What is the time taken for a flag pole to fall over and hit the ground?

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The discussion centers on calculating the time it takes for a 10-meter flag pole to topple and hit the ground after its base rusts. An initial calculation suggests a time of 1.43 seconds, assuming free fall, but participants note that the pole's fall is not a simple free fall due to its toppling motion. The forces acting on the pole and the initial conditions, such as a slight velocity at the top, complicate the scenario. It is suggested that conservation of energy principles can be applied to determine the pole's angular velocity and the time taken for the fall, leading to complex equations that may not be straightforward. Ultimately, the discussion highlights the challenges in modeling the dynamics of a toppling object.
Charlie261
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Homework Statement



The flag pole is 10m tall. The base of the pole has rusted and the pole topples over. What is the time taken for the pole to hit the ground?

s = distance = 10m
u = initial velocity = 0 m/s
v = final velocity = unknown
a = g = acceleration due to gravity = 9.81 m/s2
t = time taken



Homework Equations



s = ut + ½at2



The Attempt at a Solution



√2s/g = t = 1.43s

But this pole falling over is not in free fall. It is going to take longer. Please help.
 
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Charlie261 said:

Homework Statement



The flag pole is 10m tall. The base of the pole has rusted and the pole topples over. What is the time taken for the pole to hit the ground?

s = distance = 10m
u = initial velocity = 0 m/s
v = final velocity = unknown
a = g = acceleration due to gravity = 9.81 m/s2
t = time taken



Homework Equations



s = ut + ½at2



The Attempt at a Solution



√2s/g = t = 1.43s

But this pole falling over is not in free fall. It is going to take longer. Please help.

What is causing the pole to fall? What forces act on a perfectly vertical pole?
 
The base of the pole at ground level has completely rusted. One side of the pole at the base crumbles slightly just enough for the pole to not be verticle and so the pole is not balanced and the pole falls over due to gravity.

Another scenario would be that a person tried to balance a pole on its end. The pole is balanced and remains in the vertical for a few seconds and then loses balance and topples over. What would be the most time taken for a pole to move from the vertical to the horizontal due to the force of gravity alone?


Did not want to over complicate things but if it helps:

Assume a very small intial force at the top of the pole to make the pole not vertical.

ie. velocity of top of pole 1mm/s at a distance 1mm from the vertical position.

Assume there is no side ways movement at the base of the pole due to friction. ie rough ground.
 
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Since nobody has come up with a solution I am beginning to think that there is no equation for an object falling over.
I was also hoping to work out the velocity that the top of the pole would hit the ground.

If someone could explain the difficulties of why there is no easy solution that would help me.
 
Charlie261 said:
Since nobody has come up with a solution I am beginning to think that there is no equation for an object falling over.
Oh, there are equations all right. You can use conservation of energy to determine how angular velocity depends upon the angle from vertical. Add in initial angular velocity to get the pole away from the vertical unstable equilibrium point and you end up with a differential equation to solve for angle versus time. I believe that this leads to some variety of elliptic integral to solve, which is not exactly "introductory material".
I was also hoping to work out the velocity that the top of the pole would hit the ground.
Well, that's an easy one. Conservation of energy (center of mass height change versus rotational velocity).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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