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.