# Torque on a wheel

1. Nov 4, 2014

### Torunn

1. The problem statement, all variables and given/known data
b) What is the required torque that the engine has to provide for each wheel,
to have this acceleration?
This is the question and this is the information given.
vi = 0
vf = 100km/h
mcar = 2100 kg
mwheel = 18 kg
t= 3,4s

the wheel has a moment of inertia corresponding to a cylinder.
2. Relevant equations
Can i use the equation Ti=Fi*ri to find the torque for each wheel? the acceleration of the car is 8,17m/s^2

3. The attempt at a solution

2. Nov 4, 2014

### BvU

I don't see an acceleration :) Oh, wait: under 2) 8.17 m/s2...

Is there a part a) you need to tell us a little about, so we can understand what's going on before we try to help you ? Makes a difference if this is a motorcycle, a cheap car or a four-wheel drive !

Oh, and there are hollow cylinders and there are solid cyclinders. Different moment of inertia. Your choice ?

3. Nov 4, 2014

### Torunn

A Tesla of mass 2100 kg accelerates from rest to 100 km/h in 3.40 seconds,

and we will assume that the acceleration is constant. In the following, ignore

resistance from the air and the rolling friction of the wheels against the

surface. The wheels are rolling without sliding (no wheel-spinning here!),

and they have a diameter of 50.0 cm. A wheel weighs 18.0 kg, and has a

moment of inertia corresponding to a cylinder. The weight of the wheels is

included in the total 2100 kg. A car has 4 wheels (surprise!), and they are

taken to be the same and carry the same amount of weight.

This is all the text that comes with it :)

4. Nov 4, 2014

### BvU

Four wheel drive. There is a two wheel drive model S but that doesn't accelerate that fast.

Last edited: Nov 4, 2014
5. Nov 4, 2014

### Torunn

Yes, but can I use this equation or will it be wrong? :)

6. Nov 4, 2014

### BvU

Gives you the torque required to propel the vehicle.
Engines have to do a little bit more: they must also increase the angular speed of the wheels....

7. Nov 4, 2014

### Torunn

And how do I do that? :)

8. Nov 4, 2014

### BvU

By making good use of the relevant equations for uniformly accelerated angular motion that you collected under number 2 of the template (didn't you ? :) )

Hint: the angular euivalent of F = ma

9. Nov 4, 2014

### Torunn

If I understand you correct:
If i have the angular acceleration I can find the torque with F = ma? :)

10. Nov 4, 2014

### BvU

Not the idea. The expression for linear motion is F = ma and the angular equivalent is $\tau = I\alpha$.

11. Nov 4, 2014

### Torunn

Ok, so I find alpha by finding omega with the velocity and time?

12. Nov 4, 2014

### Torunn

T= 0.5MR^2*alpha
T = 0.5*18*0.25^2*32.7= 18.4Nm
Is this correct you think? Should I use the mass of the car or the wheel?

13. Nov 4, 2014

### BvU

expression correct, value of alpha isn't.
Wheels rotate, their inertial moment is needed; follows from their mass and assuming solid cylinder shape rotating around wheel axis.

14. Nov 4, 2014

### Torunn

ok, so I can not use:
(omegaf- omegai)/time = alpha ?
can I use at= r*alpha?

15. Nov 4, 2014

### BvU

You can use both. It's just that you forgot a $2\pi$ at some point ! :)
at = r $\alpha$ is a dimension mismatch (your first, sorry...)

It's getting worse: I was the one who inserted a $2\pi$ too many.

Last edited: Nov 4, 2014
16. Nov 4, 2014

### dean barry

Why dont you find the torque required to rotationally accelerate one wheel from rest up to its final rotation rate, multiply by four, then add to the torque required to accelerate the whole car mass at the rate mentioned ?

17. Nov 4, 2014

### Torunn

but isnt omega = v/r ? where does the 2pi fit in?

18. Nov 4, 2014

### Torunn

Dean Barry: but the question is for only one wheel, wouldnt that be wrong ? or just dont i understand it?

19. Nov 4, 2014

### BvU

Dean's plan is what we are carrying out, except indeed one forth of the car mass.
Sorry about my 2pi mishap. Lack of sleep.:(

20. Nov 4, 2014

### Torunn

ah, Ok, so then it is correct? thank u :) :)