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  #1  
Old 02-15-2000, 02:41 PM
Earl Snake-Hips Tucker Earl Snake-Hips Tucker is online now
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Not straight down, though.

If I drop a ball from 100 feet, for example, I can plug it into the familiar equations (v = 32*t and s = 16t^2) to get how fast it's moving (drag notwithstanding) when it hits the ground.

However, let's say that we have a 100-foot pole, which starts leaning.

It won't fall in a "straight" line. The tip of the pole will actually trace an arc of a circle. How does that affect the "familiar" equations ?
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  #2  
Old 02-15-2000, 03:04 PM
NanoByte NanoByte is offline
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I've been out of school too long to pull up neat equations for this in a reasonable amount of time, but the schema is to have your same gravity act on the center of gravity of the pole, which has a fixed fulcrum on its supported end. This produces a torque composed of a changing component of gravity acting on the lever arm from the fulcrum to the center of gravity, producing a fixed angular acceleration at all points on the pole.

Ray (ex-EE who never needed this, so let an ME or physicist produce the equations)
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Old 02-15-2000, 03:06 PM
frolix8 frolix8 is offline
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This problem is called a "motion in a plane" problem. There is no correct answer without getting into the distribution of weight, but if we simplify the problem by assuming all of the weight is at the top of the pole it becomes an inverted pendulum. The component of the acceleration (gravity) perpendicular to the path is g* cos(theta) where theta is the angle between the pole and the ground. For a vertical pole there is no force, it is balanced. Just before the end of the pole hits the ground the acceleration is g. I leave the rest of the problem as an exercise for the reader, and because it's too hard to type.
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  #4  
Old 02-15-2000, 03:19 PM
stolichnaya stolichnaya is offline
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Hoo daddy. I'm waiting for the infantry on the dynamics, but we have to account for some initial horizontal (in this case angular) velocity here or gravity never gets a chance to begin to work.
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  #5  
Old 02-15-2000, 03:43 PM
frolix8 frolix8 is offline
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A couple of hints:
The moment of inertia of a thin rod of mass M and length L is I = ML<supscript>2</superscript>/3

The problem is best attacked by analysing the potential energy and kinetic energy as the pole falls, rather than integrating the acceleration.
The kinetic energy is given by K=1/2(I*w<superscript>2</superscript>) where w is the angular velocity.
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Old 02-15-2000, 03:45 PM
frolix8 frolix8 is offline
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Crap, my HTML didn't work I'll try again later. Crap Crap Crap. (some of the2's should have be superscripts)
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  #7  
Old 02-15-2000, 03:56 PM
frolix8 frolix8 is offline
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Another hint: the potential energy is 1/2 MLg, if you imagine that the flag pole is allowed to rotate about its center it has the same potential energy when it is vertical as it does when it's horizontal, assuming that it is a uniform rod. If you want to look at a tapered rod it's going to get messy.

Back to HTML drawing board.
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Old 02-15-2000, 04:33 PM
Geezer Geezer is offline
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It ain't that hard. Frolix sneaked in ahead of me, but while he's fooling with HTML, here's the answer:

For a vertical pole of mass m and length h, the moment of inertia is

I = m*h^2/3

and its potential energy is

U = m*g*h/2

At the instant before it hits the ground after toppling, the (now horizontal) pole has an angular velocity w, and has converted all its potential energy into kinetic energy

T = I*w^2 /2

Since U = T, the final angular velocity is

w = sqrt(3*g/h)

The actual speed v of any point on the pole depends upon its distance r from the pivot point. Specifically,

v = r*w

For example, at the top of the pole, r=h and

v = sqrt(3*g*h)

Compare this to the result for just dropping a ball from a height h. In that case

v = sqrt(2*g*h)
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  #9  
Old 02-15-2000, 05:18 PM
frolix8 frolix8 is offline
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Thanks Geezer. It's simple enough when you make some simplifying assumptions, it could get messy for a real-world flagpole.

Is there a simple formula for the angular velocity at time T? (Oviously you can't start with the flagpole vertical).

Again, thanks2
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  #10  
Old 02-15-2000, 05:54 PM
CurtC CurtC is offline
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Now that others have brought in impressive equations, I'd like to back up a minute and address what I think was behind the OP. I wondered this exact thing in high school 25 years ago, and when I was in college I realized what I needed to know. My high-school impression of physics was that it's a bunch of equations, and you just need to know the right one for each situation.

After having a couple of years' worth (four semesters) of differential equations, I realized that what I really needed to know was how to go about solving these things, not really an equation. You can set up the problem with all the forces acting on the pole at each moment of time, and apply the rules of calculus to determine its motion. Problems like this are well-suited also for computational analysis, which I believe is much more enlightening than solving integrals.
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  #11  
Old 02-15-2000, 06:05 PM
frolix8 frolix8 is offline
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Quite right, some people have the "intuition" to set up the problems, some don't. Practice helps, but some people, even very intelligent people, don't have a knack for analysis of physical problems. If you can do it though, it's good fun. I write simulations of building damage due to explosions, the computer handles the math and I tell it how to do it. That's more fun for me than crunching the numbers. Plus I get to see it rendered in 3-D, which makes it more interesting than looking at a page of numbers.
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  #12  
Old 02-15-2000, 07:08 PM
NanoByte NanoByte is offline
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OK, so now the pole's conical and has an additional density gradient and it extends out nearly to the nearest asteroid, so that G varies. You have 5 min. Show your work. Oh yeah, there's also this sitter who needs meal halfway down.

Ray (only at half-mast today)
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  #13  
Old 02-15-2000, 08:50 PM
Geezer Geezer is offline
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Looking back at the original question, it looks like maybe Mjollnir was asking, "How long does it take for the pole to fall to the horizontal?" This is an interesting, and not altogether trivial, question. The smart-aleck answer is that it takes forever because (as has been pointed out) if the pole is vertical it is balanced and never falls. But suppose we start with the pole just a smidgen off vertical. Then how long does it take to fall? The answer depends dramatically on how small a smidgen you take!

Take p as the angle the pole makes with the vertical. Starting with an initial angle p0, how much time elapses before p=pi/2 ? If we define the zero of potential energy at p=0, then the potential energy is

U = -m*g*(1 - cosp)

As we've already seen, the kinetic energy is

T = (1/6)*m*h^2*w^2

Since the system is at rest at p=p0 the total energy remains constant at

T+ U = U(p0)

For small values of p0, U(p0) is also small, so approximately

T + U = 0

and

w = sqrt(3*g/h)*sqrt(1-cosp)

Since w=dp/dt, this is really a differential equation. It is variables separable, so it's pretty easy. Integrating it from t=0 to t=T correspopnding to p=p0 to p = pi/2 , we get the time to fall is

T = sqrt((2*h)/(3*g)) * ln(tan(Pi/8))/tan(p0/4))

Since this equation is valid only in the regime where p0<<1, it is fair to make a further simplification to

T = sqrt((2*h)/(3*g)) * ln(1.66/p0)


What about the exact solution for all values of p0? The solution above involves integrating the differential dp/sqrt(1-cosp). This is easy enough [it's sqrt(2) * ln(tan(p/4)) ]. But if you use the actual value of U(p0) for any value of p0, you have to integrate dp/sqrt(cosp0-cosp). This integral looks easy enough, but I couldn't do it in the amount of time I had to put into this.

I wrote the above before looking at the latest few posts. Coupla comments: Ray I don't think making the pole conical or any other weird shape makes it harder, it just changes the moment of inertia, which is a constant.

Frolix, I think you can extract a simple equation for w(t) if you keep the assumption that p0<<1, but I gotta scram right now.
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