# Relationship between the “angle of the floor” and the angular velocity in a banked turn?

Wel, imagine that you're in a carousel, and the floor is, let's say at $\theta=0$ so it's totally horizontal, if $\theta=90$ the floor would be vertically.

The object put above the floordoesn't move until the carousel starts.

My question is:

With what angular velocity($\omega$) an object would have to go to stay where it is knowing the angle of the floor($\alpha$), the radius of the carousel($r$) and the gravity($g$)?

What i'm asking is when you combine this two types of movements.

Something like this:

Based on the Newton's second law of motion($F=ma$)

I finally get this equation:

$$\omega = \sqrt{-\dfrac{g\tan{\theta}}{r\cos{\theta}}}$$

Being:

$g=\text{gravity}=-9,8^m/_{s^2}$

$r=\text{radius of the carousel}=1m$

$\theta=\text{angle of the floor}=x\text{ axis in the graph below this}$

$\omega=\text{angular velocity}=y\text{ axis in the graph below this}$

The graph of this function is

And if I'm right,

Is it true that this type of motion doesn't depend on his mass?

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I think the problem is that nobody understands the question. Neither do I understand the question. –  Pygmalion Apr 24 '12 at 10:30

I almost agree with your result.

For an object in a circular path of radius $r$ moving with angular velocity $\omega$ the acceleration towards the center is $r\omega^2$. This acceleration is horizontal in your top diagram, so the component of the acceleration acting along the slope (the purple arrow in your diagram) is $rw^2cos(\theta)$.

You've correctly drawn the acceleration of the object down the slope, F$_{||}$ as $g.sin(\theta)$, so just set them equal and you get:

$$rw^2cos(\theta) = g.sin(\theta)$$

$$\omega = \sqrt{\frac{g}{r} \frac{sin(\theta)}{cos(\theta)}}$$

We just disagree in that I've got $sin$ where you've got $tan$.

Note that I've just equated the accelerations in my treatment above. You could equate forces and include the mass is you want to, but the mass is the same on both sides of the equation and will cancel out, so you're correct that in this case the mass doesn't affect the result.

I suspect you've put off people from looking at your question by drawing so many diagrams, hence the lack of response. Only the top diagram is needed. Drawing a good clear diagram is often the most important part of attacking a problem like this.

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Yeah. I was wrong in the calculations. Nos I know where I have failed. Thanks! and $v$ can be written as $v=\sqrt{gr\tan{\theta}}$ and $\omega$ as $\omega=\sqrt{\frac{g\tan{\theta}}{r}}$. But this is obvious. –  Garmen1778 Apr 24 '12 at 19:27