Ball twirling on string attached to stick

[Melchoire]

So we did this little experiment; we attached a tennis ball to a string which ran through a tube and to a plastic bag with 3 tennis balls. Here's a picture

http://img7.imageshack.us/img7/6160/twirls.jpg

So in a nutshell we can represent Mg like this:
$$Mg = (m*4pi^2*R)/(T^2)$$
lol i tried my best with the latex stuff...

anyways in the image the ball is perfectly perpendicular to the tube. But in our experiment it was moved downwards somewhat due to gravity on the ball, with an angle. However I am told that this has no bearing on the results of the formula. That is the formula works even though it's spinning around on an angle.

My problem is: How do I prove this?

PS: this problem won't matter at all but M = 3m

 

[Hootenanny]

Welcome to Physics Forums.

The rotating ball is an example of what is commonly called a conical pendulum (the string traces out a conical surface). Now, let us suppose that the string makes and angle $\theta$ with the horizontal. In this case, the tension will no longer be acting parallel to the radius of the circle transcribed by the ball. Instead, it will be acting towards the pivot, which in this case is the top of the tube. So, what we need to do is look at the components of the tension. If T is the magnitude of the tension in the string, then the components of the tension will be given by

$$T_x = T\cos\theta\;\;\;,\;\;\;T_y = T\sin\theta$$

Now, what can you say about T_x and T_y, bearing in mind that the tennis ball transcribes a circle in the (x,z) plane and remains at constant height, y, throughout it's motion?

 

[Melchoire]

So the vector sum of Ty and Tx is equal to T?

 

[rock.freak667]

So the vector sum of Ty and Tx is equal to T?

Like Hootenanny said, T_x=Tcosθ and T_y=sinθ.

Now considering the horizontal component T_x, since it is spinning in a circle, T_x provides the centripetal force so Tcosθ = ?

Now the vertical tension,T, in the string balances out the weight of the mass M, to T equals what now ?

 

[Melchoire]

3m? I'm not good at this...=P

 

[rock.freak667]

3m? I'm not good at this...=P

no no

you know what is the centripetal force right? If you do then, what is the formula for centripetal force?


in the Tension-mass system (the part with the string in the tube and the mass M), I don't believe the mass M is moving up or down, so the tension T should balance out the mass M right? So what is T equal to here?

 

[Melchoire]

I don't know dude...I figure if they balance then the tension force should be equal to Mg...

 

[rock.freak667]

I don't know dude...I figure if they balance then the tension force should be equal to Mg...

Right! So T=Mg.

Since it looks like you don't know centripetal force =mω²R

so if Tcosθ= mω²R.

What is Mg equal to ?

 

[Melchoire]

Mg = mω2R?
what's that w symbol?

 

[rock.freak667]

Mg = mω2R?
what's that w symbol?

ω is called the angular velocity. Which for a circle is ω=2π/T where T=time period.

so we have

$$Mg= \frac{m \omega ^2 R}{cos \theta}$$

Now θ should be small. And what is cosθ approximately equal to for very small angles of θ?

 

[Melchoire]

So cos($$theta$$) = Mg?

 

[paul_harris77]

No, for small angles cos(theta) is approximately 1. Therefore your formula becomes Mg=m(w^2)R. That's your proof since this formula is the same as for the original case with no "dipping" of the tennis ball.