Homework Help: Fictitious forces, and motion

1. Jan 21, 2007

Brewer

1. The problem statement, all variables and given/known data

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.

Obtain an expression for the displacement r(t) for r<R, starting from the acceleration, and the initial condition r(0) = $$r_0$$.

2. Relevant equations

3. 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 thats 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).

2. Jan 21, 2007

AlephZero

The accleration is not constant. You integrated $$\frac{d^2r}{dt^2}$$ assuming it was constant.

You gave the correct acceleration in part (a):

$$\frac{d^2r}{dt^2} = \omega^2 r$$

3. Jan 21, 2007

Brewer

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.

Last edited: Jan 21, 2007
4. Jan 21, 2007

AlephZero

Yes, that's how to solve linear second order differential equations

5. Jan 21, 2007

Brewer

Never dawned on me that thats what the problem was reffering to. I naturally assumed that the acceleration was constant. Makes lots of sense now.