Calculating Time for Tree Falling at 45 Degrees: Homework Help

[mcovalt]

Homework Statement

A tree is cut with a 45 degree wedge to the center of the tree. The height from the base of the wedge to the top of the tree is 50 feet. The weight of the wood from the base of the wedge to the top of the tree is 500kg. Assuming there is no resistance when the tree starts falling, how long will it take the tip of the tree to hit the ground?

Homework Equations

I have no idea haha. I do know it will be accelerating.

The Attempt at a Solution

I know how to do a free falling problem, but the tree tipping over an angle aspect has baffled me.

Thanks for any help guys!

 

[gneill]

I suspect that the problem is indeterminate, as the tree begins in an unstable equilibrium (vertical, balanced on a line).

You would have to specify how it is set in motion (initial impulse, initial velocity, initial displacement, or something like that). Then you could develop a differential equation for the motion based upon the rotational energy and potential energy.

On the other hand, solving your problem for the final velocity of the tree tip when it hits the ground, assuming that the tree begins to fall with a negligible shove, is quite straightforward.

 

[mcovalt]

Thanks for the response!

How about instead of the tree falling with a negligible shove, it's initial "shove" is the force of gravity? Half of the gravitational force of the tree is being supported by the ground whereas the second half of the gravitational force is not being supported by the ground as it has a 45 degree angle wedge cut out underneath it.

Would I apply the gravitational force of 2500N using F=ma to find the acceleration of the net mass (500 kg)?

With that acceleration I could calculate the time it took to travel a quarter of a 50 foot radius circle?

I am just throwing this stuff out there. I need someone to make sure I am not making stuff up :)

 

[gneill]

Thanks for the response!

How about instead of the tree falling with a negligible shove, it's initial "shove" is the force of gravity? Half of the gravitational force of the tree is being supported by the ground whereas the second half of the gravitational force is not being supported by the ground as it has a 45 degree angle wedge cut out underneath it.

Would I apply the gravitational force of 2500N using F=ma to find the acceleration of the net mass (500 kg)?

With that acceleration I could calculate the time it took to travel a quarter of a 50 foot radius circle?

I am just throwing this stuff out there. I need someone to make sure I am not making stuff up :)

The weight of the tree is straight down; there's no sideways component. The only way the tree would begin to fall without some perturbation is if the undercut went more than halfway through the trunk. Then the two vertical sections of the tree on either side of the cut would be imbalanced.

If the cut goes halfway, then the tree will be critically balanced, with the center of gravity directly over the cut line (thin edge of the wedge cut).

The calculations will have to be done taking into account rotational dynamics, since the tree will be pivoting about its end (where the cut is). It's not a simple linear motion problem.

 

[mcovalt]

Ok. I'm reading up on rotational energy right now. It seems like that is the missing link. I've revised the question a little bit to give gravitational force a little bit of an x component. Here is a diagram of the new cut setup I just put together.

[PLAIN]http://img841.imageshack.us/img841/812/logwx.jpg
btw. Forgot to point this out. that 90 degree cut is -20 degrees and 70 degrees.

Before I start really studying into what I need to know to solve the problem, besides learning the concepts of rotational energy, are there any other concepts applicable to this problem?

 

[gneill]

Ok. I'm reading up on rotational energy right now. It seems like that is the missing link. I've revised the question a little bit to give gravitational force a little bit of an x component. Here is a diagram of the new cut setup I just put together.

[PLAIN]http://img841.imageshack.us/img841/812/logwx.jpg
btw. Forgot to point this out. that 90 degree cut is -20 degrees and 70 degrees.

Before I start really studying into what I need to know to solve the problem, besides learning the concepts of rotational energy, are there any other concepts applicable to this problem?

In order to solve for the time that tree reaches the ground, you're going to run into a differential equation that will need to be solved. This is because you will be able to develop equations for the torque about the pivot as a function of the angle, and thus the acceleration as a function of angle, but not directly as a functions of time.

The angular acceleration is not constant, but changes with the angle of the tree as it falls. So the "usual" equations of motion won't apply here.

 

[mcovalt]

Ok. Thanks so much! This was so helpful!

 

[mcovalt]

Could you let me know if I am setting this up right? I'm using the conservation of energies to find the velocity of the tip as a function of the angle.

M = 500kg
G = 10m/s
L = 50m (I changed feet to meters lol)

M * G * L/2 = 1/2 * I * ω²

I for a rod at pivot point is 1/3 * M * L²

Therefore:

M * G * L = 1/3 * M * L * ω²

I should just be able to solve for ω right?

***EDIT***
I found this equation applicable to my problem on another post:

$$mgl/2 = mgl\sin \phi /2 + I{\omega}^2/2$$

I fully understand the outside energies, but the inside (with the sin)... why is that there?

 

[mcovalt]

Could you let me know if I am setting this up right? I'm using the conservation of energies to find the velocity of the tip as a function of the angle.

M = 500kg
G = 10m/s
L = 50m (I changed feet to meters lol)

M * G * L/2 = 1/2 * I * ω²

I for a rod at pivot point is 1/3 * M * L²

Therefore:

M * G * L = 1/3 * M * L * ω²

I should just be able to solve for ω right?

***EDIT***
I found this equation applicable to my problem on another post:

$$mgl/2 = mgl\sin \phi /2 + I{\omega}^2/2$$

I fully understand the outside energies, but the inside (with the sin)... why is that there?

***EDIT #2***
I see that my moment of inertia would be more cylindrical than rod like.
http://hyperphysics.phy-astr.gsu.edu/hbase/icyl.html#icyl4

 

[gneill]

Could you let me know if I am setting this up right? I'm using the conservation of energies to find the velocity of the tip as a function of the angle.

M = 500kg
G = 10m/s
L = 50m (I changed feet to meters lol)

M * G * L/2 = 1/2 * I * ω²

I for a rod at pivot point is 1/3 * M * L²

Therefore:

M * G * L = 1/3 * M * L * ω²

I should just be able to solve for ω right?

***EDIT***
I found this equation applicable to my problem on another post:

$$mgl/2 = mgl\sin \phi /2 + I{\omega}^2/2$$

I fully understand the outside energies, but the inside (with the sin)... why is that there?

The sin term is there because the gravitational potential energy of the tree depends upon the angle at which the tree is leaning. The height of the center of gravity goes as the sine of the angle.

 

[jdunster]

Trees would usually be cut with a face notch about 25-30% across one side, followed by a back cut on the other side. For the purpose of the calculation assume that the tree is 2 degrees off vertical on the face notch side, and falls over under its own weight once the back notch is completed i.e. no push of gust of wind. In practice there would be a drag coefficient created by the foliage in the crown branches, but ignore that for the example. As long as the tree can fall freely off the stump the angle of the face notch is not that critical - it is simply a missing piece of support wood.

I would like to know at what angle the tree reaches terminal velocity.

 

[mcovalt]

Well, if we ignore air resistance as you suggested, the tree would exponentially accelerate as the force of gravity on the tree increases as the tree falls.

Therefore, the maximum velocity would be the instant before the tree hits the ground.