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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

imagine a long bar horizontally placed in the air, pivoting it at its centre its not going to accelerate (angular) at all since weight acts through the pivot

now, if you were to move the pivot away from the centre of mass along the bar, you would expect it to accelerate more as distance increases.

however, infact what happens is it increases as you'd expcet, then reaches a maximum at which point it begins to converge to 0 as distance increases to infinity.

if the mass of the bar is M and acceleration due to gravity g, and its moment of inertia about centre of mass is I (treating as 2D), and the distance of the pivot from centre of mass along bar horizontally is 'x' then you have the torque produced by the bar's weight equal to Mgx, larger torque = larger angular acceleration by tourque = moment of inertia × angular accleration, however moment of inertia is not constant, as you move the pivot away from the centre of mass it increases, equal to I + Mx² (Parallel axis theorem), so the angular accleration giving by the weight when the bar is horizontal is:

radians doesn't have a unit, so this can represent angular acceleration, so units atleast are fine.

differentiating with respect to x you get:

units of this: thats also fine.

equating this to 0 for the stationary points gives:

(allowing negative x to mean the other side of the centre of mass)the units here:

so thats also fine, x should be in metres, units match up fine here.so, the maximum angular acceleration that can be achieved for a horizontal bar by pivoting it and allowing its weight to act is when x = ±√(I/M) giving a value of:

units of this: so units are fine

the minimum acceleration occurs at x = 0, ±∞, for which a = 0 (loose terminology of infinity)

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this is just for horizontal position of bar, however the acceleration is proportional to cosθ where θ is angle made with horizontal so if one pivoted position gives larger acceleration at horizontal then another position, it will always have larger acceleration at any angle than the the position, so the acceleration at horizontal, and the time it would talk to swing past vertical are directly related.

at some angle theta, the acceleration at the point in time at which the bar makes that angle would be (generalising the above coincidently)

I do not think it is possible to get an exact formula for the time it would take to pass the vertical state, since acceleration and angle are dependant on eachover, perhaps someone can correct me if this is not true.

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Now, is it just me or does this seem just a tad bit weird? I dunno it makes sense thinking about it, but at the same time seems unnatural.

I need to quiz my physics teacher on this tomorrow and see if i can do an experiment to justify it

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in both cases for distance for maximum acceleration, and maximum acceleraiton, mass does not matter since moment of inertia is Kgm^2 and mass Kg, I/M and M/I are functions of area with no mass, so both of these things are only effected by the shape of the object and its density field

*Last edited by luca-deltodesco (2007-10-29 08:13:01)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

using a very small timestep in a RK4 integrator of 0.0005s ive approximated the time it would take for the 'bar' or whatever it may be to rotate through vertical from horizontal as distance of pivot from centre of mass changes.

Blue is the shape of graph given for time taken

Red is the usual acceleration at horizontal, in this case, acceleration at t = 0.

so relating this to sometihng like a pendulum, if you treat the pendulum as a rigid body its likely that any length of string linking the pivot to the ball would be far over the limit of sqrt(I/M) (this distance would probably be inside of the ball really) and so you would see the outer sides of the blue graph, where as 'x' increases 't' increases, with something like a meter ruler, you would get the whole of the blue graph (something im going to test)

supposing you had a metre ruler of dimensions 1m * 8cm (forgetting depth) the distance for the pivot to be from centre of mass to give the shortest spin time to vertical is approximately 28.95cm from middle so this should be easily investigatable supposing i find a good way of pivoting the rule about any point with minimal friction

*Last edited by luca-deltodesco (2007-10-29 08:57:29)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

heres another interesting thing:

if i start the graph for time at some other angle than horizontal, the graph for the side at which it rotates less is almost identical.

the left side is obviously quite high up and off the plot.

the blue plot is starting at 0 radians, (horizontal) the pink plot is starting at 0.499999999pi (0.5pi being the target rotation of vertical)

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

hmm, perhaps i can get some help with this, i've had a go at approximating a function for the time it will take to pass through vertical for the bar being horizontal, and i've found that this equation does it almost perfectly, the only find tuning needed is in the first constant

*Last edited by luca-deltodesco (2007-10-29 23:21:39)*

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**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

i employed the help of one of my maths teachers who came up with this:

let θ[sub]0[/sub] be the intial value of θ at t = 0, and w = 0 for t = 0. θ be the final value at some time

then the time taken to go from θ[sub]0[/sub] to θ is (taking out all the constants of the equation into a single variable 'a')

i've tried integrating this numerically ontop of my previous graphs of integrating numerically the system of equations i started with and its correct.

it can't be solved analytically, and plugged into wolfram's integrator gives:

where F is the incomplete elliptic integral of the first kind

o.O

*Last edited by luca-deltodesco (2007-11-12 01:37:03)*

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