# Accelerating a car including the moment of inertia of the wheels

• lichenguy
In summary: Otherwise, OK.They look pretty good now. For the front wheel, you should indicate that there is a torque applied by the engine in the vertical direction. Otherwise, it's unclear what the arrow with the T is supposed to represent.
lichenguy

## Homework Statement

A car accelerates from rest on a horizontal surface. The engine provides a
torque of τ = 200 Nm on each of the two front wheels. Each of the four
wheels on the car weigh m = 15 kg, have radius R = 0.35 m and can be
considered solid, uniform discs. The rest of the car (not the wheels) has a mass
of M = 1000 kg.

What is the acceleration of the car?

I = 1/2mR2
τ = F*R
τ = I*α
a = α*R

## The Attempt at a Solution

I can find the frictional force at one of the wheels:
Ff = τ/R

I can find the α of a wheel:
α = Ff*R/I
That means a = Ff*R2/I

Not much, but I'm not very good at this, really need help.
I don't know how to combine the body as well as the wheels. Anyone has some tips please? :D

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Welcome to PF!

You will need to decide how many torques act on an individual front wheel and how many act on an individual rear wheel. In the formula τ = I*α, τ is the net torque acting on a wheel.

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lichenguy
TSny said:
Welcome to PF!

You will need to decide how many torques act on an individual wheel. In the formula formula τ = I*α, τ is the net torque acting on a wheel.
Thanks for the welcome :)

I'm thinking that the two back wheels get their torque from the ground while the two front wheels get the torque from the engine.
Can i do 2*Ff/4 to find the torque at every wheel?

lichenguy said:
Thanks for the welcome :)

I'm thinking that the two back wheels get their torque from the ground while the two front wheels get the torque from the engine.
The front wheels interact with the ground, too. Draw a free body diagram for each wheel.
Can i do 2*Ff/4 to find the torque at every wheel?
No. Your free body diagrams should help.

TSny said:
The front wheels interact with the ground, too. Draw a free body diagram for each wheel.
No. Your free body diagrams should help.
I made one in paint only, for now.
If the frame is accelerating, then we have a fictitious force. If the frame is inertial, then we only have that one force from the front wheels that makes the car accelerate. The other forces sum up to 0, it's the gravitational force and normal forces.

In the accelerating frame, we should have zero net force so Ff-Ffictitious=0

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There is no need to go to the accelerating frame of the car. You can stick with the inertial frame of the earth.

Instead of drawing a force diagram for the entire car, draw a force diagram for one of the front wheels and draw a separate force diagram for one of the rear wheels.

------------------------------

An alternate approach is to use energy concepts for linear and rotational motion. This approach will yield the answer more quickly, but I don't know if you've studied these concepts yet.

lichenguy
I made pictures with torque and friction, do they look familiar?

I know 1/2*I*ω2 and 1/2*m*v2 for kinetic energy.

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Last edited:
The diagram for the front wheel needs a little clarification. There would be forces on the wheel in the vertical direction that you did not show. The friction force appears to be in the correct direction if you are thinking of the car as moving to the right.

What does the arrow labeled by T represent? Arrows on a free body diagram indicate forces. Is T a force?

The diagram for the rear wheel also has the correct direction for the friction. But, again, I'm not sure what the arrow with the T represents. There are also some vertical forces acting on this wheel.

Does the friction force on one of the front wheels necessarily equal the friction force on one of the real wheels? If not, then you should use different symbols for these forces.

lichenguy
lichenguy said:
I know 1/2*m*ω2 and 1/2*m*v2 for kinetic energy.
The first equation is not correct. The "m" should be replaced by what?

Otherwise, OK.

Do you know a formula for the work done by a torque?

lichenguy
TSny said:
The first equation is not correct. The "m" should be replaced by what?

Otherwise, OK.

Do you know a formula for the work done by a torque?
Oh, sorry, there should be an I instead of m. I edited the original post.

I fixed the wheel diagrams to be more clear now, right is positive movement.

I know power is τ*ω so work would be τ*Δθ?

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lichenguy said:
Oh, sorry, there should be an I instead of m.
...
I know power is τ*ω so work would be τ*Δθ?
OK, great. All you need now is the fundamental theorem that relates work to energy.

lichenguy said:
I fixed the wheel diagrams to be more clear now, right is positive movement.
They look pretty good now. For the front wheel, you should indicate that there is a torque applied by the engine. This could be a curved arrow labeled by τ where the arrow is clockwise or counterclockwise (you decide).

But, if you get the energy equation set up correctly, then you will not actually need these diagrams.

lichenguy
TSny said:
They look pretty good now. For the front wheel, you should indicate that there is a torque applied by the engine. This could be a curved arrow labeled by τ where the arrow is clockwise or counterclockwise (you decide).

But, if you get the energy equation set up correctly, then you will not actually need these diagrams.
I have my last pictures ready now. Thank you for the help :)

So, the theorem says ΔK=W
It's getting late, so i will try getting the energy equation tomorrow.

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lichenguy said:
I have my last pictures ready now. Thank you for the help :)

So, the theorem says ΔK=W
It's getting late, so i will try getting the energy equation tomorrow.
Ok,sounds good.

Your diagrams look correct. But I recommend using different notations for the friction forces on the front and rear wheels.

How would you solve this problem using forces and torques?

TSny said:
Ok,sounds good.

Your diagrams look correct. But I recommend using different notations for the friction forces on the front and rear wheels.
Thanks for the help, appreciated!
I ended up with: $$a = \frac {2τ} {R(M+6m)}$$

Would it be possible to use the impulse-momentum theorem?
I'm also interested in how to do it using forces, like the above guy.

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lichenguy said:
I ended up with: $$a = \frac {2τ} {R(M+6m)}$$
That's similar to what I get, but I do not agree with the factor of 6 in the denominator. Can you show the steps in your work?

Would it be possible to use the impulse-momentum theorem?
I don't see any advantage in using this theorem. It is essentially equivalent to F = ma.
I'm also interested in how to do it using forces, like the above guy.
From your diagrams, set up ∑τ = Iα for the front and rear wheels. Also, set up F = ma for the car as a whole.

lichenguy
TSny said:
That's similar to what I get, but I do not agree with the factor of 6 in the denominator. Can you show the steps in your work?

I don't see any advantage in using this theorem. It is essentially equivalent to F = ma.

From your diagrams, set up ∑τ = Iα for the front and rear wheels. Also, set up F = ma for the car as a whole.
I started with:
2τΔθ = 1/2Mv2 + 4*1/2mv2 + 4*1/2Iω2
then
v2 = 2τΔθ/(1/2)M+3m

Next i have to multiply the denominator by 2(xf-xi)=2RΔθ, that's what gives me the factor of 6. Did you get the factor of 3 earlier?

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OK. When I solved the problem I used M for the total mass of the car including the 4 wheels. But the problem clearly states that M does not include the 4 wheels . As a consequence, I had a factor of 2 where you have a factor of 6.

Your answer is correct with the factor of 6. Sorry for the mistake.

TSny said:
OK. When I solved the problem I used M for the total mass of the car including the 4 wheels. But the problem clearly states that M does not include the 4 wheels . As a consequence, I had a factor of 2 where you have a factor of 6.

Your answer is correct with the factor of 6. Sorry for the mistake.
Alright! Thanks.

About using torques and forces method. I thought this:
2Ffront - 2Frear = a(M + 4m)
But what would be the torques?

lichenguy said:
Alright! Thanks.

About using torques and forces method. I thought this:
2Ffront - 2Frear = a(M + 4m)
But what would be the torques?
The torque equations can be used to get expressions for Ffront and Frear in terms of m and α. Use your diagrams for the wheels to help set up these equations.

lichenguy
TSny said:
The torque equations can be used to get expressions for Ffront and Frear in terms of m and α. Use your diagrams for the wheels to help set up these equations.
Could it be FrearR = Iα?
Is it the same at the front?

lichenguy said:
Could it be FrearR = Iα?
Yes. Can you identify the type of force represented by Frear?
Is it the same at the front?
No, the front wheel also has the torque of the engine that you need to deal with.

It's a contact force. Frictional force, i guess.

Is it:
τ - FfrontR = Iα?

lichenguy said:
It's a contact force. Frictional force, i guess.
Yes

Is it:
τ - FfrontR = Iα?
Yes.

TSny said:
Yes

Yes.
Cool!
Thank you for all the help.

lichenguy said:
Thank you for all the help.
Sure. Good work!

Umm, guys, i just did this using:
2Ffront - 2Frear = a(M + 4m),
τ - FfrontR = Iα,
FrearR = Iα,
but it gave me ##a = \frac {2τ} {R(M+2m)}## instead of ##a = \frac {2τ} {R(M+6m)}##

Is something missing?

lichenguy said:
Umm, guys, i just did this using:
2Ffront - 2Frear = a(M + 4m),
τ - FfrontR = Iα,
FrearR = Iα,
These equations look correct.

but it gave me ##a = \frac {2τ} {R(M+2m)}## instead of ##a = \frac {2τ} {R(M+6m)}##
Check your work. The equations should yield ##a = \frac {2τ} {R(M+6m)}##.

TSny said:
These equations look correct.

Check your work. The equations should yield ##a = \frac {2τ} {R(M+6m)}##.
I made a mistake, all good now. =]

lichenguy said:
I started with:
2τΔθ = 1/2Mv2 + 4*1/2mv2 + 4*1/2Iω2
then
v2 = 2τΔθ/(1/2)M+3m

Next i have to multiply the denominator by 2(xf-xi)=2RΔθ, that's what gives me the factor of 6. Did you get the factor of 3 earlier?

Can the OP or someone else explain how that equation was simplified to get a factor of 3 in the denominator for "m"?
The original equation involves a term with rotational kinetic energy (1/2Iω2).

Was that "I" for the object's inertia converted to some kind of equivalent mass?

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NewtonianAlch said:
Can the OP or someone else explain how that equation was simplified to get a factor of 3 in the denominator for "m"?
The original equation involves a term with rotational kinetic energy (1/2Iω2).

Was that "I" for the object's inertia converted to some kind of equivalent mass?
Each wheel is treated as a uniform, solid cylinder for which I = (1/2)mR2. Also, for rolling without slipping, ω = v/R.

NewtonianAlch
TSny said:
Each wheel is treated as a uniform, solid cylinder for which I = (1/2)mR2. Also, for rolling without slipping, ω = v/R.

Thank you.

I was using I = mR2, hence why it didn't come out the same.

NewtonianAlch said:
Thank you.

I was using I = mR2, hence why it didn't come out the same.
OK. Glad it makes sense now.

## 1. What is the moment of inertia of a car's wheels?

The moment of inertia of a car's wheels refers to the resistance of the wheels to changes in their rotational motion. It is a measure of how the mass of the wheels is distributed relative to their axis of rotation. This value is important in understanding the overall dynamics of a car's movement and acceleration.

## 2. How does the moment of inertia affect the acceleration of a car?

The moment of inertia of a car's wheels plays a significant role in its acceleration. A higher moment of inertia means that more force is required to accelerate the car, as the wheels have a greater resistance to rotational motion. This is why cars with lighter wheels tend to have better acceleration.

## 3. Can the moment of inertia be changed in a car?

Yes, the moment of inertia of a car can be changed by altering the distribution of mass in the wheels. This can be done by using lighter materials for the wheels or by changing the design of the wheels to distribute the mass more evenly. However, any changes made to the moment of inertia will also affect the overall handling and stability of the car.

## 4. How does the moment of inertia affect a car's turning ability?

The moment of inertia also plays a role in a car's turning ability. A higher moment of inertia means that more force is required to turn the wheels, making the car feel less responsive and slower to change direction. This is why sports cars often have lower moments of inertia to improve their handling and agility.

## 5. How do you calculate the moment of inertia of a car's wheels?

The moment of inertia of a car's wheels can be calculated using the formula I = mr², where I is the moment of inertia, m is the mass of the wheel, and r is the radius of the wheel. This value can also be affected by the shape and distribution of the wheel's mass, so it is important to consider the design of the wheels as well when calculating the moment of inertia.

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