Required Torque for Wheel at 8.17 m/s^2 Acceleration

In summary: So the torque is mr2/4 * linear acceleration. The 18 Nm refers to the angular acceleration of the wheel itself. You need to add this to the torque for the linear acceleration.In summary, The required torque for each wheel to achieve an acceleration of 8.17m/s^2 for a Tesla Model S, with a mass of 2100kg and a wheel radius of 0.25m, is 1072Nm. This accounts for both the linear and angular acceleration.
  • #1
Torunn
19
0

Homework Statement


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
radius wheel = 0,25m
mwheel = 18 kg
t= 3,4s
the wheel has a moment of inertia corresponding to a cylinder.

Homework 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

The Attempt at a Solution

 
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  • #2
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
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
Four wheel drive. There is a two wheel drive model S but that doesn't accelerate that fast.
 
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  • #5
Yes, but can I use this equation or will it be wrong? :)
 
  • #6
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
And how do I do that? :)
 
  • #8
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
If I understand you correct:
If i have the angular acceleration I can find the torque with F = ma? :)
 
  • #10
Torunn said:
If I understand you correct:
If i have the angular acceleration I can find the torque with F = ma? :)
Not the idea. The expression for linear motion is F = ma and the angular equivalent is ##\tau = I\alpha##.
 
  • #11
Ok, so I find alpha by finding omega with the velocity and time?
 
  • #12
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
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
ok, so I can not use:
(omegaf- omegai)/time = alpha ?
can I use at= r*alpha?
 
  • #15
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...)

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

Your 18.4 is just fine.
 
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  • #16
Why don't 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
but isn't omega = v/r ? where does the 2pi fit in?
 
  • #18
Dean Barry: but the question is for only one wheel, wouldn't that be wrong ? or just don't i understand it?
 
  • #19
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
ah, Ok, so then it is correct? thank u :) :)
 
  • #21
Meaning the total torque per wheel is 1072 + 18 Nm

Done.
 
  • #22
ok, thank u :)
 
  • #23
but what about the mass of the car ? wouldn't that have anything to say about the required T ?
 
  • #24
Starting with the wheels.
Imagine one wheel is a stationary solid disc / cylinder, accelerated from rest to the final rotation rate (v/r).
You have the final rate and time so you can derive the rotational acceleration rate.
With the rotational acceleration rate you can define the torque required to achieve this.
Multiply this by four.
That takes care of the rotational inertia, now deal with the linear inertia.
 
  • #25
vebjoern said:
but what about the mass of the car ? wouldn't that have anything to say about the required T ?
Hello vebjoern, welcome to PF.
The relevant equation in the original post, T=F*r was hopefully used to find the torque for each wheel that takes care of the linear acceleration of the car.
(answer I ecxelled 1072 Nm)

In post #6, I argued that an additional torque is needed for the angular acceleration of each wheel. Much smaller (18 Nm), but from the wording of the exercise clearly not to be ignored.
 
  • #26
i understand now! thanks
 
  • #27
Have you used T= I*alpha to get 1072Nm? I have and I get 536Nm, why is that not correct? @BvU
 
  • #28
Fjompen11 said:
Have you used T= I*alpha to get 1072Nm? I have and I get 536Nm, why is that not correct? @BvU
The 1072 relates to the linear acceleration of the car as a whole. There are two ways to obtain it.
You can work out the linear force, then use T=F x d (d being wheel radius) to get the torque (at some stage also dividing by 4). Or you can use T= I*alpha.
To use the second, you need to find alpha from the linear acceleration and detemine the appriate moment of inertia. For this, the car acts as a point mass at each wheel's centre (1/4 at each). So I is just mr2, not mr2/2.
 

1. What is required torque for a wheel at 8.17 m/s^2 acceleration?

The required torque for a wheel at 8.17 m/s^2 acceleration depends on several factors such as the weight and diameter of the wheel, as well as the coefficient of friction between the wheel and the surface it is moving on. It can be calculated using the formula: torque = (mass x acceleration x radius)/2.

2. How do you determine the weight and diameter of the wheel?

The weight and diameter of the wheel can be measured using a scale and a ruler, respectively. Alternatively, if these values are not readily available, they can be estimated based on the type and size of the wheel.

3. What is the coefficient of friction and how does it affect torque?

The coefficient of friction is a value that represents the amount of friction between two surfaces in contact. It is affected by factors such as surface texture, material, and lubrication. A higher coefficient of friction means more force is required to move the wheel, resulting in a higher required torque.

4. Can the required torque for a wheel at 8.17 m/s^2 acceleration be reduced?

Yes, the required torque can be reduced by reducing the weight of the wheel, increasing its diameter, or using a surface with a lower coefficient of friction. Additionally, using a more efficient wheel design or adding gears can also reduce the required torque.

5. How is the required torque for a wheel at 8.17 m/s^2 acceleration important in real-world applications?

The required torque for a wheel at 8.17 m/s^2 acceleration is important in various real-world applications, such as designing vehicles and machinery. It helps determine the amount of power needed to move a wheel at a certain speed and acceleration, and can also affect the efficiency and performance of the system. Accurately calculating and understanding the required torque is crucial for ensuring the safe and efficient operation of these systems.

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