# Mechanics Pendulum

1. Aug 22, 2009

### Mentallic

1. The problem statement, all variables and given/known data
http://img198.imageshack.us/img198/2900/mechanicspendulum1.jpg [Broken]

2. Relevant equations

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

$$v=rw$$ (2)

$$F=ma$$ (3)

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

3. 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

Last edited by a moderator: May 4, 2017
2. Aug 22, 2009

### 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 $$\omega^{2}=\frac{g}{l\cos(\alpha)}$$

Last edited: Aug 22, 2009
3. Aug 22, 2009

### Mentallic

Thanks, the Tx 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:

4. Aug 23, 2009

### 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.

5. Aug 23, 2009

### 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?

6. Aug 23, 2009

### ideasrule

There's this:

And this:

7. Aug 23, 2009

### 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.