1/2*m*v^2= 1/2*(55+24)*7^2=1935.5 J

Rotation kinetic energy of 2 wheels= 2*(1/2)*I*W^2

where I is the moment of inertia for the wheel and I=MR^2= 3*0.33^2=0.3267 kg m^2

and W is the angular velocity of the wheel and W=Linear velocity/Radius= 7/0.33=21.21 rad/s

Hence, rotational KE= 2*1/2*0.3267*(21.21^2)= 146.98 J

Percentage =146.98/1935.5*100= 7.59%