As we know purely rotational motion occurs if the object moves in a circle about a single line. This line is called the axis of rotation. Also, we know that rolling is a type of motion that combines rotational motion and translational motion of that object. So, the rolling motion differs from pure rotational motion by the translational motion.
On the contrary side, the rolling motion differs from the pure linear motion by the rotational motion.

As we know the momentum of inertia is determining how easy or hard it is to change its rotation. So, if the object with a large moment of inertia (increasing the moment of inertia) is difficult to start or stop rotating.

Solve for the case of mass equals $m$:

As the problem mentions the moment of inertia depends linearly on the mass and on the distance squared

$$
begin{align*}
I_{1} &= m ~ r^{2}
end{align*}
$$

Solve for the case of mass equals $2m$:

As the problem mentions the moment of inertia depends linearly on the mass and on the distance squared

$$
begin{align*}
I_{2} &= 2m ~ r^{2} \
&= 2 ~ I_{1}
end{align*}
$$

So, if the mass of the object is doubled, then the moment of inertia will be double.

According to table $8.1$, The moment of inertia of a disk is equal to $I_{d} = dfrac{1}{2} m ~ r^{2}$ and the moment of inertia of a solid sphere is equal to $I_{s} = dfrac{2}{5} m ~ r^{2}$. So, the moment of inertia of the disk is greater than the moment of inertia of a solid sphere. So, the disk is hard to rotate. But the solid sphere is easy to rotate. So, the sphere reaches the bottom before the disk. Therefore, the solid sphere is the winner.

Since the wheels roll without slipping, then the angular velocity is only the affecting velocity. So, the instantaneous speed of the wheels at the bottom is equal to zero.

As the problem mentions that the mass and the radius of the hoop and the disk is the same. According to table $8.1$, the moment of inertia of the hoop is $I = m ~ r^{2}$ and the moment of inertia of the disk is $I = dfrac{1}{2} m ~ r^{2}$. So, the moment of inertia of the hoop is twice the moment of inertia of the disk. As the textbook mentions that the rotational kinetic energy is given by

$$
begin{align*}
K.E_{rot} &= dfrac{1}{2} ~ I ~ omega^{2}
end{align*}
$$

So, the rotational kinetic energy of the hoop is twice the rotational kinetic energy of the disk. Therefore, the rotational kinetic energy of the hoop is greater than the rotational kinetic energy of the disk.

The speed at the top of the wheels of the car is $v = 46 mathrm{~m/s}$.

$textbf{Required: }$

Finding the speed of the car and its occupants.

In order to evaluate the tangential speed of the rotating wheels at the center, we use the following relation:

$$
begin{align*}
v_{c} &= r ~ omega
end{align*}
$$

In order to evaluate the tangential speed of the rotating wheels at the top of the wheels of the car, we use the following relation:

$$
begin{align*}
v_{t} &= d ~ omega \
&= 2 r ~ omega \
&= 2 ~ v_{c}
end{align*}
$$

Rearrange and solve for the speed of the car at the center of the wheels of the car and its occupants:

$$
begin{align*}
v_{c} &= dfrac{ v_{t} }{2} \
&= dfrac{ 46 mathrm{~m/s} }{ 2 } \
&= 23 mathrm{~m/s}
end{align*}
$$