The discussion focuses on calculating the acceleration of a car considering the moment of inertia of its wheels. The car has an engine torque of τ = 200 Nm applied to the front wheels, each weighing m = 15 kg with a radius R = 0.35 m, while the rest of the car has a mass of M = 1000 kg. The final formula derived for acceleration is a = 2τ / R(M + 6m), which incorporates both the rotational and translational dynamics of the system. Participants emphasize the importance of using correct free body diagrams and torque equations to arrive at the solution.
PREREQUISITESStudents in physics or engineering disciplines, automotive engineers, and anyone interested in the dynamics of vehicle motion and the interplay between rotational and translational forces.
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.