Time of Fall for a Pencil Dropped at an Angle

[Satvik Pandey]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

I[/B]nitially the rod makes 0.001 rad with vertical.
So it makes 89.947 deg with horizontal.

Torque 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?

 

[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".

 

[Orodruin]

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.

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?

 

[voko]

Can you think of something that multiplied with the RHS of your equation makes the RHS on this form?

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

 

[Orodruin]

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

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 ...

 

[ehild]

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

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

 

[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?

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

Oh! I missed 2 in denominator.

I would name the angle enclosed with the vertical as theta. In that case, $\omega^2 =\frac {3g}{l}\left(\cos(\theta)-\cos(\theta_i)\right)$...(1)
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 $\omega^2=\frac{3g}{l}(1-cos(\theta))$...(2)
Here is the next trick: $1-cos(\theta)=2 \sin^2(\theta/2)$
Can you continue from here?

ehild

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

 

[ehild]

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?

Yes.

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

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

 

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

I got (-)ve sign inside root?

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?

 

[Satvik Pandey]

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

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?

 

[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).

 

[Satvik Pandey]

Is equation in #post 10 correct?

 

[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

 

[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?

 

[ehild]

Yes, it looks right. Integrate.

 

[Satvik Pandey]

Yes, it looks right. Integrate.

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

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?

 

[ehild]

Yes it will be correct. I do hate sec and cosec. How much nicer it would be $2 ln(\tan(\Phi/4) )\big | _{0.001}^{\pi/2}$

http://www.wolframalpha.com/input/?i=int(dx/sin(x/2)) Evaluate.

 

[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?

 

[ehild]

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

ehild

 

[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!

 

[ehild]

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

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

 

[voko]

It looks nonsense.

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.

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.

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).

 

[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

 

[voko]

The result of Satvik and mine are about the same

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

But it deviated from vertical by almost 0.2°-s.

The experiment I suggested was illustration for the problem, not a precise measurement.

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.

 

[ehild]

Stand the pencil on the peak :)

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

 

[Orodruin]

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

I now did it independently without neglecting the contribution from the initial angle in the integral (corresponding to having zero velocity as the initial condition rather than the small velocity given by the fall from exactly zero angle). I obtained 0.66 s, which seems close enough to what ehild and satvik got. The dependence on the initial angle is logarithmic in the angle with a prefactor of roughly 0.08 s. Thus, to get 280 s extra you would need to be roughly a factor $e^{280/0.08} \simeq 10^{1520}$ closer to absolute vertical ... (of course assuming the idealised situation) Due to the logarithmic behaviour, it does not really matter so much exactly how close to vertical you start out - a ten times smaller angle still gives a time less than a second. (If you experiment with it as ehild suggests, you will end up with the pen falling in a time which is typically less - but not much less - than a second.)

 

[ehild]

Thank you Orodruin!

ehild

 

[Satvik Pandey]

Thank you ehild,Orodruin and voko for helping me!:)