Mechanics Pendulum Homework: Solve Horizontal & Vertical Components

[Mentallic]

Homework Statement

http://img198.imageshack.us/img198/2900/mechanicspendulum1.jpg

Homework Equations

w=\frac{d\theta}{dt} (1)

v=rw (2)

F=ma (3)

a=\frac{v^2}{r} (4)

The Attempt at a Solution

I'm guessing that the horizontal component is using formula (4) where r=l.sin(\alpha) and v is found from formula (2) such that v=l.sin(\alpha).w

Thus, a=\frac{(l.sin(\alpha).w)^2}{l.sin(\alpha)}=l.sin(\alpha).w^2

However, for the vertical component, I'm unsure how to begin. Oh and I'm not certain if I'm solving the horizontal correctly either, so don't hesitate to scold my mistakes

 

[zcd]

It's not really a pendulum problem as the mass doesn't oscillate through an energy valley. Consider an angle θ such that it lies on the circle around which the mass revolves. It can be said that \omega=\frac{d\theta}{dt}.

The net force F_{net,y} on the mass must be zero, and a free body diagram of the mass will note that F_{net,y}=0=T_{y}-mg.

For the mass to revolve in a circle, there must be a centripetal force T_{x}=m\frac{v^{2}}{r}=m\omega^{2}r; in this case, r=l\sin(\alpha) and x component of tension can be rewritten as T_{x}=\omega^{2}lm\sin(\alpha).

Noting that \vec{T}=T_{x}\vec{i}+T_{y}\vec{j}, T_{x}=T\sin(\alpha) and T_{y}=T\cos({\alpha}). This gives T_{x}=T_{y}\tan(\alpha).

After some substitution, mg\tan(\alpha)=\omega^{2}lm\sin(\alpha) and \omega^{2}=\frac{g}{l\cos(\alpha)}

 

[Mentallic]

Thanks, the T_x makes a lot of sense now!
But I'm unfamiliar with the notations you've used here and so I couldn't follow it from there-on:

Noting that \vec{T}=T_{x}\vec{i}+T_{y}\vec{j}, T_{x}=T\sin(\alpha)

 

[zcd]

I just separated it into components to form a right triangle. From the right triangle, you can see how each component is related to the other component and the tension force vector itself.

 

[Mentallic]

Oh ok I see. While I still can't figure out what that vector notation is meant to represent (somehow, a right triangle), I can see how you got T_x=T_y.tan\alpha

Ok but now, what did you substitute and into which equations?

 

[ideasrule]

There's this:

F_{net,y}=0=T_{y}-mg.

And this:

T_{x}=\omega^{2}lm\sin(\alpha).

 

[Mentallic]

Aha now it all makes sense! So for most questions like these, to resolve the horizontal and vertical components, I should find the horizontal in terms of m,\omega,l,\alpha and then the vertical in terms of the horizontal tension force.