Does Increasing Ramp Angle Double Ball Speed?

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Homework Help Overview

The discussion revolves around the physics of a ball rolling down a frictionless incline and the relationship between the angle of the ramp and the speed of the ball. The original poster questions whether doubling the angle of the incline results in doubling the speed of the ball.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the angle of the incline and the speed of the ball, questioning the correctness of the original poster's mathematical expression. There are suggestions to consider the acceleration along the incline and the dependence of speed on height.

Discussion Status

The discussion is active, with participants questioning the clarity and correctness of the original poster's equations. Some guidance is offered regarding the acceleration of the ball and the factors influencing speed, but no consensus has been reached on the simplest way to demonstrate the relationship between angle and speed.

Contextual Notes

There is uncertainty regarding whether the ball is rolling or sliding, and the implications of this distinction on the analysis are being considered. The original poster's approach is noted as complex, and participants are seeking simpler explanations.

opticaltempest
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A ball is rolled down a frictionless incline at some angle [tex]\theta[/tex] below the horizontal. If you increase the angle of the incline by a factor of two (make the ramp steeper downward by twice as much), does the ball roll down at double the speed?

Here is what I said:

No. because of the following relation

[tex]\[<br /> \begin{array}{l}<br /> Speed = \sqrt {\left( {v_i \cos \left( \theta \right)t} \right)^2 + \left( {h + v_i \sin \left( \theta \right)t + \frac{1}{2}gt^2 } \right)^2 } \\ <br /> 2Speed \ne \sqrt {\left( {v_i \cos \left( {2\theta } \right)t} \right)^2 + \left( {h + v_i \sin \left( {2\theta } \right)t + \frac{1}{2}gt^2 } \right)^2 } \\ <br /> \end{array}<br /> \][/tex]

Where [tex]v_i[/tex] is the initial velocity of the ball, [tex]\theta[/tex] is the angle of the ramp below the horizontal, [tex]h[/tex] is the initial height of the ball, [tex]g[/tex] is the acceleration due to gravity, and [tex]t[/tex] time.because

[tex] \[<br /> \begin{array}{l}<br /> \cos (\theta ) \ne \cos (2\theta ) \\ <br /> \sin (\theta ) \ne \sin (2\theta ) \\ <br /> 0 < \theta < \frac{\pi }{2} \\ <br /> \end{array}<br /> \][/tex]Is this a correct way to show that increasing the downward angle by a factor of two does not double the speed of the object rolling down the incline?
 
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Why not just write an expression for the acceleration along the incline as a function of angle? (Are you concerned with something rolling, or did you just mean sliding without friciton?)

I assume you mean speed as a function of time as it goes down the incline. The speed at the bottom will only depend on the height.
 
It is speed as a function of time, and I guess the object could be rolling or sliding with no resistance.

Even though my answer isn't the simpliest, is it still a correct way to show that increasing the downward angle by a factor of two does not double the speed?

What is the simplest way to show that the increasing the downward slope of the ramp by a factor of two does not double the speed?
 
opticaltempest said:
Even though my answer isn't the simpliest, is it still a correct way to show that increasing the downward angle by a factor of two does not double the speed?
I don't understand your equation, since you have speed on one side but distance on the other.

What is the simplest way to show that the increasing the downward slope of the ramp by a factor of two does not double the speed?
Imagine a block sliding down a frictionless slope making an angle [itex]\theta[/itex] with the horizontal. What's its acceleration down the incline?
 

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