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.