Homework Help: Time required by rod to fall

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1. Oct 6, 2014

Satvik Pandey

1. The problem statement, all variables and given/known data
Estimate the time it takes a pencil of length L=10 cm to hit the ground if initially the pencil makes an angle of 0.001 rad with the vertical. Take the gravitational acceleration to be 10m/s2

Details and assumptions

Assume that the pencil is just a thin solid rod with uniform mass distribution along its length. Also to simplify the problem, assume that the pencil's contact point with the ground (the tip) is fixed.

2. Relevant equations

3. The attempt at a solution

I
nitially the rod makes 0.001 rad with vertical.
So it makes 89.947 deg with horizontal.

T
orque acting along point O is $mgcos\theta \frac { l }{ 2 }$ [I will change the units of l at end]
So

$\frac { dL }{ dt } =mgcos\theta \frac { l }{ 2 }$

or $I\frac { d\omega }{ dt } =mgcos\theta \frac { l }{ 2 }$

or $I\frac { { d }^{ 2 }\theta }{ { dt }^{ 2 } } =mgcos\theta \frac { l }{ 2 }$

But I don't know how to solve this second order differential equation.

So I tried this

$I\omega \frac { d\omega }{ d\theta } =mgcos\theta \frac { l }{ 2 }$

or $I\frac { { \omega }^{ 2 } }{ 2 } =mgsin\theta \frac { l }{ 2 } +C$

At $t=0$ $\omega=0$ and let $\theta ={ \theta }_{ i }$

So $C=-mg lsin{\theta}_{i}$

So $I\frac { { \omega }^{ 2 } }{ 2 } =mgsin\theta \frac { l }{ 2 }-mg lsin{\theta}_{i}$

or $\frac { m{ l }^{ 2 } }{ 3 } \frac { { \omega }^{ 2 } }{ 2 } =mgsin\theta \frac { l }{ 2 } -mgsin{ \theta }_{ i }l$

or ${ \omega }^{ 2 }=\frac { 3\left\{ gsin\theta -2gsin{ \theta }_{ i } \right\} }{ l }$

or $\frac { d\theta }{ dt } =\sqrt { \frac { 3\left\{ gsin\theta -2gsin{ \theta }_{ i } \right\} }{ l } }$

or $\frac { \sqrt { l } }{ \sqrt { 3\left\{ gsin\theta -2gsin{ \theta }_{ i } \right\} } } d\theta =dt$

or $\sqrt { \frac { l }{ 3g } } \int _{ { \theta }_{ i } }^{ 0 }{ \frac { d\theta }{ \sqrt { \left\{ sin\theta -2sin{ \theta }_{ i } \right\} } } } =t$

Am I right till here?
How to solve this integral?

2. Oct 6, 2014

voko

You have a sign error in the first equation, which then results in negative kinetic energy and other weirdness. Remember, the usual convention is "counter-clockwise is positive".

3. Oct 6, 2014

Orodruin

Staff Emeritus
Just to add to what voko already said. Do you remember what we did in the chain-slipping-off-the-table thread? (see post #68 in that thread: Chain hanging from a table) That was a slightly more complicated differential equation, but we found a function we could multiply with to make both sides of the form $\dot{y}(t)f(y(t))$, where $f$ is a function. Can you think of something that multiplied with the RHS of your equation makes the RHS on this form?

4. Oct 6, 2014

voko

I believe Satvik did just that, albeit with a typo in LaTeX.

5. Oct 6, 2014

Orodruin

Staff Emeritus
You are right. I guess I was confused by him first saying he did not know how to solve it and then going on to solve it ...

6. Oct 6, 2014

ehild

You made a mistake: the integration constant is C=-mgl/2 sin(thetai)

I would name the angle enclosed with the vertical as theta. In that case, $\omega^2 =\frac {3g}{l}\left(\cos(\theta)-\cos(\theta_i)\right)$
The initial theta is very small and the problem text want you to estimate the time of fall. So you can approximate cos(theta)=1. (Actually, it is 0.9999995 )
So you have $\omega^2=\frac{3g}{l}(1-cos(\theta))$
Here is the next trick: $1-cos(\theta)=2 \sin^2(\theta/2)$
Can you continue from here?

ehild

7. Oct 7, 2014

Satvik Pandey

I have a confusion. Can't we assume clockwise as a positive?
I mean does't it depends upon us whether to choose clockwise positive or anti clockwise positive?
Oh! I missed 2 in denominator.

If $cos\theta_{i}$ is 1
putting this value in eq(1) yields
$\omega^2 =\frac {3g}{l}(cos(\theta)-1)$
whereas your eq(2) is $\omega^2=\frac{3g}{l}(1-cos(\theta))$

8. Oct 7, 2014

ehild

Yes.

I think I missed a minus sign. Check. $\omega^2 =-\frac {3g}{l}(cos(\theta)-cos(\theta_i))$

9. Oct 7, 2014

Satvik Pandey

If we assume clockwise direction (-)ve then
$\tau =-mgsin(\theta )\frac { l }{ 2 }$

or $I\omega d\omega =-mgsin(\theta )\frac { l }{ 2 } d\theta$

or $I\frac { { \omega }^{ 2 } }{ 2 } =mgcos\theta \frac { l }{ 2 } +C$

on putting the value of $C$

I got

${ \omega }^{ 2 }=\frac { 3g }{ l } (cos\theta -1)$

As ehild said I used this $1-cos(\theta)=2 \sin^2(\theta/2)$

${ \omega }^{ 2 }=-\frac { 3g }{ l } (1-cos\theta )$

or ${ \omega }=\sqrt { -\frac { 3g }{ l } 2{ sin }^{ 2 }\left( \frac { \theta }{ 2 } \right) }$

If I choose clockwise direction as +ve then I got

${ \omega }=\sqrt { \frac { 3g }{ l } 2{ sin }^{ 2 }\left( \frac { \theta }{ 2 } \right) }$

It looks good to me.
So should I choose clockwise +ve?

10. Oct 7, 2014

Satvik Pandey

I got ${ \omega }=\sqrt { \frac { 3g }{ l } 2{ sin }^{ 2 }\left( \frac { \theta }{ 2 } \right) }$
if I choose clockwise direction as +ve. Whereas in #post1 voko asked me to choose clockwise direction as -ve?

11. Oct 7, 2014

voko

One can use any convention as long as it makes sense, but the standard convention in mathematics and physics is that angles are measured counter-clockwise. If one chooses a non-standard convention, there must be a good reason for that, and I do not see any in this case.

In the original notation, where $\theta$ was measured from the horizontal, the correct equation should be $$I {d^2 \theta \over dt^2} = - {mgl \over 2} \cos \theta .$$

If now the angle - I will call it $\phi$ - is measured from the vertical, still with the convention "counter-clockwise is positive", then the equation is $$I {d^2 \phi \over dt^2} = {mgl \over 2} \sin \phi .$$

The second equation can be trivially obtained from the first by noticing that $$\theta = {\pi \over 2} + \phi.$$ So there is really no major difference in using $\phi$ vs. $\theta$, but one should stick with one choice and use it consistently (which is not the case in this thread, I am afraid).

12. Oct 7, 2014

Satvik Pandey

Is equation in #post 10 correct?

13. Oct 8, 2014

ehild

It is correct, but to avoid confusion use $\Phi$ instead of $\theta$. If you choose anti-clockwise direction positive for the angles, you need to do the same with the torque. So the torque which rotates the rod in the clockwise direction is negative.
In your original post, theta was measured from the positive x axis in anticlockwise direction. But the torque turned the rod in clockwise direction, so it should have been taken with a minus sign. You missed that minus in the first equation. The correct equation is $I \ddot {\theta}=-mg l/2 \cos (\theta)$ as voko pointed out in Post#11.

If you calculate with the angle from the vertical, and you stick to anticlockwise as positive direction, you can draw the falling rod to the other side of the vertical, and then both the torque and the angle are positive.
The last advice: Always indicate the direction of the angles in your drawings.

ehild

14. Oct 8, 2014

Satvik Pandey

Thank you ehild and voko!

So $\frac { d\phi }{ dt } =\sqrt { \frac { 6g }{ l } } sin\left( \frac { \phi }{ 2 } \right)$
or $\frac { d\phi }{ sin\left( \frac { \phi }{ 2 } \right) } =\sqrt { \frac { 6g }{ l } } dt$
or $\int _{ 0.001 }^{ \pi /2 }{ cosec\left( \frac { \phi }{ 2 } \right) } =\sqrt { \frac { 6g }{ l } } t$

I can integrate the term in LHS.But am I right till here?

15. Oct 8, 2014

ehild

Yes, it looks right. Integrate.

16. Oct 8, 2014

Satvik Pandey

$\int _{ 0.001 }^{ \pi /2 }{ cosec\frac { \phi }{ 2 } } d\phi$

Let $\frac { \phi }{ 2 } =x\quad So\quad \frac { 1 }{ 2 } =\frac { dx }{ d\phi }$

or ${ d\phi }=2dx$

or $2\int { cosec(x) } dx=2log{ | cosec(x)- cot(x)| }$

or $2log{ { | cosec\frac { \phi }{ 2 } - cot\frac { \phi }{ 2 } | } }_{ 0.001 }^{ \pi /2 }$
Is this right?

17. Oct 8, 2014

ehild

18. Oct 8, 2014

Satvik Pandey

I got LHS $14.82$

So $14.82=\sqrt { \frac { 6g }{ l } } t$
L=0.1m

$t=\frac{14.82}{\sqrt{60g}} \approx 0.61sec$
Is this right?

Last edited: Oct 8, 2014
19. Oct 8, 2014

ehild

I got slightly different value for the LHS, but your result is about correct.

ehild

20. Oct 8, 2014

Satvik Pandey

I got this question online on the internet and answer of this question on that website is 280.71
There is no unit even!

21. Oct 8, 2014

ehild

It looks nonsense. Try it with your pen, let it fall from vertical position (it never will be exactly vertical). It takes less than 1 second to fall.

ehild

22. Oct 9, 2014

voko

No, it does not. A rod positioned vertically will stay vertical indefinitely. By varying the deviation from the vertical position, the tumble time can be made arbitrarily long.

This experiment is flawed. It is not possible, with bare hands, to position such small objects at such small angles; even measuring this is difficult. Moreover, at such small angles the lateral dimensions and the shape of the pen are important, because at least certain shapes will have a stable equilibrium (it is possible to make a typical pencil stand on its flat end).

23. Oct 9, 2014

ehild

The result of Satvik and mine are about the same, and hopefully, the maths was correct. Such big difference 0.6 s and 280.71 if it is seconds (unit is not given) the missing unit and the number of the significant digits .suggest that the online source was unreliable.

I know that the rod will stay in the vertical position forever if it is truly vertical. But it deviated from vertical by almost 0.2°-s.

The experiment I suggested was illustration for the problem, not a precise measurement. And it is best to do with a pen or pencil which has rounded end. That corresponds to the situation in the problem, where the thin rod was able to rotate about a fixed hinge.
I always encourage my pupils to do experiments, even then, when they do not quite copy the set-up described in the problem.

ehild

24. Oct 9, 2014

voko

Hardly surprising, after you gave Satvik virtually every step of the derivation.

Again, your experiment cannot demonstrate that an ideal rod can spend arbitrary time tumbling down. The pencil from the problem, assuming the typical diameter 7 mm, will be stable when tilted within 0.07 rad from the vertical, which is 70 times greater than the angle given in the problem. Even without making any assumptions on its shape, no real rod can be used to analyse the behaviour of the ideal rod at angles this small.

25. Oct 9, 2014

ehild

Stand the pencil on the peak :)

It would be nice if you checked satvik's derivation. and the numerical result.

Last edited: Oct 9, 2014