Accelerating a car including the moment of inertia of the wheels

[NewtonianAlch]

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(x_f-x_i)=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?

 

[TSny]

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]

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.

 

[TSny]

Thank you.

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

OK. Glad it makes sense now.