# Circular Motion.

• Jun 23rd 2010, 04:18 AM
Sunyata
Circular Motion.
A particle, attached by a light, inelastic string of length l to a fixed point O, describes with uniform speed a horizontal circile. The string is just capable of supporting n times the weight of the particle without breaking. Prove that the period T of the circular motion cannot be less than $\displaystyle 2 \pi * \sqrt\frac{l}{ng}$

I've worked it out but where it indicates l, i only have h and i have no idea how to change it to l. so basically i just resolved the forces by Newton's second law and then converted the 2 equations into tan x and then equated it.
• Jun 23rd 2010, 04:40 AM
Ackbeet
• Jun 23rd 2010, 04:48 AM
Sunyata
oh sorry. the perpendicular height from O.
• Jun 23rd 2010, 04:52 AM
Ackbeet
Why should there be any h? I mean, the point of support, O, and the plane of motion are all in the same plane, right? And that plane is perpendicular to gravity, correct? Unless there's more to the problem than you've stated?
• Jun 23rd 2010, 04:54 AM
Sunyata
Hmmm really? For these "conical pendulum" questions, they're always drawn as cones so i presumed there would be a perpendicular height, radius and hyptonesus that is the length of the string?
• Jun 23rd 2010, 04:55 AM
Ackbeet
Please state the original problem in full, from the book, without omitting a single word. Thanks!
• Jun 23rd 2010, 04:56 AM
Sunyata
That's the question.
• Jun 23rd 2010, 04:58 AM
Ackbeet
If the OP states the original problem accurately, then I must confess I am completely mystified by post #5. What does "conical pendulum" have to do with anything here?
• Jun 23rd 2010, 05:00 AM
Sunyata
That's the topic. Mechanics - Conical Pendulum. Oops, is this the wrong subforum?
• Jun 23rd 2010, 05:17 AM
Ackbeet
Oh, I see it now. Sorry about that. Brain lapse. Hang on for a bit. Let me think about it for a second.
• Jun 23rd 2010, 05:32 AM
Ackbeet
Ok, let me throw out a few relevant equations. Perhaps you missed one.

$\displaystyle T\le n m g$.

$\displaystyle T\cos(\theta)=mg$

$\displaystyle T\sin(\theta)=\frac{mv^{2}}{r}$

$\displaystyle 2\pi r=v t$

$\displaystyle h = \ell\cos(\theta)$.

$\displaystyle r=\ell\sin(\theta)$

I was able to derive the result from these equations.
• Jun 24th 2010, 04:58 AM
Ackbeet
Subset of those equations.
You can actually derive the result from the following subset of the equations I listed before:

$\displaystyle T\le n m g$

$\displaystyle T\sin(\theta)=\frac{mv^{2}}{r}$

$\displaystyle r=\ell\sin(\theta)$

$\displaystyle 2\pi r=v t$.

Maybe that will joggle something in your mind.