How Do Incline Angles Affect the Speed of a Rolling Ball?

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Homework Help Overview

The problem involves analyzing the speed of a ball rolling down three inclined planes with angles of 30, 45, and 60 degrees, while maintaining a constant height. The focus is on understanding the relationship between incline angles and the resulting speed of the ball.

Discussion Character

  • Exploratory, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the conversion of potential energy to kinetic energy and question how the angles affect the speed at the bottom of the incline. Some express uncertainty about resolving the problem without specific equations.

Discussion Status

Some participants have offered insights suggesting that the speed at the bottom of the incline will be the same for all angles, assuming no energy loss. Others have confirmed the reasoning presented, but the discussion remains open to further exploration of the implications of incline angles.

Contextual Notes

Participants are working under the assumption that height remains constant and are considering the effects of different angles on the speed of the ball without additional factors such as friction or air resistance.

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[RESOLVED] Speed in relation to angles

Homework Statement


You leave a ball roll down 3 different inclined plans with the same height. They each have 30, 45 & 60 degrees of incline respectively. Compare the sped of each of these plans

Height = constant, doesn't change
Angles of plans = 30, 45, 60

Homework Equations


None that I know of.

The Attempt at a Solution


I honestly do not see a way how to resolve this problem, if anyone can just shed any light, I know we're working in energy at the moment but I cannot find a way to get information with just an angle.
 
Last edited:
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That would be correct. Potential energy at the top will be kinetic energy at the bottom.

They will take different times to reach the bottom ... but at the bottom they will be going at the same speed. (Assuming there is no slipping at the steeper angle or bouncing on impact at the bottom etc.)
 
I _think_ I got it, this is what I did

[tex]mgh = 1/2mv^{2}[/tex]

m = m therefore removed.

[tex]gh = 1/2v^{2}[/tex]

take 1/2 on other side becomes 2

[tex]2gh = v^{2}[/tex]

take the ^2 and make it sqrt the other side

[tex]\sqrt{2gh} = v[/tex]

Therefore we conclude that the only variables that matter in it's speed is the gravity and height, which is the same in all 3 problems therefore it will arrive at the same speed because gravity and height are constant?

Thanks in advanced if I'm correct. :)
 
That's correct.
 
LowlyPion said:
That's correct.

Thank you very much again, I really appreciate it! :)
 

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