A bicycle racer is going downhill at $\mathbf{11}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$, when to his horror, one of his $\mathbf{2}\mathbf{.}\mathbf{25}\mathbf{-}\mathbf{kg}$wheels comes off as he is $\mathbf{75}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{m}$above the foot of the hill. We can model the wheel as thin-walled cylinder $\mathbf{85}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}$in diameter and ignore the small mass of the spokes. (a) How fast the wheel is moving when it reaches the foot of the hill if it rolled without slipping all the way down? (b) How much kinetic energy does the wheel have when it reaches the bottom of the hill?

Initial Speed of the bicycle $\left(v_i\right)=11.0\frac{\text{m}}{\text{s}}$.

Mass of one of the wheels $\left(m\right)=2.25kg$.

Diameter of the cylinder $\left(D\right)=85.0cm$.

The radius of the cylinder $\left(R\right)=42.5cm$.

Initial height $\left(h_i\right)=75.0m$.

Consider the total kinetic energy which is the sum of the rotational kinetic energy about its center of mass, the translational kinetic energy of the center of mass and gravitational potential energy.

$$\begin{gathered} E=K_{tr}+K_{rot}+U \\ \mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{I\omega }}^{\mathbf{2}}\mathbf{+}\mathbf{mgh}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\dots \dots ..\left(1\right) \end{gathered}$$

Consider the formula for the velocity at the center as:

$v=R\omega$ ……. (2)

Here, $\omega$is the angular speed about its center of mass and R is radius of wheel.

Consider the formula for the moment of inertia:

$I=m{R}^2$ …… (3)

The rotational kinetic energy is determined as:

$$\begin{gathered} K_{rot}=\frac{1}{2}m{R}^2{\omega }^2 \\ {K}_{rot}=\frac{1}{2}m{R}^2{\left(\frac{v}{R}\right)}^2 \\ {K}_{rot}=\frac{1}{2}m{v}^2 \end{gathered}$$

Hence, in this case, we get the same expression as one of the kinetic energies due to translation.

Thus, from $\left(1\right)$, we get,

$$\begin{gathered} E=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2+mgh \\ \Rightarrow E=m{v}^2+mgh \end{gathered}$$

This energy is conserved.

Therefore, the initial and final state are related as:

$m{v}_i^2+mg{h}_i=m{v}_f^2+mg{h}_f$ ……. (4)

Let us take $h_f=0$for final speed.

Then $\left(4\right)$becomes,

$$\begin{gathered} m{v}_i^2+mg{h}_i=m{v}_f^2+0 \\ \Rightarrow v_i^2+g{h}_i=v_f^2 \\ \Rightarrow v_f=\sqrt{{\left(11.0\right)}^2+\left(9.80\right)\left(75.0\right)} \\ \Rightarrow v_f=29.3 m/s \end{gathered}$$

Hence, the velocity of the wheel is $v_f=29.3\frac{\text{m}}{\text{s}}$.

The kinetic energy at the bottom of the wheel is,

$$\begin{gathered} E=\frac{1}{2}m{v}^2+\frac{1}{2}m{v}^2 \\ \Rightarrow E=m{v}^2 \\ \Rightarrow E=\left(2.25\right){\left(29.3\right)}^2 \\ \Rightarrow E=1.93\times {10}^{3}J \end{gathered}$$

Hence, the kinetic energy at the bottom of the wheel is, $E=1.93\times {10}^{3}J$.