How do frequency and mass affect centripetal force in lab experiments?

[n3w ton]

Homework Statement

Please go to: http://web.dsbn.edu.on.ca/~andre.hu...0DC3EE/Lab Centripetal Force .pdf?Plugin=Loft

Evaluation Question (E)

Homework Equations

The Attempt at a Solution

Well I was thinking that as the frequency of revolution of the stopper increases, this must mean that either more force tension/Fc is being applied to make the stopper spin faster, thus resulting in faster revolutions and increasing the frequency.

In addition, I think that as the mass and radius both decrease, the frequency also increases.

Though, I'm sure of why and how it would increase the accuracy of the as the frequency of revolution of the stopper increases?

(I think that as the stopper becomes more 'horizontal' it will be spinning in a imaginary flat surface, and this will cancel out gravity and the theoretical normal force, though this is just an idea)

Thank you for your help in- advance!

 

[billVancouver]

Hi n3w ton,

I don't blame you for struggling with this problem; the wording of it is very awkward. I will take my best stab at it, but I make no promises that I have correctly read your instructor's mind:

The key point is (probably?) in the experimental method where you're told to measure the time the ball takes to make 20 orbits, in order to measure its frequency of rotation. At constant radius, as the frequency increases, so does the velocity of the ball; and the faster the ball is moving, the harder it is to stop your stopwatch after *exactly* 20 orbits, thus degrading your accuracy for very high frequencies.

This is certainly true; but apologies in advance if it isn't what was meant by the opaque wording of this problem.

Hope this helps,
Bill Mills

 

[MrWarlock616]

They've said that the metal masses provide the necessary centripetal force, say F_c

$F_c = \frac{m v^2}{r}= mv\omega$

$F_c = 2 \pi n m v$

Thus, the frequency $n = \frac{F_c}{2\pi\ m v}=\frac{F_c}{4\pi^2 m r n}$
$\therefore n=\frac{1}{2 \pi}\sqrt{\frac{F_c}{m r}}$

$\therefore n \propto \sqrt{F_c}$

$\therefore n \propto \frac{1}{\sqrt{r}}$


So, as the metal masses increase, F_c increases, thus the frequency of revolution increases. As the radius decreases, frequency increases. The mass of the ball is constant, as stated. I don't understand what you mean by accuracy...Accuracy of what?
I hope this helped.