What Causes a Sledge to Accelerate on Rails in Physics?

AI Thread Summary
The discussion centers on the physics of a sledge accelerating on rails, identifying centrifugal force as the fictitious force responsible for this acceleration, quantified by the equation a = ω² * r. The challenge arises in deriving the displacement r(t) for r < R, starting from the acceleration, with the initial condition r(0) = r₀. A misunderstanding occurs when the user assumes constant acceleration while integrating, leading to confusion about the variable r appearing on both sides of the equation. The correct approach involves recognizing that the acceleration is not constant and applying the method for solving linear second-order differential equations, resulting in the expression r = r₀e^(ωt). This clarification helps resolve the initial confusion regarding the relationship between r and the acceleration.
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Homework Statement


q4.jpg


The first question asks:
Which (fictitious) force will cause the sledge to accelerate along the rails? Give the modulus and direction of the sledge at distance r<R from the centre.

The next question asks:
Obtain an expression for the displacement r(t) for r<R, starting from the acceleration, and the initial condition r(0) = r_0.



Homework Equations





The Attempt at a Solution


a) Centrifugal force is the fictious force that causes the sled to accelerate. The acceleration is given by \omega^2 * r.

b) Now this is where I get most stuck.

So far I have said that a = \frac{d^2r}{dt^2}, so I integrated twice and ended up with the basic kinematic equation r = ut + \frac{1}{2}at^2 (which I would have guessed, but integrated as I assumed that's what the question was hinting at).
From here I said that u = 0, and was left with r = \omega^2rt^2 + r_0, but I'm a little confused about this. Surely r cannot be on both sides of the equation at the same time. As I've been writing this I've also thought about using \frac{v^2}{r} instead of \omega^2r, but on further thought I end up with the same problem of having r on both sides of the equation (it would now be multiplied by the constant term), as well as having the v term that is never mentioned in the question itself (I know I probably could use it, but further questions relate to omega so I have this gut feeling that I should stick with omega in this question).

Any thoughts about this?
 
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The accleration is not constant. You integrated \frac{d^2r}{dt^2} assuming it was constant.

You gave the correct acceleration in part (a):

\frac{d^2r}{dt^2} = \omega^2 r
 
I just had the thought, that going from \frac{d^2r}{dt^2} = \omega^2r that you could say that r=e^{at} and put it back into the equation. From this and the initial conditions it emerges that r=r_{0}e^{\omega t}.

I think that makes a little more sense to me.
 
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Yes, that's how to solve linear second order differential equations :approve:
 
Never dawned on me that that's what the problem was reffering to. I naturally assumed that the acceleration was constant. Makes lots of sense now.
 
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