Calculating the power developed by the motor

1. Aug 21, 2014

NewtonianAlch

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

2. Relevant equations
A few basic conversions/equations:

50km/h = 13.89m/s
ω = 13.89/0.3 = 46.3 rad/s

Power = Torque x Angular Velocity

3. The attempt at a solution

Solution is already given above in the picture, but I am confused by it. Is FL calculated correctly? I thought the angle of 30° should be at the vertex between F=mg and FL but it appears in the solutions they have taken it at the vertex of F = mg and Fv

2. Aug 21, 2014

Andrew Mason

I am having difficulty understanding the solution. Where did you get this question and the solution?

This car should be rolling down the incline by itself so I am not sure why the motor would be doing any work or exerting any torque.

Since we are dealing with a rolling wheel the relevant coefficient of friction is the static coefficient μs. But that gives the maximum static force (before sliding occurs), not the actual force. You just assume that the wheel rolls so the static friction force is sufficient to keep the tires from slipping. The torque on the wheels, therefore, is just FL x R = .3FL

AM

3. Aug 22, 2014

haruspex

Yes, there are definitely a few things wrong with this question.
A static friction coefficient of 0.3 would not be enough to stop it sliding down the hill. (Maybe that's why it's going downhill!) But then, it would be accelerating downhill, wheels slipping, and the power from the engine unknowable.
So let's suppose it means the coefficient of rolling resistance is 0.3, and the car is moving uphill. I think that makes the solution OK. (Note that rolling resistance is more a function of the tyres and axles than of the road surface: http://en.wikipedia.org/wiki/Rolling_resistance#Primary_cause.) If that's what was meant, the question setter needs to take a physics course.
No, the text is right. Look at the diagram. The angle between road and horizontal is 30, so that between horizontal and Normal force is 60, and that between Normal force and vertical is again 30.

4. Aug 22, 2014

NewtonianAlch

I am not sure where exactly the question came, it was part of our tutorial questions -- I would assume the lecturer made it up. It's not really a straight out "physics" course as such, it's about electric motors.

I don't think he's trying to say the coefficient of friction is enough to stop it rolling. I think rather that the car is driving downwards intentionally and the coefficient of friction between the road and tyres is 0.3. I'd think that at a 30 degree incline the motor would no longer be operational and the driver would be on the brakes instead, but that's a different story.

5. Aug 22, 2014

NewtonianAlch

Leaving aside some of the interesting technicalities of the question, would the working out be correct? I kind of did it differently and come to a slightly different answer and I just want to make sure the solution itself is correct.

6. Aug 22, 2014

dean barry

What if its generating power ?
Then the power being generated = 4,410 * 13.89 = 61,250 Watts

7. Aug 22, 2014

NewtonianAlch

But you still need to take into account the frictional force.

What I did was:

F = mg = 8829N
Fn (normal force) = cos 30 x 8829 = 7646N
Ff (forward force down slope) = sin 30 x 8829 = 4414.5N

Then the frictional force is: 0.3 x 4414.5N = 1324.35N

Fr (resultant force) = 4414.5N - 1324.35N = 3090.15N

Now the Power = Force x Velocity = 3090.15 x 13.89 = 42,922W

I do not understand why he's calculated the frictional force from the normal force, can someone explain?

Either way, my solution would mean the Torque = 3090.15N x 0.3m = 927.05Nm

or his solution Torque = 635.38Nm

A heck of a lot of torque for an electric motor in a car

8. Aug 22, 2014

NewtonianAlch

Nvm, I see what you mean. The power generated is independent of the friction on the road.

9. Aug 22, 2014

dean barry

Would the rolling resistance co-efficient be more relavent here ?

10. Aug 22, 2014

haruspex

No, let me try again.
If the coefficient of friction between tyre and road is only 0.3 then the car will slide, accelerating down the hill, rather than travel at a steady speed.
Friction between tyre and road does not slow a car down - it's what is required for a car to get traction. If the tyres are rolling without slipping then there's no work done against friction; there's no relative motion of the surfaces in contact, so there's no movement in direction of the force.

For both of these reasons, I suggest that the author means rolling resistance, not friction.
The work that is done (on a level road) is against rolling resistance. As mentioned, this is mainly the work done in changing the shape of the tyre as it rolls along.

If the car is travelling downhill at a steady speed, despite some work being done by the engine, then the rolling resistance must be so great that without the engine power it would not move. But this is not the case here: a slope of 30 degrees would be enough to overcome a rolling resistance coefficient of 0.5, let alone 0.3.

So I cannot make any sense of the question if the car is supposed to be going downhill. Let's try the uphill hypothesis.
On that presumption, there are two possibilities for the "0.3 coefficient of friction".
a) It should be rolling resistance. In this case, the work done by the engine is in overcoming rolling resistance and ascending the hill.
b) The coefficient should have been stated as rather more, enough to ensure rolling without slipping. In this case, the work done by the engine is only in ascending the hill.