Since static friction helps in the movement of rolling motion, not opposes it, Why do we say we need more torque to get the car to move from rest if we have the rolling friction coefficient static and the torque = mass * acceleration * wheel radius, surely the mass and radius do not change. So, why do we need more torque at the starting of movement than at regular acceleration when the car is already moving?
[Physics] Why does a car need more torque when accelerating from rest
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The static friction will be pointing forwards not backwards. It comes into existence to counteract sliding when the torque pulls the wheel around - the wheel tries to push the ground backwards and static friction is the ground's equal and opposite response.
That static friction is then the only force on the wheel. Newton's 2nd law says that the wheel will then accelerate forward.
Update
Since this means that there is translational acceleration $a$, there must be angular acceleration $\alpha$ of the wheel as well. The well spins faster and faster while the car goes faster and faster. This is because of the so-called geometric bond between wheel and ground:
$$a=R\alpha$$
You can't have $a$ without $\alpha$ in this case.
This means that the torques do no balance out, which seemed to be the assumption that caused your whole confusion. The applied torque from the engine will not equal the torque from friction. It will be a bit bigger in order to also cause some angular acceleration. You must use $$\sum \tau =I\alpha$$ And not $$\sum\tau=0\;.$$
This equation along with the geometric bond and Newton's 2nd law should be enough equations for you to solve for the static friction.
In order to allow your robot to continue moving up the hill we must apply an external torque (which you have labelled as $M$ in your case) to drive the wheel up the ramp. For your case it may be easier to look at it instead as a sum of torques given as \begin{equation} \sum \tau = I \beta \end{equation}
Where $I$ is the moment of inertia of your object, and $\beta$ is the angular acceleration (typically it is written as $\alpha$ but since you use that as your angle, I will instead use $\beta$).
Now any torques on an object are just forces on an object that are being exerted away from its pivot point (or in your case the center of mass which would be the center of the wheel). First we must define a positive direction (whichever way we define it won't change the outcome). I will define clockwise as positive.
Let's pretend you didn't have the external torque of the motor, $M$. The only force in your situation that is acting away from the pivot is the frictional force. In general the torque is given as $\tau = rF \sin \theta$, where $\theta=90^\circ$ since your radial vector is perpendicular to your frictional force vector. In addition, it is necessary to know the sign of the torque. Since your frictional force would cause a counter clockwise rotation (opposite what we defined as positive) then it will be a negative torque. So our sum of torques becomes
\begin{equation} -r F_{tr} = I \beta \end{equation}
Thus giving,
\begin{equation} \beta = -\frac{r F_{tr}}{I} \end{equation}
The negative in the equation is very informative. It tells us that the wheel would have a negative angular acceleration if there was no external torque. In layman's terms this simply means the wheel will be slowing down, which makes sense that if we have no driving mechanism, something rolling up a hill will just slow down. You can always view this in terms of your linear acceleration as well, $\displaystyle \beta = \frac{a_x}{r}$.
If we now include the external torque into our sum of torques,
\begin{equation} M-r F_{tr} = I \beta \end{equation}
Then you can see that your required torque would have to be
\begin{equation} M = r F_{tr} + I \beta \end{equation}
We could put this in a more usable form. I'll assume your wheel could be modelled as a solid cylinder which has a moment of inertia $I = \frac{1}{2} m r^2$. And using, the linear form of our acceleration mentioned above, then your required torque would be
\begin{equation} M = r F_{tr} + \frac{m r a_x}{2} \end{equation}
Also, in regards to your last statement about the friction pointing up the hill whether it was rolling up or down. Remember, that for a rolling object that doesn't slip, the point at which it makes contact with the surface MUST have zero velocity (which means it cannot have any net acceleration). So in your case, whether the wheel is going up or down the hill, you will always have what you call the dynamic component of gravity pointing down the ramp. In order to prevent the bottom of the wheel contacting the surface from slipping (accelerating in the direction of that force component) the frictional force MUST exist to counteract that force and keep the bottom of the wheel stationary.
EDIT for clarification: Sum of forces is doable as well. However, in your method you forgot that the motor of your robot is applying an external torque means its applying an external force on the wheel as well. So your sum of forces in your x direction should reflect that. So your sum of forces would be more like
\begin{equation} F_{ext} - F_d + F_{tr} = m a_x \end{equation}
Otherwise, your robot would just roll down the hill since there was nothing forcing it to proceed up the hill.
Best Answer
There are two factors I know of. First, and less important, overcoming static friction requires more force than kinetic friction. This applies to all of the internal parts that have to get going/moving past each other, and probably to the rolling friction of the tires.
The major reason, though, has to do with how cars based on internal combustion engines work. See, an internal combustion engine can only supply torque and power when it's already moving. That's why you need an electric starter motor to get the engine going when you start the car. Now think about if the engine were linked directly to the wheels by gears - that would mean if the car is stopped, the engine isn't running. To get over this problem cars have a clutch inside of them that transmits the torque from the engine to the gear box and drive shaft. When the clutch is fully engaged, all of the torque and power are transmitted. As it is in the process of engaging, though, only part of the power is transmitted. This is especially important when starting from rest because it is that partial engagement that allows the wheels to come up to a speed that the engine can supply torque at without stopping.
So, bottom line, the engine needs to be able to supply more torque at low rotation rates in order to get the car moving because not all of the power is being transmitted to the drive train by the clutch.
With an electric motor this is not a problem - they can supply 100% of their torque even at zero rotation.