# Rings sliding on rods

1. May 1, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
I am not sure about this but as the string is inextensible, the component of velocity of the rings along the string should be equal. O' moves downwards and O moves upwards (not sure but it looks so). Hence, $v_1\cos \alpha=v_2\cos \alpha \Rightarrow v_1=v_2$ but this is apparently wrong.

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2. May 1, 2013

### voko

There are a couple of equations that relate the positions of the rings with the length of the string and the distance between the poles. Write them down, use them to find the velocity of O.

3. May 1, 2013

### Saitama

Position from where? Should I fix the co-ordinate system with origin at A?

4. May 1, 2013

### haruspex

Suppose the string has length s and the rods are distance h apart. When O'A' = x, what is the distance OA?

5. May 1, 2013

### Saitama

$OA=x+\sqrt{s^2-h^2}$?

6. May 1, 2013

### voko

How come OA grows infinitely as x increases?

7. May 1, 2013

### Saitama

See attachment.
$OA=OB+BA \Rightarrow OA=x+\sqrt{s^2-h^2}$

Why is it wrong?

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8. May 1, 2013

### voko

How come BA is fixed? Surely as x grows, it should get smaller, because the string is inextensible.

9. May 1, 2013

### judas_priest

It cannot be v1=v2.
Here's why:
Assume the distance between the two poles as 'x' and the string length 's', the distance to be traveled by O' to reach the height of O will be √(s^2-x^2). Now, as s tends to x (i.e the ring O' starts to reach height of the ring O, it will already have a velocity, whereas O will remain to be at rest till O' lowers and starts pulling O.
Therefore, V1>V2
You can try using trigonometric ratios to prove the same.
i.e
Sin(90-alpha) = √(s^2-h^2)/(s^2)

10. May 1, 2013

### haruspex

Because s is not equal to the hypotenuse, OO'. The string runs from O to A' via O'..

11. May 1, 2013

### Saitama

Yep, I realised it before I went to sleep.

This time I get $OA=x+\sqrt{(s-x)^2-h^2}$.

Differentiating w.r.t time,
$$\frac{d(OA)}{dt}=v_2=\frac{dx}{dt}-\frac{(s-x)}{\sqrt{(s-x)^2+h^2}}\frac{dx}{dt}$$
$dx/dt=v_1$ and $s-x=h/\sin \alpha$, using these
$$v_2=v_1\left(1-\frac{1}{\cos \alpha}\right)=-v_1\frac{2\sin^2 \alpha/2}{\cos \alpha}$$

Is this correct?

I still don't understand what's wrong with my attempt in the first post.

12. May 2, 2013

### haruspex

A minus turned into a plus in the middle there, but it must have been a transcription error because it came out right in the end.
I didn't understand your reasoning there, so I can't say where it was wrong.

13. May 2, 2013

### Saitama

Thank you haruspex!