Optimizing Time for a Ball on a Sloped Path: Arc or Chord?

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Homework Help Overview

The discussion revolves around a physics problem involving a ball sliding down two different paths—a chord and an arc—under the influence of gravity. The objective is to determine which path allows the ball to reach the lowest point in the shortest time, while adhering to constraints that prohibit the use of certain known results, such as those related to pendulum motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various methods to analyze the motion of the ball, including the use of Newton's laws and the concept of piecewise linear approximations. Some express difficulty in avoiding known results about pendulums, while others suggest alternative paths of reasoning involving multiple chords.

Discussion Status

The conversation is ongoing, with participants exploring different approaches to derive the time taken for the ball on the arc without relying on pendulum equations. Some guidance has been offered regarding the use of equations of motion and energy conservation, but there is no explicit consensus on the best method to proceed.

Contextual Notes

Participants note that the problem constraints include a requirement to compute the time for the arc path, despite some suggesting it may not be necessary to answer the original question. There is also mention of the teacher's instructions, which influence the direction of the discussion.

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Homework Statement


We let a ball with mass m to slide down under the influence of his weight the path defined by the chord or the path defined by the arc of the circle until the ball reach the lowest point, as it seems from the picture. In which path the ball will make the shorter time to reach the lowest point?

Homework Equations





The Attempt at a Solution


At first to compute the time at the path of the arc the problem is the same with the period T of a pendulum which is T=2\pi\sqrt{\frac{L}{g}} and we are going to have t=\frac{T}{4} but we must not use any known fact about pendulum. I am trying to use Newton's Law F=ma but it's still difficult, I can't find a way to express sinw with L. And even if I find it I don't think I can compute time because the acceleration is not constant.
 

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There are not too many options. Either you use the pendulum result, or derive it. Another option would be to prove that piecewise linear curves approximating the circle give the lesser time of descent the closer they approximate the circle.
 
Our teacher told us that it can be solved without using the pendulum. I have tried lot of things but can't find it.
 
Consider the path connecting the same point that consists of two equal chords. Which path is faster: one-chord or two-chord? Then consider the path of four equal chords and so on.
 
voko said:
Consider the path connecting the same point that consists of two equal chords. Which path is faster: one-chord or two-chord? Then consider the path of four equal chords and so on.

I know how to comptute the time in chord but I can't compute the time in the arc.
 
You do not need to compute the time in the arc to answer the question in the problem.
 
voko said:
You do not need to compute the time in the arc to answer the question in the problem.

I know that, but our teacher told us to compute it anyway.
 
Your options are listed in #2.
 
voko said:
Your options are listed in #2.

How I will derive it?
 
  • #10
You derive it by solving the equation of motion, which you could get from the laws of Newton or conservation of energy.
 
  • #11
voko said:
You derive it by solving the equation of motion, which you could get from the laws of Newton or conservation of energy.

I used the small angle approximation where sin\theta\approx\theta and I'm here: a(t)=-\frac{L}{g}\theta(t) where a(t) is the angular momentum in a given time t. How I will find now that T=2\pi\sqrt{\frac{L}{g}}?
 
  • #12
Are you sure a(t) is the angular momentum? Isn't it the angular acceleration?
 
  • #13
voko said:
Are you sure a(t) is the angular momentum? Isn't it the angular acceleration?

Sorry, is the angular acceleration. How I find T now?
 
  • #14
You have differential equation ## \frac {d^2\theta} {d\theta^2} + \frac L g \theta = 0 ##. Solve it.
 

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