The discussion revolves around a physics problem involving the acceleration of a car, taking into account the moment of inertia of its wheels. The scenario describes a car accelerating from rest on a horizontal surface, with specific parameters for the wheels and the car's mass. Participants are exploring how to combine the dynamics of the car's body and its wheels to determine the overall acceleration.
The discussion is ongoing, with participants providing feedback on each other's diagrams and reasoning. Some guidance has been offered regarding the setup of free body diagrams and the application of torque equations. There is a recognition of differing interpretations regarding the factors involved in the equations, but no consensus has been reached on a single approach.
Participants are working within the constraints of the problem statement, which specifies the mass of the car excluding the wheels. There are discussions about the assumptions made regarding forces and torques, as well as the need for clarity in the diagrams representing the system.
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?
Each wheel is treated as a uniform, solid cylinder for which I = (1/2)mR2. Also, for rolling without slipping, ω = v/R.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?
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.
OK. Glad it makes sense now.NewtonianAlch said:Thank you.
I was using I = mR2, hence why it didn't come out the same.