Torque Plus Power In Relation to Velocity

Click For Summary

Homework Help Overview

The discussion revolves around a physics problem involving torque, power, and velocity in the context of a car driving up an incline. The problem presents a scenario where the maximum torque and power output of an engine are given, along with the car's mass and incline angle, and asks for the maximum sustained speed under certain assumptions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between torque, power, and velocity, with some attempting to derive relevant equations. Questions arise regarding the definition of force in this context and the role of the incline angle in the equations.

Discussion Status

The discussion is ongoing, with participants clarifying concepts related to force and torque. Some have suggested that the force in question is related to the engine's torque output, while others are seeking to understand how to apply the equations correctly in the context of the incline.

Contextual Notes

Participants are navigating assumptions about friction and the ideal conditions of the problem, including the implications of the incline angle on velocity and force components.

postfan
Messages
259
Reaction score
0

Homework Statement



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.

(A) v = 2P/(mg)
(B) v = 2P/(√3mg)
(C) v = 2P/(µmg)
(D) v = τω/(mg)
(E) v = τω/(µmg)

Homework Equations

The Attempt at a Solution


I don't even know how to start on this. I'm supposing that the first step is finding a relationship between the torque, power, and velocity , but I don't know how to do that. Thoughts?
 
Physics news on Phys.org
postfan said:
relationship between the torque, power, and velocity
Can you write a relationship for two of the three variables you've listed?
 
P=F*v*cos(theta).
 
And "F" is what?
 
Force
 
postfan said:
Force
Well, yes, I think we all guessed that, but which force?
 
The force is the one that the car provides.
 
postfan said:
The force is the one that the car provides.
The car's engine provides a torque, not a force.
I could guess you mean the propulsive force up the hill provided by friction, but then I don't know where the cos(theta) comes from.
 
I looked up the torque-velocity relation ,was I supposed to derive it somehow?
 
  • #10
postfan said:
I looked up the torque-velocity relation ,was I supposed to derive it somehow?
If you mean P=Fv, no. Just explain exactly what F is in the present context and how cos theta comes into it.
 
  • #11
F is the force produced from the power of the engine and the cos theta comes from the angle of the incline, the bigger the angle the less velocity.
 
  • #12
postfan said:
F is the force produced from the power of the engine and the cos theta comes from the angle of the incline, the bigger the angle the less velocity.
You are still not explaining what you mean by "the force from the engine". Could you point to it on a diagram? As I wrote, the engine produces a torque, not a force.
When you apply a standard equation like P=Fv, you need to understand what relationship those entities must have for the equation to be applicable. In this case, F is a force applied to and driving the motion of an object, and v is the velocity of the object in the direction of that force. (And both need to be constant.)
Assuming you mean the v as given in the question, that's the velocity up the plane. If your F, when you have defined it, is acting up the plane also then it's going to be P=Fv, no role for theta.
 
  • #13
Ok so F is the component of the weight that is parallel to the incline, is that right?
 
  • #14
postfan said:
Ok so F is the component of the weight that is parallel to the incline, is that right?
Yes!
Small correction -it's equal and opposite to the component of the weight parallel to the plane.
 
  • #15
Ok, so now what?
 
  • #16
postfan said:
Ok, so now what?
So what is the component of the weight parallel to the incline, in terms of the given data?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
8
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
19
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 97 ·
4
Replies
97
Views
6K