Question

A bicycle has wheels of radius 0.330 m. Each wheel has a rotational inertia of 0.0800 kg⋅m2 about its axle. The total mass of the bicycle including the wheels and the rider is 75.0 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

Given data:

* The radius of the wheel is r = 0.33 m.

* The rotational inertia of the wheel is,

`I=0.8kgm^2`

* The mass of the wheel and rider is m = 75 kg.

Solution:

The rotational kinetic energy of the wheel is,

`K_r=(1)/(2)I\omega^2`

where,

`\omega\text{ is the angular velocity of the wheel}`

The linear velocity of the wheel in terms of the angular velocity is,

`\begin{gathered} v=r\omega \\ \omega=(v)/(r) \end{gathered}`

Thus, the rotational kinetic energy in terms of the linear velocity is,

`\begin{gathered} K_r=(1)/(2)I*((v)/(r))^2 \\ K_r=(1)/(2)(Iv^2)/(r^2) \end{gathered}`

For the two wheels of the bicycle, the rotational kinetic energy is,

`\begin{gathered} K_r=2*(1)/(2)(Iv^2)/(r^2) \\ K_r=(Iv^2)/(r^2) \end{gathered}`

The linear kinetic energy of the bicycle is,

`K_t=(1)/(2)mv^2`

Thus, the total kinetic energy of the bicycle is,

`\begin{gathered} K=K_t+K_r \\ K=(1)/(2)mv^2+(Iv^2)/(r^2) \end{gathered}`

The ratio of rotational kinetic energy and total kinetic energy is,

`\begin{gathered} (K_r)/(K)=((Iv^2)/(r^2))/((1)/(2)mv^2+(Iv^2)/(r^2)) \\ (K_r)/(K)=(2I)/(mr^2+2I)*((1)/(2)*(v^2)/(r^2))/(((1)/(2)*(v^2)/(r^2))) \\ (K_r)/(K)=(2I)/(mr^2+2I) \end{gathered}`

Substituting the known values,

`\begin{gathered} (K_r)/(K)=(2*0.08)/(75*(0.33)^2+2*0.08) \\ (K_r)/(K)=\frac{0.16}{8.17+\text{0}.16} \\ (K_r)/(K)=0.019 \end{gathered}`

Thus, the rotational kinetic energy of the wheels is 0.019 times the total kinetic energy of the bicycle.