# 2014 F = MA #15 Torque Output Engine

1. Jul 29, 2014

### Jzhang27143

1. The problem statement, all variables and given/known data

The maximum torque output from the engine of a new experimental car of mass m is τ . The
maximum rotational speed of the engine is ω. The engine is designed to provide a constant power
output P. The engine is connected to the wheels via a perfect transmission that can smoothly
trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of
static friction between the wheels and the road is µ.

What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no
frictional losses and assume µ is large enough so that the tires do not slip.

2. Relevant equations

v_cm = rω, P = τω

3. The attempt at a solution

I know that P = τω = τv_cm /R. However, I do not quite understand how to find the torque output of the engine. The answer is v = 2P/mg. After rearranging the answer, I found that P = mg sin θ R * v/R so the torque output is mg sin θ R, the torque due to gravity. Under this assumption, the instantaneous axis of rotation must be at the point of contact. How would I know that from the problem itself? Is it something to do with how the engine works?

2. Jul 29, 2014

### Nathanael

This is true of all rolling motion (with no sliding)

It makes more sense when you imagine the velocity vectors of every point on the wheel

Edit:
The image I've attatched shows it for 3 points on the wheel (sorry it's the best I could find)

It should be explained in most (introductory) physics books in the "rolling/rotation" chapter(s)

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• ###### Rolling.jpg
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Last edited: Jul 29, 2014
3. Jul 30, 2014

### Jzhang27143

Ok, that part makes sense to me now. However, how would I know to consider the axis of rotation to be at the point of contact instead of the center of mass in this question? I think that these two setups would lead to two different answers because rotation about the center of mass involves torque due to friction.

4. Aug 1, 2014

### dean barry

Calculating top speed

If you can calculate the engine power, great, because then the power at the drive wheels will be the same (as you have no losses), then you can involve the equation :

power (watts) = force overcome * velocity

Note : just a thought, the peak torque and peak power dont usually happen at the the same rpm

Ive added a picture, its not directly exact, but it shows the force curve derived from a constant power situation (as yours is), but the resistance forces are air drag and rolling resistance, whereas yours is a gravity based force.

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5. Aug 1, 2014

### dean barry

The second thumbnail is a 5 speed engine showing the drive forces through the gears, note the engine is revved beyond the torque peak in top gear (to peak power rpm), this coincides with top speed.
Its an anomoly worth noting, though not important in this case.
Dean

6. Aug 1, 2014

### Jzhang27143

Wait so this is asking for the maximum sustained speed. Therefore, the net force must be 0 so f_static + Mg sin theta = F_engine. The correct answer for the max speed requires Mg sin theta = F_engine so f_static = 0. Why can we let the value of static friction go to 0? The problem states that mu is large enough so that the wheels dont slip.