Torque is connected with force through the next equation:

$$
tau=Fcdot rcdot sin (theta)
$$

All cases have equal force so we will focus on the differences in distance and angle.

Between cases $A$ and $B$

The angle is equal, but distance from the pivot point is greater in case $A$ which further means that the torque will be as well.

The distance is equal, but angle is greater in case $B$ which further means that the torque will be as well.

The angle is greater in case $D$, but distance is significantly greater in case $C$, which overall means that the torque will be greater in case $C$

The angle relative to the direction of the force and distance between the force and the pivot point is equal in both cases.
The distance between the force and the pivot point is equal as well. This means that the torque will be equal as well.

$$
Delta omega=frac{taucdot t}{I}
$$

Where $Delta omega$ is a change in angular velocity, $tau$ is a torque applied to the object, $t$ is time and $I$ is a moment of inertia.

We can affect two of those factors, torque applied and object’s moment of inertia. Therefore by changing torque applied to the object or moment of inertia of the object we can effectively change its angular frequency.

It is done by placing a wheel on vertical shaft and adding weights until the wheel is horizontal.

This is equivalent to moving the center of the mass of the wheel because we actually are moving a center of mass. The only difference is that the center of mass of the wheel by itself is always at the same place, but center of mass of the whole assembly consisting of wheel and weights is changing as we add weights. The next equation explains the location of the center of mass in assemblies:

$$
x_{cm}=frac{x_1cdot m_1+x_2cdot m_2}{m_1+m_2}
$$

Where $x$ is a location of center of mass of each part and $m$ is a mass of that part. $x_{cm}$ is the location of the center of mass of the assembly.

That torque can be calculated as:

$$
tau=(mg)cdot r
$$

Where $tau$ is torque and $r$ is $textbf{horizontal}$ distance from the axis of rotation (which is between those two wheels the truck is balancing on).

Tangential velocity can be expressed through equation:

$$
v=omega r
$$

Since we know that their linear, tangential velocities are equal, we can set up an equation:

$$
omega_1 cdot r_1=omega_2 cdot r_2
$$

From this we can finally get the relation between their angular velocities:

$$
boxed{frac{omega_1}{omega_2}=frac{r_2}{r_1}}
$$

– Wheel with mass at the hub

– Wheel with unifrom mass

– Wheel with mass at the rim.

The greater the moment of inertia, the smaller the angular acceleration which can be shown through the equation:

$$
alpha=frac{tau}{I}
$$

To sum it up, the $textbf{wheel with mass at the hub}$ will have the greatest angular acceleration and the $textbf{wheel with the mass at the rim}$ will be having the least.

To do so, we need to write down equation that describes torque:

$$
tau=Fcdot r
$$

Where $tau$ is torque, $F$ is a force exerted on the object and $r$ is a distance between the direction on which the force acts and the axis of rotation.

Looking at that equation, we can see that for forces that act on the axis of rotation the distance $r$ is equal to zero, which further means that the torque caused by those forces is also equal to zero.

$$
theta=frac{s}{r}
$$

Inserting values we get:

$$
theta=frac{1.5}{2.5}
$$

Angular displacement is:

$$
boxed{theta=0.6,,rm{rad}}
$$

$$
omega=frac{v}{r}
$$

Inserting values we get:

$$
omega=frac{23}{0.45}
$$

Angular velocity of the tire is:

$$
boxed{omega=51.1,,rm{rad/s}}
$$

$$
theta=frac{s}{r}
$$

From this equation we can express distance traveled by those points:

$$
s=theta cdot r
$$

Inserting values we get:

$$
s=128cdot frac{2cdot pi,,rm{rad}}{360,,^{o}}cdot 22cdot 10^{-2},,rm{m}
$$

$$
boxed{s=0.491,,rm{m}}
$$

$$
omega=(1880,,rm{rev/min})cdot (2cdot pi,,rm{rad})cdot ( frac{1}{60},,rm{s/min})
$$

$$
boxed{omega=196.9,,rm{rad/s}}
$$

$$
theta=omega cdot t
$$

Inserting values we get.

$$
theta=196.9cdot 2.5
$$

$$
boxed{theta =492.3,,rm{rad}}
$$

$$
theta =492.3,,rm{rad}
$$

$$
begin{align*}
omega_1 &=475,,rm{rev/min}\
omega_2 &=187,,rm{rev/min}\
t &=4,,rm{s}
end{align*}
$$

Angular acceleration can be calculated as:

$$
begin{align*}
alpha &=frac{omega_2 -omega_1}{t}\
alpha &=frac{187-475}{4} cdot frac{2 pi}{60}\
end{align*}
$$

Finally:

$$
boxed{alpha=-7.54,,rm{rad/s^2}}
$$

$$
begin{align*}
r &=9,,rm{cm}\
omega &=2.5,,rm{rad/s}\
r_2 &=7,,rm{cm}
end{align*}
$$

In order to calculate the velocity of a point $7$ cm away from the center we only need to know angular velocity and its distance from the center:

$$
begin{align*}
v &=omega r_2\
v&=2.5cdot 0.07
end{align*}
$$

Finally:

$$
boxed{v=0.175,,rm{m/s}}
$$

$$
begin{align*}
omega_1 &=328,,rm{rev/min}\
omega_2 &=542,,rm{rev/min}\
d&=0.43,,rm{m}
end{align*}
$$

a) Ratio of centripetal accelerations can be calculated as:

$$
begin{align*}
a_c &=romega^2\
frac{a_{fast}}{a_{slow}} &=frac{r omega_2}{romega_1}\
frac{a_{fast}}{a_{slow}} &=frac{542^2}{328^2}\
end{align*}
$$

$$
boxed{frac{a_{fast}}{a_{slow}} =2.73}
$$

b) Ratio of velocities can be calculated as:

$$
begin{align*}
v &=romega\
frac{v_{fast}}{v_{slow}} &=frac{romega_2}{romega_1}\
frac{v_{fast}}{v_{slow}} &=frac{542}{328}
end{align*}
$$

$$
boxed{frac{v_{fast}}{v_{slow}} =1.65}
$$

b) $frac{v_{fast}}{v_{slow}} =1.65$

$$
begin{align*}
r &=0.215,,rm{m}\
omega &=542,,rm{rev/min}
end{align*}
$$

Centripetal acceleration can be calculated as:

$$
begin{align*}
a_c &=romega^2\
a_c&=0.215cdot (542cdot frac{2pi}{60})^2
a_c&=frac{693}{9.8}\
end{align*}
$$

$$
boxed{a_c=70.7,,rm{g}}
$$

First we need to convert acceleration to SI units:

$$
a=0.35cdot 10^6,,rm{g} cdot 9.8,,rm{(m/s^2)/g}=3.43cdot 10^6,,rm{m/s^2}
$$

Now we have to do the same for the distance:

$$
r=2.5,,rm{cm}=2.5cdot 10^{-2},,rm{m}
$$

$$
a=romega ^2
$$

From which we can get:

$$
omega=sqrt{frac{a}{r}}
$$

$$
omega =sqrt{frac{3.43cdot 10^6}{2.5cdot 10^{-2}}}
$$