# 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.