91Ó°ÊÓ

${\mathit{E}}_{\mathbf{i}}\mathbf{+}\mathit{W}\mathbf{=}{\mathit{E}}_{\mathbf{f}}$ (1)

Where is initial and final energies respectively.

is work done

${\mathit{U}}_{\mathbf{g}}\mathbf{=}\mathit{m}\mathit{g}\mathit{y}$ (2)

Where ${\mathit{U}}_{\mathbf{g}}$is gravitational potential energy.

m and y is position and mass respectively.

The transitional kinetic energy of an object is

${\mathbf{K}}_{\mathbf{T}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$ (3)

Where is mass of object and is speed relative to given coordinate system.

${\mathit{K}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{I}{\mathit{Ó\neg }}^{\mathbf{2}}$ (4)

Where l is moment of inertia and$\mathit{Ó\neg }$ is angular speed.

role="math" localid="1663810181777" ${\mathit{v}}_{\mathbf{C}}\mathit{M}\mathbf{=}\mathit{Ó\neg }\mathit{R}$ (5)

Where$\mathit{Ó\neg }$ areR angular speed and radius of rigid object respectively.

Here we have given that total weight of the vehicle and rider is fixed.

From equation (1) we have

$E_i+W=E_f$

Now, the initial energy of the system is only the gravitational potential energy since the vehicle starts from rest. The work done on the system is zero, ignoring air resistance and friction. The final energy of the system is only kinetic. So

$U_{g,1}+0=K_R+K_T$ (6)

Where$U_{gi}$ is initial gravitational potential energy

$K_R,K_T$are rotational and transitional kinetic energies respectively.

Now, put the values of $U_{gi},K_R, K_T$in equation (6). We get,

$Mgh=4\left(\frac{1}{2}I{\mathrm{Ó\neg }}^2\right)+\frac{1}{2}M{v}_c{m}^2$

Where l is the moment of inertia of each wheel for which we can substitute$xm{R}^2$

Where$x=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ when the wheel is a disk and x=1 when wheel is a hoop.

Now,$\mathrm{Ó\neg }$ can be found in terms of$v_{cm}$ from equation (5)

So, above equation becomes

$$\begin{gathered} Mgh=2xm{R}^2{\left(\frac{v_{cm}}{R}\right)}^2+\frac{1}{2}M{v}_{cm}^2 \\ 2Mgh=4xm{v}_{cm}^2+Mv{v}_{cm}^2 \\ {v}_{cm}^2=\left(4xm×M\right)=2Mgh \\ {v}_{cm}=\sqrt{\frac{2Mgh}{\left(4xm×M\right)}} \\ {v}_{cm}=\sqrt{2gh\left(\frac{M}{\left(4xm×M\right)}\right)} \end{gathered}$$

Now, here we haveM of the vehicle and the rider is constant.

So, we can change in x and m .

Since x and m are in denominator, making them smaller, increases the speed of the vehicle at the bottom of the hill.

Therefore light and solid wheels should be used.

Note that the final speed does not depend on the radiusR of each wheel, so it doesn’t matter if the wheels are massive or small.

Hence, light solid wheels should be used for a greater speed at the bottom of the hill.