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:

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

So, the speed of the car at the center of the wheels of the car and its occupants is $23 mathrm{~m/s}$.

The radius of the baseball is $r = 0.12 mathrm{~m}$. The baseball’s mass is $m = 0.57 mathrm{~kg}$. Let the baseball to be a hollow sphere.

$textbf{Required: }$

Finding the moment of inertia of the baseball.

As the textbook mentions that the moment of inertia of the hollow sphere is given by

$$
begin{align*}
I &= dfrac{2}{3} ~ m ~ r^{2} \
&= dfrac{2}{3} times 0.57 mathrm{~kg} times left( 0.12 mathrm{~m} right)^{2} \
&= 5.472 times 10^{-3} mathrm{~kg cdot m^{2}}
end{align*}
$$

So, the moment of inertia of the baseball which is equal to the moment of inertia of the hollow sphere is $5.472 times 10^{-3} mathrm{~kg cdot m^{2}}$.

The angular speed of the disk of pizza is $omega = 7.2 mathrm{~rad/s}$. The moment of inertia of the pizza dough is $I = 6.3 times 10^{-6} mathrm{~kg cdot m^{2}}$.

$textbf{Required: }$

Finding the rotational kinetic energy of the disk of the pizza dough.

As the textbook mentions that the rotational kinetic energy is given by

$$
begin{align*}
K.E_{rot} &= dfrac{1}{2} ~ I ~ omega^{2} \
&= dfrac{1}{2} times 6.3 times 10^{-6} mathrm{~kg cdot m^{2}} times left( 7.2 mathrm{~rad/s} right)^{2} \
&= 1.633 times 10^{-4} mathrm{~J}
end{align*}
$$

So, the rotational kinetic energy of the disk of the pizza dough is $1.633 times 10^{-4} mathrm{~J}$.

The moment of inertia of the ball is $I = 1.6 times 10^{-8} mathrm{~kg cdot m^{2}}$. The angular speed of the ball is $omega = 8.2 mathrm{~rad/s}$.

$textbf{Required: }$

Finding the angular momentum of the ball.

As the textbook mentions that the angular momentum is given by the product of an object’s momentum of inertia and its angular speed

$$
begin{align*}
L &= I ~ omega \
&= 1.6 times 10^{-8} mathrm{~kg cdot m^{2}} times 8.2 mathrm{~rad/s} \
&= 1.312 times 10^{-7} mathrm{~kg cdot m^{2}/s}
end{align*}
$$

So, the angular momentum of the ball is $1.312 times 10^{-7} mathrm{~kg cdot m^{2}/s}$.