1. The problem statement, all variables and given/known data
The maximum torque output from the engine of a new experimental car of mass m is τ . The
maximum rotational speed of the engine is ω. The engine is designed to provide a constant power
output P. The engine is connected to the wheels via a perfect transmission that can smoothly
trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of
static friction between the wheels and the road is µ.
What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no
frictional losses and assume µ is large enough so that the tires do not slip.
2. Relevant equations
v_cm = rω, P = τω
3. The attempt at a solution
I know that P = τω = τv_cm /R. However, I do not quite understand how to find the torque output of the engine. The answer is v = 2P/mg. After rearranging the answer, I found that P = mg sin θ R * v/R so the torque output is mg sin θ R, the torque due to gravity. Under this assumption, the instantaneous axis of rotation must be at the point of contact. How would I know that from the problem itself? Is it something to do with how the engine works?
The maximum torque output from the engine of a new experimental car of mass m is τ . The
maximum rotational speed of the engine is ω. The engine is designed to provide a constant power
output P. The engine is connected to the wheels via a perfect transmission that can smoothly
trade torque for speed with no power loss. The wheels have a radius R, and the coefficient of
static friction between the wheels and the road is µ.
What is the maximum sustained speed v the car can drive up a 30 degree incline? Assume no
frictional losses and assume µ is large enough so that the tires do not slip.
2. Relevant equations
v_cm = rω, P = τω
3. The attempt at a solution
I know that P = τω = τv_cm /R. However, I do not quite understand how to find the torque output of the engine. The answer is v = 2P/mg. After rearranging the answer, I found that P = mg sin θ R * v/R so the torque output is mg sin θ R, the torque due to gravity. Under this assumption, the instantaneous axis of rotation must be at the point of contact. How would I know that from the problem itself? Is it something to do with how the engine works?
0 commentaires:
Enregistrer un commentaire