In order to calculate angular displacement, we need to calculate how many $textbf{rotations}$ each hand makes in a period of $textbf{one }$hour:

$textbf{Second hand will make 60 complete rotations}$. One each minute and there are 60 minutes in one hour.

$textbf{Minute hand will make 1 complete rotation}$ in a period of one hour,

$textbf{Hour hand will make 1/12 complete rotations }$since it will move for one out of twelve hours marked on a clock in a period of one hour, which means it will make $frac{1}{12}$ complete rotations in one hour.

In 1 hour, the second hand will have 60 complete rotations in the clockwise direction:

$$
Delta_{theta s}=60cdot (-2cdotpi )=-377,,rm{rad}
$$

In 1 hour, the minute hand will have 1 complete rotation in the clockwise direction:

$$
Delta_{theta m}=1cdot (-2cdotpi )=-6.28,,rm{rad}
$$

In 1 hour, the hour hand will have $frac{1}{12}$ complete rotations in the clockwise direction:

$$
Delta_{theta h}=frac{1}{12} cdot (-2cdotpi )=-0.524,,rm{rad}
$$

$$
Delta_{theta s}=-377,,rm{rad}
$$

$$
Delta_{theta m}=-6.28,,rm{rad}
$$

$$
Delta_{theta h}=-0.524,,rm{rad}
$$

We need to calculate diameter of the wheels. The information that we are given is linear acceleration:

$$
a=1.85,,rm{m/s^2},
$$

and angular acceleration:

$$
alpha =5.23,,rm{rad/s^2}
$$

Just as angular velocity and linear velocity are connected, so are angular acceleration and linear acceleration through the next equation:

$$
a =alpha cdot frac{d}{2}
$$

From the previous equation we can get:

$$
d=frac{2cdot a}{alpha}
$$

When we insert values we get:

$$
d=frac{2cdot 1.85}{5.23}=0.707,,rm{m}
$$

$$
d=0.707,,rm{m}
$$

Truck from the previous problem is towing a trailer with diameter of wheels of:

$$
d=0.48,,rm{m}
$$

Wheels on the truck have diameter of:

$$
d_t=0.707,,rm{m}
$$

Linear acceleration of the truck is:

$$
a_t=1.85,,rm{m/s^2}
$$

And angular acceleration of the wheels on the truck was:

$$
alpha_t=5.23,,rm{rad/s^2}
$$

a) While towing a trailer, between a truck and trailer is a fixed connection maintaining always constant distance between them. This means that displacement of the truck is equal to that of the trailer. Which further means that so is linear velocity as well as linear acceleration. That means that:

$$
a=a_t=1.85,,rm{m/s^2}
$$

b) Since we know that their linear acceleration is equal, we can use the equation that connects linear and angular acceleration in order to compare them:

$$
a=alpha cdot 0.5cdot d
$$

$$
alpha =frac{a}{0.5cdot d}
$$

First, lets compare linear accelerations:

$$
a=a_t
$$

Inserting previous equation we get:

$$
alpha cdot 0.5cdot d=alpha_t cdot 0.5cdot d_t
$$

By solving equation we get:

$$
frac{alpha}{alpha_t}=frac{d_t}{d}
$$

Inserting value we get:

$$
frac{alpha}{alpha_t}=frac{0.707}{0.48}=1.47
$$

$$
a=1.85,,rm{m/s^2}
$$

$$
frac{alpha}{alpha_t}=1.47
$$

Angular velocity can be expressed as:

$$
omega =frac{v}{0.5cdot d}
$$

And distance traveled ($s$) can be expressed through number of revolutions ($n$) as:

$$
s=ncdot dcdot pi
$$

With constant distance we get:

$$
n=frac{s}{dcdot pi}
$$

From the previous equations we can conclude that by $textbf{increasing diameter}$, both angular velocity and number of revolutions will $textbf{decrease}$.