91Ó°ÊÓ

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.

$$\begin{gathered} E=K_{tr}+K_{rot}+U \\  \mathbf{E}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{\pm õÓ\neg }}^{\mathbf{2}}\mathbf{+}\mathbf{mgh}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}……..\left(1\right) \end{gathered}$$

Consider the formula for the velocity at the center as:

$v=R\mathrm{Ó\neg }$ ……. (2)

Here, $\mathrm{Ó\neg }$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:

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

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,

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

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,

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

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,

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

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