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The experiment setup looks (roughly) like:

enter image description here

First, a weight (W2), is hung to simulate centripetal force when the platform is not rotating.

Then its removed (W2 is now hanging to one side as shown in the bottom image) and we adjust the speed of rotation until W1 is aligned again. Then we measure the time taken for 20 revolutions.

This experiment is repeated for varying radius (the distance between the 2 vertical columns). Then I'm asked to plot a graph of radius vs Period^2. But I don't get why Period^2?

Then when the experiment is repeated for constant radius, but varying W2, I'm asked to plot a graph from radius to Period^(-2), why powers of 2 and -2?

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up vote 1 down vote accepted

In the experiment you are finding the angular velocity needed to create a force on $W_1$ of magnitude $W_2g$ (where $g$ is the acceleration due to gravity). For an object moving in a circle of radius $r$ and angular velocity $\omega$ the acceleration is given by $a = r\omega^2$ so the force is $F = W_2g = W_1r\omega^2$. Since you're keeping the force constant, you can rearrange the equation to give:

$$ r = \frac{W_2g}{W_1} \frac{1}{\omega^2} = \frac{W_2g}{W_1} \frac{\tau^2}{4\pi^2} $$

where $\tau$ is the period and I've used the equation $\omega = 2\pi/\tau$. So $r$ is proportional to $\tau^2$ and a graph of $r$ against $\tau^2$ should give a straight line.

Now you can probably see why in the second experiment you graph $W_2$ against $\tau^{-2}$. If you take the equation above and rearrange it to get $W_2$ as a function of $\omega$ you get:

$$ W_2 = \frac{W_1r}{g} \omega^2 = \frac{W_1r}{g} \frac{4\pi^2}{\tau^2} $$

so $W_2$ is proportional to $\tau^{-2}$.

Response to comment:

The acceleration is given by $a = r\omega^2$ where $\omega$ is the angular velocity. $\omega$ is related to the tangential velocity $v$ by $\omega = v/r$. If you substitute for $\omega$ in the expression for the acceleration you get:

$$ a = r\omega^2 = r \left( \frac{v}{r} \right)^2 = \frac{v^2}{r}$$

So you are quite correct that $a = v^2/r$.

Response to response to comment

If you take my first equation and rearrange it to get $\tau^2$ in terms of $r$ you get:

$$ \tau^2 = \frac{4\pi^2 W_1}{W_2 g} r $$

so if you graph $\tau^2$ against $r$ the gradient will be $(4\pi^2 W_1)/(W_2 g)$. This is the same result as in the Google doc, $m(2\pi N)^2/F_c$, but that document uses $m$ to denote $W_1$ and $F_c$ for $W_2g$. Also the period I've used, $\tau$, is the period for one revolution so in their equation $N$ is one.

When you're doing the calculation it's certainly easiest to use SI units from the beginning so you don't make a mistake with units along the way. If you want to put your experimental data into a Google docs spreadsheet I'd be happy to have a look at it. Remember to include the values for $W_1$ and $W_2$ as well as the values for $r$ and $\tau$ (remember to say how many revolutions you timed).

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I thought the equation was $a = \frac{v^2}{r}$? not $a = rv^2$? Or am I looking at the wrong equation? – Jiew Meng Oct 16 '12 at 13:42
@JiewMeng: I've edited my answer to respond to your comment – John Rennie Oct 16 '12 at 13:56
Hmm how do I get the experimental centripetal acceleration from this equation? – Jiew Meng Oct 16 '12 at 14:55
@JiewMeng: remember Newton's law $F = ma$. The force on $W_1$ when it's hanging vertically is $W_2g$, so the acceleration is just $a = F/m$ or in this case $W_2g/W_1$. – John Rennie Oct 16 '12 at 15:14
That will be the theoratical accelearation? I was asked the experimental. I'm looking at…. They say that the centripetal force, is simply the slope of the line of $T^2$ vs $r$ – Jiew Meng Oct 16 '12 at 15:31

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