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Old Jul 16th 2009, 07:16 AM   #1
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How do I solve this question?

One end of a string of length 100 cm is fixed and the other end has a bob of mass 100g attached to it. A conical pendulum is formed when the bob moves in a horizontal circle. If l is the length of the string and $\displaystyle \theta $ is the angle between the string and the vertical, show that the number of revolutions made by the bob per minute is

$\displaystyle n = \frac{30}{\pi}\sqrt{\frac{g}{l cos \theta}} $

The string breaks when the tension in it exceeds 3.0 N. Calculate the maximum value for the angle $\displaystyle \theta$

Last edited by mark1950; Jul 16th 2009 at 07:21 AM.
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Old Jul 16th 2009, 08:41 PM   #2
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Please state what difficulties you have encountered first.

You better draw a diagram first and name all forces in diagram. Then resolve the forces in components and you would find some equations you need.

Then you should find the quantity which relates the equations with the number of revolutions.
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Old Jul 25th 2009, 12:20 PM   #3
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Well, we know that

$\displaystyle T = 2\pi\sqrt{\frac {L}{g}}$

and

$\displaystyle \frac {60}{T} = n$

$\displaystyle n = \frac {60}{2\pi\sqrt\frac{L}{g}} $

$\displaystyle n = \frac {30}{\pi} \sqrt{\frac{g}{L}} $

Now, here's where I'm troubled. If $\displaystyle \theta$ is truly
the angle between the string and the vertical plane, then
$\displaystyle \theta= 90^\circ$ and that would make our equation
invalid because $\displaystyle n$ would be undefined (Remember,
$\displaystyle \cos 90^\circ = 0$ and we know that nothing can be
divided by zero).

But it can easily shown that $\displaystyle L = l \sin \theta$

Making the equation:

$\displaystyle n = \frac {30}{\pi} \sqrt{\frac{g}{l \sin \theta}}$


Now, for the tension.

$\displaystyle F_t = F_c - F_g = \frac{mv^2}{r} \sin \theta - mg \cos \theta $

$\displaystyle \frac {F_t}{m} = \frac {v^2}{r} \sin \theta - g \cos \theta $

Now just compare $\displaystyle \frac{mv^2}{r}$ and $\displaystyle mg$

Since $\displaystyle \frac {v^2}{r} > g$

It is at a maximum when the coefficients are the the greatest.

Making $\displaystyle \sin \theta = 90^\circ$

I replaced this with my earlier proof as it's easier to read through.

...I think

Last edited by Deco; Jul 26th 2009 at 08:55 PM.
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Old Jul 25th 2009, 10:49 PM   #4
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Is 2pi sqrt( L / g ) a valid formula for a conical pendulum? Think.
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Old Jul 26th 2009, 08:41 PM   #5
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I obtained the desirable answer, how could this be wrong?
It makes perfect sense.

A pedulum's period is unaltered if it's going in a straight line
or a circle (they both complete one cycle), assuming it
travels the same speed in one dimension
as before. We can assume that here.
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Old Jul 27th 2009, 04:42 AM   #6
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Well the answer has cos in the denominator while you have sin !

So the answer is not right.

And just because a method gives a right answer, it does not mean it is right.

And the period T of a conical pendulum works out to be

T = 2 pi sqrt (Lcos theta / g ).

Maybe your understanding of a conical pendulum is different.

And theta is < 90 for a conical pendulum; else the problem reduces to circular motion in a horizontal plane.
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Old Jul 27th 2009, 01:29 PM   #7
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A conical pendulum is formed when the bob moves in a horizontal circle.
And theta is < 90 for a conical pendulum; else the problem reduces to circular motion in a horizontal plane.
That was implied.

My equation could involve cosine if theta represented the angle between
the vertical plane and the plane normal to the conical pendulum.

T = 2 pi sqrt (Lcos theta / g )
Can you please demonstrate how you came to this result?

And just because a method gives a right answer, it does not mean it is right.
True. Let me try to demonstrate why I think I'm right.

Let's say that a pendulum swings side to side in a period of 3 seconds,
covering 30 cm.

Now, lets take that same pendulum and let it spin horizontally,
it would still cover 30 cm from side to side, but would also move along
another plane, unaltering the period....I'm sorry if that doesn't make sense.
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Old Jul 28th 2009, 07:41 PM   #8
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Maybe we use a simpler approach
Let X be the tension, r is the radius of circle, l be the length of string

$\displaystyle Xcos\theta=mg$---------(1)
$\displaystyle Xsin\theta=mr\omega^2$
$\displaystyle Xsin\theta=mlsin\theta\omega^2$
$\displaystyle X=ml\omega^2$
$\displaystyle X=ml(\frac{2\pi}{T})^2$--------(2)

Solving (1) and (2)
$\displaystyle
T=2\pi \sqrt{\frac{lcos\theta}{g}}
$
Since T represents time required in 1 revolution. For revolutions per minute=60/T, which is the answer provided
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