# Homework Help: Approximation for a slipped pendulum

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1. Dec 31, 2017

### Pushoam

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Applying conservation of potential energy,

$mgL (1 - \cos{ \theta_0}) = mg(L + \delta ) (1 - \cos{ \theta_1})$

$\cos{ \theta_1} - \cos{ \theta_0} = \frac { \delta - \delta \cos{ \theta_1}} L$

Taking the approximation,

$( \theta_0 - \theta_1)(\theta_0+ \theta_1) =\frac { 2 \delta } L$

$\theta_0 -\theta_1 = \frac { \delta }{ L \theta_0}$

$\theta_1 = \theta_0 \left (1 - \frac { \delta }{ L {\theta_0}^2} \right )$

Is this correct?

2. Dec 31, 2017

### haruspex

Always try to avoid assuming work is conserved. In the present case it surely is not. What else can you use?

3. Dec 31, 2017

### Pushoam

I could not get it.

The force doing work is the gravitational force, which is a conservative force. So, the total mechanical energy of the system remains conserved.

Now, at angle $\theta_0$ and $\theta_1$, kinetic energy of the system is 0. So, there potential energy should be equal.

Is this correct?

4. Dec 31, 2017

### haruspex

Only until the string slips. It will not be conserved during the slip.

5. Dec 31, 2017

### Pushoam

But, the slip is for negligible time so can't we ignore it.
Can't we just take the system as on the left side the length is L and on the right side, the length is L + $\delta$ and so on?
I am just considering the two final points.

6. Dec 31, 2017

### haruspex

Can't ignore what?
There is no reason to suppose that any of the GPE lost during the slip will turn into horizontal KE.
What will be conserved?

7. Dec 31, 2017

### haruspex

By the way, in your attempt using conservation of energy the line after "taking the approximation" is wrong. You seem to have approximated the cos(θ1) on the right to zero instead of to 1-½θ12.

8. Dec 31, 2017

### Pushoam

I had taken $\delta - \delta \cos \theta_1 \approx \delta$ as $\delta$ itself is very smalll and $\delta \cos \theta_1$is smaller than $\delta \cos \theta$. I realised that this appeoximation is not useful here.
Taking the approximation,

$( \theta_0 - \theta_1)(\theta_0+ \theta_1) =\frac { {\theta_1}^2 \delta }{2 L}$

$\theta_0 -\theta_1 = \frac { \delta \theta_0 }{ 2L }$ Taking $\theta_1 = \theta_0$ in RHS

$\theta_1 = \theta_0 \left (1 - \frac { \delta }{2 L} \right )$

9. Dec 31, 2017

### Pushoam

Any force due to slipping will occur in radial direction. So, slipping will not affect angular momentum of the system about pivot and its kinetic energy.

10. Dec 31, 2017

### haruspex

There is no law of conservation of KE, but presumably you are thinking of the horizontal component of linear momentum.
So yes, horizontal linear momentum and angular momentum are candidates, but can they both be conserved? Won't they lead to different conclusions?

11. Jan 1, 2018

### haruspex

Further to my preceding post...
It was not at all clear to me which of horizontal linear momentum and angular momentum is the "more conserved", so I bit the bullet and launched into a thorough analysis. For this, I assumed that the string went slack at a very small angle before reaching vertical, and while travelling at speed u. The mass then followed a parabolic trajectory until the string became taut again at length L+δ, that being at some other very small angle past the vertical.
Writing out all the equations and discarding terms of order higher than (small angle)2 (and taking δ/L to be of that order or smaller, which it is) I got a match on one of the options.

Next I will check what each of the two conservation assumptions yields.

12. Jan 3, 2018

### Pushoam

I didn't understand this part. So, I am not able to write the equations. Please give me some more insights.
Now, what I have understood is:

The string slips when it passes through vertical. So, the forces due to slip acts on the string in the vertical direction mostly. So, the momentum in the horizontal direction does not get noticeably affected by these forces.

So, if there were no slip, then at the right end of the swing $m v_x = m \sqrt{ Lg \cos{\theta_0} } \cos{\theta_0}$ .....(1)

As, $\frac { mv^2}L = mg \cos{\theta_0}$ .....(2), $v_x = v \cos{\theta_0}$ .....(3)

Similarly,

With slipping,

$mv'_x = m \sqrt{ (L+\delta )g \cos{\theta_1} }\cos{\theta_1}$ .....(4)

$mv_x = mv'_x$ .....(5)

Using ( 1), (4), (5),

$\sqrt{ Lg \cos{\theta_0} } \cos{\theta_0} = \sqrt{ (L+\delta )g \cos{\theta_1} }\cos{\theta_1}$ .....(6)

Further calculation doesn’t give the answer which matches anyone of the given options.

Is this correct so far?

13. Jan 3, 2018

### haruspex

I do not think you are expected to go through the pages of algebra I went through. Rather, you are expected to choose the appropriate conservation law. But as I wrote, it was not clear to me which one to use. Angular momentum and linear horizontal momentum cannot both be preserved since they lead to different answers.

Slipping "as it goes through the vertical" is quite misleading. It is clear that δ must be an order smaller than the horizontal distance traversed during the slip. It is reasonable to consider that the string is slack during the slip, but there will be a large force (for a very short time) as the string becomes taut again.

Can we make a clear case for horizontal linear momentum being conserved?
If the tug as it goes taut occurs after the vertical then that may have a large negative x component, so instead suppose that occurs precisely at the vertical. But then, the start of the slip must be before the vertical. In the absence of slip, the string would have been accelerating the mass in the +x direction at that stage, so some gain in horizontal linear momentum was forgone during the slip.

What about angular momentum? The string tension does not affect that, but gravity does. This gets messy. We want the time integral of gravity's torque, and the torque changes sign through the vertical.

On balance, I would guess that angular momentum is the better bet, and indeed that answer matches the result from my full analysis.
As remarked, it is affected.

I do not understand your eqn 1. Do you mean 1-cos(θ0)? Shouldn't there be a factor 2 in there? And what is vx, as distinct from v? Isn't your v horizontal velocity already?

14. Jan 3, 2018

### Pushoam

No, I do not mean $1 - \cos{\theta_0}$. Why should there be a factor of 2?

$v_x$ is the horizontal component of velocity at the right end of the swing if there were no slipping.

If there were no slipping, then

$\frac { mv^2}L = T - mg \cos{\theta_0}$ .....(2), $v_x = v \cos{\theta_0}$ .....(3)

I missed the tension force.

If tension force is different at the two ends of the swings due to slipping, then I will have to calculate it before applying the principle that the horizontal momentum remains almost unaffected due to slipping. But, you said that we cannot take change in horizontal momentum due to slipping as negligible.

So, the above approach turns out to be wrong.

It is sure that angular momentum does not get affected due to slipping. Even then the calculation is not easy.

You get the answer by applying the fact that the angular momentum doesn’t get changed by slipping. Right?

Do you suggest me to use the above information to calculate the answer?

15. Jan 3, 2018

### haruspex

It looked like an energy equation, as in your post #1, but now I think you intend a centripetal force equation. However, I still do not understand your v and vx. There is no velocity at either end of the swing.
Yes.