Calculating the velocity decrease going up an incline

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SUMMARY

This discussion focuses on calculating the angular velocity of a car ascending a 45-degree incline while maintaining a speed of 60 mph. The key equation derived is (1/2)(mv^2) = mgR(1-cosx), which allows for determining the velocity at any angle x on the incline. It is established that at a steep incline, the fixed engine output may not suffice to overcome gravitational forces, leading to a decrease in acceleration as the incline steepens. The discussion emphasizes the importance of analyzing force vectors and energy conservation in this context.

PREREQUISITES
  • Understanding of basic physics concepts, particularly forces and energy conservation
  • Familiarity with vector decomposition in physics
  • Knowledge of power output measurements (e.g., horsepower, kilowatts)
  • Experience with mathematical equations related to motion and forces
NEXT STEPS
  • Study the principles of energy conservation in mechanics
  • Learn about vector decomposition and its applications in physics
  • Explore the effects of incline angles on vehicle dynamics
  • Review force diagrams related to inclined planes and their implications on motion
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Automotive engineers, physics students, and anyone interested in vehicle dynamics and the effects of inclines on performance will benefit from this discussion.

Niaboc67
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I know there are many factors that go into this. But in general how would one go about calculating the angular velocity for a car going up an incline. What I want to know in particular is if a car say is driving at a consistent speed of 60mph on an even plane and then hits an incline up wards at 45* going the speed the person is going at what rate is the car's acceleration decreased. I'd think it must be a gradient going slower and slower as the incline increases in steepness. If someone could explain this and then give a mathematical equation that would be greatly appreciated.

Thanks
 
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Assuming its a circular incline going steeper and becomes vertical
As the car covers angle x in the incline
Conserve energy(neglecting friction)
##(1/2)(mv^2) =mgR(1-cosx)##
From there you can find v(x)

Can you make the question more clearer
 
Well, if you discount the rolling resistance of the tires/drive train, at the fixed output power of the engine when traveling 60 MPH, the force vectors are all in the same horizontal plane: The force of air resistance is matched by the engine's output. For a given object, air resistance is proportional to velocity squared.

On the inclined plane, you must break the velocity vector into the horizontal and vertical components. (At 45 degrees , the horizontal and vertical speed components are equal, and both equal the velocity divided by the square root of 2.)

On an inclined plane, the fixed engine output must be sufficient to lift the car's mass against the force of gravity. Since overcoming the drag at 60 MPH would be on the order of 20 HP/15KW, it is entirely possible that this 'fixed' output would not be enough to lift the car at all, unless the speed dropped to near zero. (in real life, given most vehicles gear reduction range, it's probable that a car would not be able to go up this steep an incline at all.) At 45 degrees, the force vector due to gravity is split into the horizontal and vertical components by dividing mass by sqrt(2), also. And, the air resistance vector, which is parallel to the plane, would need to be included needs to be included in the solution as well.

I really think you should take a look at the force diagrams and explanations at the lesson on inclined planes here. Diagrams are helpful in understanding the forces involved.
 

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