Another Physics Problem

[Borodog]

Consider a physical pendulum, a rigid rod of length L and mass m. Where do you have to place the pivot point to maximize the period? You may restrict yourself to small angles.

[Borodog]

Whoops; typo. Minimize the period; maximize the frequency.

[danny]

L/sqrt(12) above the midpoint.

[Borodog]

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L/sqrt(12) above the midpoint.

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Bingo.

[arahant]

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L/sqrt(12) above the midpoint.

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Bingo.

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Well that's surprising...care to sketch out the reason for those of us who don't wish to reprise hs physics?

[Borodog]

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L/sqrt(12) above the midpoint.

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Bingo.

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Well that's surprising...care to sketch out the reason for those of us who don't wish to reprise hs physics?

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Later, If I have the time. It's a straightforward application of torque and calculation of the proper moment of inertia.

[Siegmund]

Mathematicians who wander into this thread will find it helpful to recall that a uniform distribution on [0,1] has variance 1/12... and there's a reason we refer to a variance as a second moment.

[arahant]

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Mathematicians who wander into this thread will find it helpful to recall that a uniform distribution on [0,1] has variance 1/12... and there's a reason we refer to a variance as a second moment.

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Ah, ty...that's good enough.

[Borodog]

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Mathematicians who wander into this thread will find it helpful to recall that a uniform distribution on [0,1] has variance 1/12... and there's a reason we refer to a variance as a second moment.

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whoosh
------->
<font color="white">..</font> /images/graemlins/shocked.gif

[arahant]

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Mathematicians who wander into this thread will find it helpful to recall that a uniform distribution on [0,1] has variance 1/12... and there's a reason we refer to a variance as a second moment.

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whoosh
-------&gt;
<font color="white">..</font> /images/graemlins/shocked.gif

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Really? It's the same math, isn't it?

[Borodog]

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Mathematicians who wander into this thread will find it helpful to recall that a uniform distribution on [0,1] has variance 1/12... and there's a reason we refer to a variance as a second moment.

[/ QUOTE ]

whoosh
-------&gt;
<font color="white">..</font> /images/graemlins/shocked.gif

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Really? It's the same math, isn't it?

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Damned if I know. I calculated the moment of inertia I of the rod about some point a distance x from one end, then stuck it into the torque equation (torque = |r x F| = I alpha), substituted in for what the lever arm and whatnot are, recovered an equation of the form theta'' = -omega^2 theta (where the prime denotes a time derivative), and maximized omega^2(x).

[Borodog]

I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[gumpzilla]

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I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[/ QUOTE ]

Moment of inertia is the integral of dm over r^2, right? Add all the mass points weighted by distance r^2 from the pivot? Well, that looks an awful lot like variance if you replace pivot by "mean."

[ApeAttack]

Here we go. I spent an hour pouring over this problem... stupid algebra.

For a physical pendulum, period = T = 2*pi*sqrt[I/(m*g*D)]

where I = moment of inertia,
m= mass of pendulum,
g = grav constant,
D = distance from pivot pt to center of mass


In general, I = integral from 0 to M {r^2 dm}

where r = distance from pivot.

dm = density * dr = M/L * dr

We need to adjust the limits of integration. The system is described as follows.

I imagined the bar on its side when setting up my system. The pivot pt is at zero and the short end is in the 'negative' side of the x-axis.
The center of mass (with is at 1/2 the length of the bar) is 'D' distance below it.
The short end of the pendulum is at -L/2 + D from the pivot.
The long end of the pendulum is at +L/2 + D from the pivot.


So the limits of integration now go from -L/2 + d to L/2 + d

Integrate {M/L r^2 dr}

I = M/L * r^3/3 and evaluate at the limits.

M/3L = Y (to make life easy)

I = Y * [ (L/2 + D)^3 - (-L/2 + D)^3 ]

You can show (with a crapload of algebra) that,

I = Y*2*( (L/2)^3 + 3*(L/2)D^2)


Back to the period,

T = 2*pi*sqrt[I/(m*g*D)]
dT/dD = 0 = d(I/D)/dD (I've set the derivative to zero to find the minimum and I've already eliminated the constants variables... also, sorry for the bad choice of variable, 'D')

d(I/D)/dD = d[ (L/2)^3 /D + 3*(L/2)*D] / dD = 0
-L^3/8D^2 + 3*L/2 = 0
D = L/sqrt(12)

I may have messed up a sign somewhere in the derivation, but it would be cancelled out later in the derivative step.

To show that it is a maximum, do a second derivative test.

[Borodog]

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I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[/ QUOTE ]

Moment of inertia is the integral of dm over r^2, right? Add all the mass points weighted by distance r^2 from the pivot? Well, that looks an awful lot like variance if you replace pivot by "mean."

[/ QUOTE ]

How exactly is that relevant? Isn't the mean of a uniform distribution on [0:L] at L/2, which is not where the pivot that minimizes the period is, which is at L/sqrt(12) above the midpoint, as already mentioned? The moment of inertia of a uniform rigid rod of length L and mass m about it's midpoint is indeed mL^2/12, but what does that have to do with the problem?

[Borodog]

Ape,

Good enough.

[Neuge]

[ QUOTE ]
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I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[/ QUOTE ]

Moment of inertia is the integral of dm over r^2, right? Add all the mass points weighted by distance r^2 from the pivot? Well, that looks an awful lot like variance if you replace pivot by "mean."

[/ QUOTE ]

How exactly is that relevant? Isn't the mean of a uniform distribution on [0:L] at L/2, which is not where the pivot that minimizes the period is, which is at L/sqrt(12) above the midpoint, as already mentioned?

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You're confusing the terminology. The rod midpoint is the mean (1/2), and the moment of inertia is the variance (1/12). The variance of a uniform distribution on [a,b] is (b-a)^2/12 (obv L^2/12 for a rod on [0,L]), which is called the second moment. Multiply that by mass to get its moment of inertia, mL^2/12.

[Borodog]

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I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[/ QUOTE ]

Moment of inertia is the integral of dm over r^2, right? Add all the mass points weighted by distance r^2 from the pivot? Well, that looks an awful lot like variance if you replace pivot by "mean."

[/ QUOTE ]

How exactly is that relevant? Isn't the mean of a uniform distribution on [0:L] at L/2, which is not where the pivot that minimizes the period is, which is at L/sqrt(12) above the midpoint, as already mentioned?

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You're confusing the terminology. The rod midpoint is the mean (1/2), and the moment of inertia is the variance (1/12). The variance of a uniform distribution on [a,b] is (b-a)^2/12 (obv L^2/12 for a rod on [0,L]), which is called the second moment. Multiply that by mass to get its moment of inertia, mL^2/12.

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I'm not confusing anything. He said "that looks an awful lot like variance if you replace pivot by 'mean.'" You can't replace "pivot" by "mean" because the pivot is not where the "mean" would be. The "mean" is in the middle, and the pivot ain't.

I'm still waiting for this to get me to where the pivot has to be to minimize the period, since the moment of inertia you've calculated is incorrect for the problem.

[Neuge]

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I'm not confusing anything. He said "that looks an awful lot like variance if you replace pivot by 'mean.'" You can't replace "pivot" by "mean" because the pivot is not where the "mean" would be. The "mean" is in the middle, and the pivot ain't.

I'm still waiting for this to get me to where the pivot has to be to minimize the period, since the moment of inertia you've calculated is incorrect for the problem.

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Ah, sorry. I misread your post, or his, I'm not sure which.

Either way, changing the pivot point from the midpoint makes this a non-uniform distribution because it changes its mean from L/2. You have to recalculate the probability density function (the mass isn't evenly distributed around the midpoint anymore) from which you determine the second moment. Once you have the probability density function, determine the moment generating function (http://en.wikipedia.org/wiki/Moment-generating_function) to calculate the variance (i.e. the moment of inertia about the pivot point).

This is the equivalent of what you are doing when you integrate r^2 over the length of the rod to determine the moment of inertia.

[jason1990]

According to ApeAttack's post, the period is

T = C sqrt{ E[(t - X)^2]/(t - E[X]) },

where t is the position of the pivot, X is uniform on (0,L), and C is some constant. In general, for any random variable with a second moment, the minimum of

E[(t - X)^2]/(t - E[X])

over all t &gt; E[X] occurs when t is equal to E[X] + \sigma, where \sigma is the standard deviation of X. To see this, let m = E[X] and set s = t - m. Then the above expression can be rewritten as

E[(s + m - X)^2]/s = (s^2 + \sigma^2)/s.

Elementary calculus shows that this is minimized when s = \sigma.

[gumpzilla]

[ QUOTE ]
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I would really like to see the physics behind this second moment/variance of a uniform distribution stuff, if anyone can produce it.

[/ QUOTE ]

Moment of inertia is the integral of dm over r^2, right? Add all the mass points weighted by distance r^2 from the pivot? Well, that looks an awful lot like variance if you replace pivot by "mean."

[/ QUOTE ]

How exactly is that relevant? Isn't the mean of a uniform distribution on [0:L] at L/2, which is not where the pivot that minimizes the period is, which is at L/sqrt(12) above the midpoint, as already mentioned? The moment of inertia of a uniform rigid rod of length L and mass m about it's midpoint is indeed mL^2/12, but what does that have to do with the problem?

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Then it's quite possible that the whole thing is just a coincidence. The comments that the variance of the unit interval is 1/12 fits in with the latter fact, not the minimization problem, I would assume.

[jason1990]

I do not think it is a coincidence. Try the problem under the assumption that the rod has a variable density f(x) which is symmetric about L/2. I suspect the optimal pivot location will be s units above the midpoint, where s is the standard deviation of a random variable whose density is proportional to f.

Edit: In fact, I believe the symmetry assumption on f can be dropped. If m is the center of mass and s^2 is the moment of inertia about m, then at least one of the points m + s and m - s will actually be on the rod. These points should give the optimal pivot location(s). Note that my physics knowledge is pretty meager, so it would be nice if a physicist checked these claims.

[Borodog]

[ QUOTE ]
According to ApeAttack's post, the period is

T = C sqrt{ E[(t - X)^2]/(t - E[X]) },

where t is the position of the pivot, X is uniform on (0,L), and C is some constant. In general, for any random variable with a second moment, the minimum of


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I am so totally not seeing where this is comming from.

[jason1990]

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According to ApeAttack's post, the period is

T = C sqrt{ E[(t - X)^2]/(t - E[X]) },

where t is the position of the pivot, X is uniform on (0,L), and C is some constant. In general, for any random variable with a second moment, the minimum of


[/ QUOTE ]

I am so totally not seeing where this is comming from.

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From ApeAttack's post:

[ QUOTE ]
For a physical pendulum, period = T = 2*pi*sqrt[I/(m*g*D)]

where I = moment of inertia [about the pivot pt, t],
m = mass of pendulum,
g = grav constant,
D = distance from pivot pt to center of mass

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The density of the rod is f(x) = m/L, so

I = \int_0^L (t - x)^2 f(x)dx = m E[(t - X)^2], and
D = t - (\int_0^L x f(x)dx)/m = t - E[X].

Hence,

T = (2pi/sqrt{g}) sqrt{ E[(t - X)^2]/(t - E[X]) }.

[Borodog]

Ah, thanks. Got it.

[danny]

jason,

You are correct, it is not a coincidence, and the symmetry requirement on the distribution is not needed.

If we define the mass distribution as m*f(x), then we can solve the problem using the Lagrangian equations of motion.

If the pivot point is a distance d above the center of mass, then the potential energy can be expressed as:

V = -m*g*d*cos(theta)

The kinetic energy can be expressed as the sum of two terms, the kinetic energy of the motion of the center of mass, (m*d^2)*(dtheta/dt)^2/2, and the kinetic energy of the rotational motion about the center of mass, I*(dtheta/dt)^2/2, where I is the moment of inertia about the center of mass.

T = (m*d^2 + I)*(dtheta/dt)^2/2

As mentioned above, the moment of inertia about the center of mass is just the total mass times the second central moment of the mass distribution, otherwise known as the variance (=sigma^2).

T = m*(d^2 + sigma^2)*(dtheta/dt)^2/2

The Lagrangian is then formed by L=T-V, and the equations of motion can be found.

m*(d^2 + sigma^2)*(d^2theta/dt^2) = -m*g*d*sin(theta)

For small theta, this is simple harmonic motion with frequency given by

omega^2 = g*d/(d^2 + sigma^2)

Maximizing this with respect to d yields d = sigma.

[Borodog]

Danny,

Nice work.

Edit to add: Although I think Lagrangian dynamics might be a bit of overkill for the problem. /images/graemlins/wink.gif