# Horizontal Motion

1. Mar 18, 2006

### GregA

This question I was working on last light though I've solved it (by doing as the book has shown)...is giving me major cause for concern... I'm going to have to slightly re-word the books question so as to explain it without a diagram:

A particle P of weight 2kg is fastened to a length of extensible string AB of natural length 0.4m, and modulus of elasticity 12N. A is fixed and B rotates in a horizontal circle. Given that the extention of the string is 0.1m find,
a) the tension in the string
b) the angular speed of P

Firstly I shall define $$\theta$$ as the angle at A and $$L$$and $$M$$ length of string and the mass of P respectively .

I find (a) merely by plugging what I know into $$T = \lambda \frac {x}{a}$$ and getting an answer of 3N
(b) is the problem there is only 3N of tension in the string and this is not going to balance 2g...so my first inclination to resolve vertically, find the angle, find the radius and then solve fails when I end up with $$cos\theta = \frac {2g}{3}$$...this answer is greater than 1
To reach the books answer I had to express it as follows:
$$F = ma$$
$$T = mr \omega^2$$
$$Tsin \theta = MLsin \theta \omega^2$$
after cancelling $$sin \theta$$ on both sides The wole thing simplifies to
$$\omega = \sqrt {\frac {T}{ML}}$$
and by plugging values: $$\omega = \sqrt {3}$$ which is what the book wants.

My major cause for concern is that the tension is not resolved vertically and the book makes no statement that it is resting on a horizontal surface...and even then, a mass of 2kg is going to want to stretch that string by just over 0.6m even at rest...by rotating as well it will want to stretch the string some more. But in spite of this I was able to just blindly use a formula and get an answer...an answer I am certain is wrong and makes me worry about other answers I may get in the future. Am I just talking BS and someone can point this out to me?

Last edited: Mar 18, 2006
2. Mar 18, 2006

### Hootenanny

Staff Emeritus
You've done this wrong. If you resolve the weight of the object so that is it parallel with the string you get;
$$2g\cos\theta = 3$$
$$\theta = \cos^{-1}\frac{3}{2g}$$

Can you see?

3. Mar 18, 2006

### GregA

Thanks for the reply Hootenanny and though I can see what you're saying...I don't know where my line of reasoning breaks down then because:
3N is the tension in the string (the actual physical part of the string...the hypoteneuse if you will) and some of this 3N has to balance 2gN you suggest however I should express it as
$$2gcos \theta =3$$ as opposed to $$3cos \theta =2g$$ ? ( I know that with your suggestion I can find a value for $$\theta$$ but I just can't see how it suits the problem to use this if you know what I mean)

Last edited: Mar 18, 2006
4. Mar 18, 2006

### Hootenanny

Staff Emeritus
I'm saying that instead of resolving the tension vertically, you should resolve the weight of the particle parallel to the tension. Tension can only be applied by the force parallel to the string/rod. If you resolve the string/tension, this force is not parallel to the string, this is why the cosine of your angle is greater than one.

If you resolve the particle parallel to the string, you can then use trig to find the radius of the horizontal circle and then as the solution states $F = mr\omega^2$ to find the angular velocity.

5. Mar 18, 2006

### GregA

I see...thanks Hootenanny (stupid book lets me work in ignorance )