Acceleration of Car: How to Find the Equation for Acceleration | Help Needed

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To find the acceleration of a car considering its mass and the moment of inertia of its tires, the relationship between torque, friction, and acceleration must be established. The torque supplied by the engine (τ) is related to angular acceleration (α) and moment of inertia (I) through the equation τ = Iα. The rolling constraint connects linear acceleration (a) to angular acceleration by a = rα. By incorporating the friction force (F = ma), the equation can be rearranged to τ - Fr = Iα, leading to the formula a = τ/(m + I/r²). Understanding this relationship helps clarify how the moment of inertia affects the car's acceleration.
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Homework Statement


I have a car with a mass (m) and tires having moment of inertia (I). I need to find an equation for the acceleration of the car from what is given. thanks in advance for the help.


Homework Equations


Not sure what all is relevant here but, M=I\alpha F=ma


The Attempt at a Solution


I really have no idea how to even begin this. My attempt got me a=I\alpha/m and I am pretty sure this isn't correct
 
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welcome to pf!

hi flasheddread! welcome to pf! :smile:

(have an alpha: α :wink:)

the only external force on the car is the friction, F

(so the external torque is Fr)

you'll also need the rolling constraint, a = rα :wink:

(you'd better assume it's four-wheel drive … two-wheel drive gets rather complicated)​
 
:) thanks

I get that friction is the only external force here. I guess I'm just having trouble understanding how to get the acceleration for the car using the moment of inertia.

I start off by saying the Fr=Iα, which is the torque created by the tires that will move the car. Then i need to relate that to the acceleration of the car so i say Fr=ma?? Now this is where I think I'm messing up. I say that Iα=ma and rearranging that, I can say that a=Iα/m.

I know this is completely wrong, since having a bigger moment of inertia for the tires would mean the car would accelerate slower, and this equation proves the opposite. Or would this be correct because the acceleration is on determined by how much force is placed on the tires to rotate? Maybe I'm just thinking about this too hard, because it doesn't seem that difficult.

Any help please??
 
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Alrighty I think I might have it now!

The engine supplies a torque to the tires, which in turn gives them an angular acceleration.
τ=Iα
α=a/r

so, a= τr/I

This makes much more sense. I have no idea what I was thinking earlier. I'm pretty sure this is correct this time, :smile: and if not I might just drop out of college. :cry:
 
wait a mo! :rolleyes:

you haven't used either the mass of the car, or the friction force F

don't forget that there are two torques on the wheel (why do you keep saying "tires"? :redface:), the known applied torque τ from the engine and the unknown torque from the friction from the road (which you'll have to eliminate) :wink:
 
Oh wow, yea that's just what kind of day I was having yesterday. :frown: Sorry for being so difficult and thanks for sticking around for me.

Alright so about this friction force. Would we eliminate it by saying the friction force is equal to the mass times acceleration of the car? If so and I am doing this correctly we get.

τ-Fr=Iα, which then goes to τ-mar=Ia/r

And from here its just manipulating it to solve for a.
 
hi flasheddread! :wink:
flasheddread said:
Alright so about this friction force. Would we eliminate it by saying the friction force is equal to the mass times acceleration of the car?

yup, the https://www.physicsforums.com/library.php?do=view_item&itemid=39" on both axles), so F = ma :smile:
If so and I am doing this correctly we get.

τ-Fr=Iα, which then goes to τ-mar=Ia/r

And from here its just manipulating it to solve for a.

exactly :smile:

btw, note that if you fivide by r, you get τ = (m + I/r2)a,

so effectively we can ignore the friction if we pretend that the wheels have an extra "rolling mass" of I/r2 :wink:
 
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Wouldn't torque be divide by r as well?
 
  • #10
oops!

oops! :redface:

yes … τ/r = (m + I/r2)a :smile:
 
  • #11
Sweet, thanks so much for your help! I will keep rolling mass in memory; it sounds pretty useful. :smile:
 

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