[Physics] Free fall into circular motion

accelerationforceshomework-and-exerciseskinematicsnewtonian-mechanics

If I'm on a roller coaster free falling from height $h$ and then suddenly start going into horizontal motion with a radius $r$ of turn what is the $g$-force I experience?

I worked out the equation like this but am not sure if it is correct:

  • (1) instant velocity of free-fall $v=\sqrt{2 g h}$
  • (2) uniform circular motion acceleration $a = \frac{v^2}{r}$
  • (3) $g$-force $ gf = \frac{a}{g} = \frac{v^2}{g r}$

My doubts are:

  • I don't know if I can use uniform circular motion equation since $v$ is not constant
  • Where is the g-force directed towards? The center of the turn?

Best Answer

There are two accelerations involved: The gravitational acceleration $g$ that points down, and the centripetal acceleration $a_r = \frac{v^2}{r}$ that points along the radius vector of the curve. The component of the gravitational acceleration that is tangential to the curve does not contribute to the g-force as it accelerates the cart and us in this direction. We do feel the component of the gravitational acceleration that points along the radius vector given by $a_g = g \text{sin}\phi$ where $\phi \in [0,\frac{\pi}{2}]$ is the angle that starts with zero when you enter the curve and ends with $\frac{\pi}{2}$ when you enter the horizontal part. We can add the two accelerations as they are parallel:

$$a_{tot} = a_g +a_r$$

We now observe: $ v = \sqrt{2g(h+r\text{sin}\phi)}$ and so

$$a_r = \frac{v^2}{r} = \frac{2g(h+r\text{sin}\phi)}{r}$$

thus:

$$a_{tot} = g \text{sin}\phi + \frac{2g(h+r\text{sin}\phi)}{r} = 3g \text{sin}\phi + \frac{2gh}{r}$$

The g-force is $ a = \frac{a_{tot}}{g}$ thus

$$a = 3 \text{sin}\phi + \frac{2h}{r}$$

So the g-force is no constant but depends on where you are in the curve (on $\phi$). And as sin is monoton increasing over $[0,\frac{\pi}{2}]$ so is the g-force. it reaches it's maximum as it enters the horizontal part.

What you might also wanto observe is that if you build a rollercoaster like this people would run screaming, as the g-force is not continues. There is a jump as you enter the curve and a big jump as you go onto the horizontal. And basically at these discontinuities the g-force is infinite :)