Maths on Friday

P.S. This was inspired by a near miss with a toilet seat

was anybody hurt?

Is one degree enough to have the moment of enertia offset enough it to make it fall?

I say no!

Now if we were talking about toilet seat instead of a steel bar........

I'd say that 1° would be plenty to get it moving (remember we're neglecting friction). In fact in an ideal situation it would move if the rod were anything other than vertical.

http://www.youtube.com/watch?v=c86QENS2Uq4

Can take longer then you think with no more near misses.

Can take longer then you think with no more near misses.

But will I have time to finish the job?

I have a "slow closer" at home - tho' it moves pretty swiftly for the first 60° or so. The "near miss" was at work.

You need one of these....

Automatic toilet seat.... one wave = lid opens....another wave = seat opens...auto close in 15 sec...

You need a secondary one that lifts the seat and TO PUT IT DOWN AFTER YOUR DONE USING IT.

This is going to save a lot of marriages.

Toilet automation can be uncomfortable.

eg the toilets at Changi airport autoflush when a client gets up off the seat however they do have a tendency to false trigger and initiate a flush when one leans forward just a bit.....not fun. You can all picture it for yourselves.

Anyhow, default position for a toilet seat is "raised" and should be held there by a detent. No more problem.

42 henways

Calculating this seems like too much work.

However, it stimulated some thought and a search. Chimneys break around the middle when they fall, as demonstrated in this video: http://myweb.lmu.edu/gvarieschi/chimney/toy2m-forweb.mov

Here's why that is.

So your bar will flex in the middle as it falls (imperceptibly of course, if shaped as you've drawn it.)

Imagine what would happen if you had a four foot long piece of uncooked spaghetti.

Had you asked for the rotational velocity of your steel bar, I could have stolen this solution (suitably modified).

Perhaps extra credit could be given for times starting from .25 degrees, 1 degree, 10 degrees and 45 degrees from vertical.

yeah.... do your own homework....

I think the mass is not important, but we do need to know the length.

There are two forces at play here, one is gravity pulling down and another is the force of the pivot point acting on the lower end of the rod. Since the rod rotates, a portion of the downward gravitational force is converted into rotational torque, effectively slowing down the rod's fall time. However, the top of the rod end will actually be traveling faster (angular velocity) than a normal free falling object.

The angular acceleration, a = (3/2)(g/L) cos(rho). This assumes that the initial value of rho is 89° before release.

However, we are solving for the point in time when the rod is horizontal or rho = 0.

If we have the acceleration we can rearrange for t from v = a * t = {(3/2)(g/L)cos(rho)} t, v is the angular velocity.

So, the initial angle is 1.571 radians. the time to fall should be equal to a final angle of 0° with the total arc = 1.571 radians. so when rho final = 0 = (1.571) - (1/2)(angular acceleration) t^2

Rearrange the equation and solve for t:

-1.571 = (1/2)(angular acceleration) t^2

t = sqrt[(1.571*2)/(angular acceleration/2)]

t = sqrt[3.142((2/3)(g/L)]

Probably screwed up something...

The 89° arc makes this a little awkward.

I used a stick about 40 cm (.4 m) long, and it took about 1 sec to fall from nearly vertical. The calculation using your formula yields 7 sec.

So there must be an error there, but I'm on vacation and have my brain mostly turned off. Too dumb and lazy right now to review your work and offer any helpful suggestions.

/Good try, though.

"The 89° arc makes this a little awkward."

That's cool - call it 90° - just imagine someone gave it a miniscule shove (call it a perturbation) to get it started.

Confess I haven't studied your answer (will try tomorrow) - but I suspect the solution invovles some fancy calculus (as you've indicated there is a component of "G" acting, depending on the (varying) angle).

A' first-order' approach could ignore the moment of inertia of the bar (among other things). Think it's a d³θ/dθ³ problem (AKA a 3-pipe job).

Ewe silly bugger JohnDG - don't think you can do it at all without considering the moment of inertia somewhere along - tho' the mass does cancel out before you have to do any arithmetic.

If we model this as the gravitational force acting on the mass at the centroid of the bar or toilet seat or whatever, then the moment induced at the pivot point is going to increase with time- in other words, I don't think you can consider angular acceleration as a constant in this case. The force due to gravity would be constant, but the moment arm would be increasing as the angle decreased from 89º to 0º.

It is too late on Friday night to work through the dynamics of this...and the wife awaits...

a clock maker from the 17th century could answer this question without your confusing numberic jibberish.

save it for nasa. sorry, i just remembered, nasa is'nt interested either.

Ok, I'm looking at the initial and final potential and rotational kinetic energies (PE and RKE).

The initial PE is m g h, where h=half the length of the bar. Call the length of the bar L.

Then PE_i = m g (L/2).

The final PE is zero.

The initial RKE is zero. The final RKE_f = 1/2 (m (L²/3) ω_f²)

Equating RKE_f to PE_i and cancelling terms I get:

1/6 L² ω_f² = g L/2; so ω_f = √ (3 g/L)

Then assuming the average ω is half the final ω_f, and since the rotation is through θ=1/2 pi radians, and further assuming that L=0.4 m.

t = θ⁄ω

I get t = (pi/2)/[0.5(√(3×9.8⁄0.4)], or t = 0.366 sec.

Of course, I could be wrong, too. Likely at this late at night (~ 11 PM) I've made a mistake somewhere.

A quick check with a .6m bar shows it drops in about 1.2 second ±.20 seconds.

The same calculation with your formula is .45 seconds. While much closer than mine, it seems that it still is wrong.

If the arc is 90° from vertical I still get t = SQRT[ {(pi)(2/3)(L/g)} ], which is .55s seconds for my bar.

Someone will spank us with the right answer.

My Excel model comes up with 0.81 second for a 0.6m bar .

Stop press! The starting angle is very important: I get 1.13 seconds (0.6m bar) if I start from 90° (and give it a shove to start it moving ).

That sounds good!

Yes, starting angle makes a huge difference.

Get a toilet seat made of wood or plastic,missing the bowl could start rusting in steel.

Bazzer

I guess you need to take moments of the force of gravity acting on the center of gravity about the fulcrum. This will give the force of the turning moment which will increase in proportion to the sine of the angle from the vertical. From which the incremental acceleration and velocity can be calculated. and hence the time per degree

A bit of integral calculus will given the total time. But I'm pretty sure the air pressure under the seat will have a deceleration effect that increases with angle (otherwise parachutes would not work) so we will need to factor in the area of the seat.

My maths is too rusty to do the sums myself.

Assuming length 0.4m (about right for a toilet seat) I make it 0.865 sec. Had to use numerical integration.

I did an Excel model with 0.5° steps, and got 0.382 seconds for a 0.4m bar. Fully prepared to be wrong! (Tho' it's pretty close to USBport's answer).

New result: 0.540 seconds (forgot that the force is acting at (L/2), not L). So I'm somewhere between Usbport's and Codemaster's results.

Never mind - nearly time to go to the pub! .

I'd like to post my workings, but I can't get the hang of copying images, it seems overly complicated. But it's not hard to write the equation of motion. The neat bit is when you get dω/dt as a function of θ, put dω/dt = dθ/dt*dω/dθ = ω*dω/dθ. That integrates OK to give ω² = 3g/L*(cos 1° - cos θ) but the next step to get t as a function of θ needs numerical integration.

For various initial angles, I (with help of Mathcad) get time, sec

1° - 0.865

0.1° - 1.25

0.01° - 1.62

0.001° - 2.00

0.0001° - 2.38

0.00001° - 2.76

0.000001° - 3.06

For smaller angles, Mathcad overloads. But until that point, the increase in time with decreasing angle is surprisingly slow, so if anybody can fault the maths.......? BTW, for different lengths, time varies as √L.

On the subject of toilet seats, the ones that irritate me are when the seat hits the cistern before going overcentre, so it doesn't stay up at all. It's quite acrobatic holding it up with one foot while doing the biz!

Agree up to 0.00001°. VB6 claps out if the counter in a for...next loop is greater than 16,777,215 (= 2²⁴-1) so I can't go any further (without pulling some tricks) - and anyway the run time is getting excessive.

May see if VB10 is any better tomorrow (don't have it loaded on my home pc).

OK, good to see 2 different approaches give similar results

0.00001° - 2.76

0.000001° - 3.06

For smaller angles, Mathcad overloads. But until that point, the increase in time with decreasing angle is surprisingly slow, so if anybody can fault the maths.......?

I can't fault the maths. I agree that the increase in time with decreasing angle is slower than I would have guessed. However, I tried balancing a broom handle quite carefully in my shop, and even given its longer length, it was still hard to get it to "just stand there" for more than 2 seconds before getting into the "obviously falling" stage.

What if you do cos1º≈1,cosΘ≈1-Θ²/2.then for so small angles log(Θ1/Θ2)appears..

Good idea, I'll have a look at that when I have a minute.

I've re-worked my Excel model in VB (and corrected an error in the calculations). For a 0.4m bar falling from 89°, with 10000 (or more) increments I get 0.865 sec. I reckon you're right.

1)mg½lsinΘ=Idω/dt , ω=dΘ/dt I≈(ml²)/3; 2)mg½lωsinΘ=Iωdω/dt ;3)mgcosΘ+I½ω²=const ; 4)√((k-mgcosΘ)/I)=dΘ/dt;5)Δt=∫(dt/dΘ)dΘ=∫1/(√....)dΘ;this last eq.must be calc.by numerical methods..no for saturday..

The time taken to fall will be dependent on the length ie (height ) of the arm.

It will also fall faster if you've had a bevvy or 6.

If you are holding a can of beer in your, err, other hand, then it will be faster than you really want and will lead to a panicked solo sword fight.

Also,get your willy out of the way before it slams shut,Owwww.

Bazzer.

Heck this is easy!

Time to fall (not looking) = 0

Time to fall (watching) = about one second

Time to fall (watching with baited breath) = Forever + one day

See how easy that was?

This is horribly complex for a weekend problem. I'm just going to print the thread off, and show it to Mrs K as justification for leaving the toilet seat up .

I think with help of GRAPHCALC i got the results:Δt(10º-90º)=1.839√(l/g), that means .455 sec for l=.6m, and .372sec for l=.4m.That's using: Δt=(√(l/3g))∫dΘ/√(1-cosΘ), cosΘo≈1. For case 0‹Θ‹10º: Δt=.816•(√(l/g))•ln(Θ1/Θ2), so Δt(5º-10º)=.18√l=.11sec for.4m and .14sec for .6 m.Anyway Δt is infinite if you start in 0º not only improper integral and is natural because it could stay in that pos.for ever (therically).-

This whole problem goes away if we could just convince our ladies not to install fluffies on the toilet seat when there is a resident male...

Hey! That's just a shameless attempt to get a GA!

I'd give him one, except the problem goes away if the seat is held back by elastic bands.

about .5 sec before you realize you may never have kids again.

The numerical integration gives following results:

start angle 1° time to reach 90° 1.06 s end rotation speed 6.89 rad/s

start angle 2° 0.92 s and final velocity 6.88 rad/s

Time is only function of gravity and bar length (assuming that mass is uniform over the length).

Bar length is the CG distance so uniform mass distribution isn't an issue. If it is bottom heavy then the resultant arm length is shorter and vice versa,

What bar length? Back-calculating I make it ~ 0.6m.

I think the bar length is supposed to model a hypothetical dunny lid.

Original post said a rectangular bar, inspired by a toilet seat. Time of fall doesn't mean much without giving the length it's based on.

When I first posted, I hadn't done any "back of envelopes" or even thought much about it. I simplified it to a steel bar, to try to avoid distractions relating to density, air resistance and whatever else could have been thrown in (despite my direction to neglect friction).

As I hadn't thought about it, I didn't know (at that stage) whether or how the length, mass or anything else would affect the time to fall. Having scratched out some sums, I now have a reasonable idea of what's going on (which probably should've been obvious, but wasn't initially - I claim a "senior moment" _{never blame the beer} ).

I did ask for assumptions made to be declared (this would clearly include the length).

We've had 400mm suggested as a reasonable length (relative to the toilet seat which really did trigger the thoughts), and I think 600mm came in from AH's experiments.

Maybe I should've dropped the toilet seat reference. Just thought some context would make it a bit more of a "real world" problem.

Don't drop the toilet seat (!), the 'added value' of humerous situations has held my attention - not that it's helped me come up with a solution, but I'm having a fun time pondering all this. For reasons best unexplained, I've found myself distracted by roller coasters (with people falling out), pendulums, and assorted futile paths.

If AH doesn't mind, could we stick with the 40cm (ie mass at a radius of 20 cm) so as to avoid confusion ?

Problem with that is it's not equivalent to a bar. For mass m and length L, MoI = m*(L/2)² = m*L²/4, but for a bar MoI = m*L²/3.

Well now you've scrambled my eggs !

There's a nice simplicity to the question, which is much more intriguing than first inspection would suggest. The type of thing that I'd have expected to be in standard texts - but it isn't (not directly as John posted).

Really?

Yes, really Not sure what point you're making

My ignorance perhaps. I don't get the division by 3 instead of 2 bit.

Does that follow from MoI = m*(L/√3)² perhaps?

That's right, MoI of a rod about one end m*(L/√3)² = m*L²/3. (don't usually bother with the (√3)² = 3 part ). Not hard to work out if you have a smattering of calculus. For a point mass at distance L/2 it's m*(L/2)² = m*L²/4.

This is a pendulum on its side.

The travel of the falling pendulum is arrested abruptly approx quarter way through its travel if starting from nearly vertical.

The time to reach that point would be quarter of the natural pendulum period (loss-less pendulum)

pendulum period

L=0.4m, g=9.8ms^-2

gives 1.269 seconds

Quarter of that is 0.32secs. for L =0.6 we get 0.39secs

Notes:

• 89°/358° is little less (not much) than 1/4 for 1° start offset, (good enough for this illustration)
  
• if we started at say 5° then the quotient would be 85°/350°, 10° would be 80°/340°.....
• If you want to go with minute start offsets of fractional degrees then the quotient just asymptotes to 1/4 any way.
• if you dropped the lid from say 1° above horizontal then the quotient would be 1°/182° and this extreme asymptotes to zero.
  

..... is only an approxmation for small angles.

It was that (roughly 5^o) limitation the left me headed for the roller coasters - have the thing enter a (circular, not clothoid) loop with just enough velocity to reach the top.

Starting angle 2° from vertical.Equations are: J= M*l^2/3 valid which ever comments are ONLY for an uniform mass distribution which gives dM=M/l*dx and J= ∫dM*x^² = M/l*∫x²*dx. If not J= ∫dM*x² and has another equation.The second equation d²φ/dt²=1.5*g/l*sin(φ) can be for easier integration written asdω/dt=1.5*g/l*sin(φ) and dφ/dt=ω.

Angle in radian versus time in s for start between 0.005 and 0.035 rad. Graph is limited at π/2 which corresponds to 90°. In this graph the angle is measured with respect to vertical. The 1st graph had a shift of 90°.
Evolution of rotation speed for same start values in rad/s.

Nice - was it a maths package? Too hot and too far away from my desk to verify your calcs - on holiday in Montpellier, France.

I take it from what everyone's come back with that there's no analytical solution.

I'll desist from calling you a sun-lounging ***. No matter which way I slice this, I can't see an easy way to solution. As good as the input from everybody is, I have a nagging feeling that we miss something. The complexity from such a seemingly straightforward question is amazing.

It's a good one for a school teacher to use. The students will not get it right, but wow will they learn a lot along the way. Thanks, John, it's a superbly provoking question.

May we move on to some further ananytics - towels on sun loungers at 7 a.m, .