# Homework Help: Accelerating a car including the moment of inertia of the wheels

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1. Nov 5, 2017

### TSny

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.

2. Nov 5, 2017

### lichenguy

Could it be FrearR = Iα?
Is it the same at the front?

3. Nov 5, 2017

### TSny

Yes. Can you identify the type of force represented by Frear?
No, the front wheel also has the torque of the engine that you need to deal with.

4. Nov 5, 2017

### lichenguy

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

Is it:
τ - FfrontR = Iα?

5. Nov 5, 2017

Yes

Yes.

6. Nov 5, 2017

### lichenguy

Cool!
Thank you for all the help.

7. Nov 5, 2017

### TSny

Sure. Good work!

8. Dec 3, 2017

### lichenguy

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?

9. Dec 3, 2017

### TSny

These equations look correct.

Check your work. The equations should yield $a = \frac {2τ} {R(M+6m)}$.

10. Dec 4, 2017

### lichenguy

I made a mistake, all good now. =]

11. Jul 7, 2018

### NewtonianAlch

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?

Last edited: Jul 7, 2018
12. Jul 7, 2018

### TSny

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

13. Jul 7, 2018

### NewtonianAlch

Thank you.

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

14. Jul 7, 2018

### TSny

OK. Glad it makes sense now.