# Mechanics 2

1. Aug 23, 2009

### Mentallic

1. The problem statement, all variables and given/known data
http://img21.imageshack.us/img21/3430/mechanics3.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
For (i) to find the horizontal component I used (2) and (4) to get $$a=rw^2$$ and $$r=l.cos\alpha$$ hence $$a=l.cos\alpha.w^2$$
But I have no idea for the vertical component. I just keep guessing at that one.

I think I'll need the answer to (i) before I can answer (ii)

Last edited by a moderator: May 4, 2017
2. Aug 23, 2009

### zcd

$$r=l\sin(\alpha)$$ if you draw out the triangle.

In this case of uniform circular motion, the horizontal component of force would be the centripetal force keeping the object in circular motion.

Don't forget that both tensions contribute to the centripetal force, so $$T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha)$$.

The vertical net force on P would have a magnitude equal to the vector sum of the vertical components of the two tension forces and gravity. Although not explicitly stated by the problem, the vertical net force is probably 0, so $$F_{net,y}=0=T_{1y}-T_{2y}-mg$$ (with up as positive).

Btw, most of these look like they should be in the physics section o.o

3. Aug 23, 2009

### Mentallic

Ahh yes thanks for correcting me on that.

So then the force F=ma is equivalent to the tension's of each string holding it in place? Ahh that makes sense

So $$F_x=T_{1x}+T_{2x}=lm\omega^{2}\sin\alpha$$

This too makes sense

It's always assumed zero for our course, unless otherwise stated.
Why is it the negative of $$T_{2y}$$ and not the positive? Maybe because its significance is that it keeps the mass pulled downwards?

Actually I've already learnt basics of motion in physics, but we were studying this in maths. We also haven't learnt about vectors, so please bare with me here.

4. Aug 23, 2009

### Mentallic

Uhh I'm a little worried what I tried is again, incorrect.

$$T_{1y}=l.cos\alpha$$ yes?

But also, $$T_{2y}=l.cos\alpha$$ by the looks of it.

Hence, mg=0 ? which doesn't sound right...

5. Aug 23, 2009

### ideasrule

Looks can be deceiving.

6. Aug 23, 2009

### ideasrule

...but in this case they're not. lcos(a) is half the height of the apparatus, not the tension. To use cos(a), you can express Ty in terms of the total tension, T:

T1y=T1cos(a)
T2y=T2cos(a)

7. Aug 23, 2009

### Mentallic

So uh...

$$T_{1y}-T_{2y}-mg=cos\alpha(T_1-T_2)-mg=0$$

$$cos\alpha(T_1-T_2)=mg$$ (1)

$$T_{1x}+T_{2x}=sin\alpha(T_1+T_2)=lmw^2sin\alpha$$

$$T_1-T_2=lmw^2$$ (2)

I'm unsure what to do now. I'm thinking both tensions are supposed to be solved simultaneously, but I wouldn't be able to eliminate both tensions with just 2 equations.

8. Aug 23, 2009

### zcd

The problem was asking for forces in vertical and horizontal directions, and that seems to have finished the job. If you want to solve for $$|\vec{T}_{1}|$$ or $$|\vec{T}_{2}|$$, note that:
$$T_{1y}=T_{1}\sin(\alpha)$$
$$T_{1x}=T_{1}\cos(\alpha)$$
$$T_{1y}=T_{1x}\tan(\alpha)$$
and proceed with the algebra

9. Aug 24, 2009

### Mentallic

Then if $$T_{1y}=T_{1x}tan\alpha , T_{2y}=T{2x}tan\alpha$$

and $$T_{1x}+T_{2x}=lm\omega^2sin\alpha$$

then $$cot\alpha(T_{1y}+T_{2y})=lm\omega^2sin\alpha$$

So the best I can do with this is: $$T_{1y}+T_{2y}=lm\omega^2sin\alpha tan\alpha$$

When you say proceed with the algebra, sure I have no problem doing that for other topics, but in this which is very new to me, all I'm seeing is a big mess where I have a bunch of equations that don't make sense to me... Could you please direct me to which equations I need to be solving for? Maybe $F_x$ with something else?

10. Aug 24, 2009

### zcd

Instead of grouping into the same directional component (x in that case), you could try to reduce it to components of the same tension force. For example:
$$T_{1x}-T_{2x}=mg\cot(\alpha)$$
$$T_{1x}+T_{2x}=lm\omega^{2}\sin(\alpha)$$
system of equations

$$T_{1x}=\frac{1}{2}(mg\cot(\alpha)+lm\omega^{2}\sin(\alpha))$$
$$T_{1}=\frac{1}{2}(mg\sec(\alpha)+lm\omega^{2}\tan(\alpha))$$
is that what the question is asking for?