With what speed does the tip of a 25 meter high tree hit the ground?

In summary, the conversation discusses a physics problem involving the speed of a falling tree and the conversion of potential energy to rotational energy. The equation ## mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2}) ## is used to solve the problem, and the resulting speed is found to be around 19 meters per second. However, the formula used in the link provided contradicts this solution and gives a different speed of 27 meters per second.
  • #1
spsch
111
21
Homework Statement
With what speed does the tip of a 25-meter high tree hit the ground?
Assume the tree has a moment of inertia of a long thin rod.
Relevant Equations
## mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2}) ##
Hi, I am still working on the same exams problems, the second last one is giving me a bit of a headache after I found this https://physics.stackexchange.com/q...r-an-upright-rigid-body-to-fall-to-the-ground which contradicts my solution.
Can I assume that when the tree hit's the ground all of it's potential energy is converted to rotational energy?
I was thinking this is why they mention the moment of inertia being that of a long thin rod.

So I went with:
## mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2}) ##

and got: v= ## \sqrt {3*g*l} ##
 
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  • #2
spsch said:
Problem Statement: With what speed does the tip of a 25-meter high tree hit the ground?
Assume the tree has a moment of inertia of a long thin rod.
Relevant Equations: ## mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2}) ##

Hi, I am still working on the same exams problems, the second last one is giving me a bit of a headache after I found this https://physics.stackexchange.com/q...r-an-upright-rigid-body-to-fall-to-the-ground which contradicts my solution.
Can I assume that when the tree hit's the ground all of it's potential energy is converted to rotational energy?
I was thinking this is why they mention the moment of inertia being that of a long thin rod.

So I went with:
## mg* \frac {l}{2} = \frac {1}{6} *m*l^2 * (\frac {v^2} {l^2}) ##

and got: v= ## \sqrt {3*g*l} ##

That looks correct. I'm not sure where the contradition is.
 
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Likes spsch
  • #3
PeroK said:
That looks correct. I'm not sure where the contradition is.
Hi, Thanks. I used the formula from that link and got a different speed. Around 19, whereas my solution gets about 27 meters per second.
I probably made a mistake in plugging in the numbers.

Thank you so much for checking!
 

Related to With what speed does the tip of a 25 meter high tree hit the ground?

1. What is the formula for calculating the speed of a falling object?

The formula for calculating the speed of a falling object is v = √(2gh), where v is the speed, g is the acceleration due to gravity (9.8 m/s²), and h is the height of the object.

2. Does the mass of the object affect its speed when falling?

No, the mass of the object does not affect its speed when falling. The speed is only dependent on the height and the acceleration due to gravity.

3. How long does it take for the tree to hit the ground?

Using the formula t = √(2h/g), where t is the time and g is the acceleration due to gravity, it would take approximately 2.59 seconds for the tree to hit the ground.

4. Is air resistance a factor in calculating the speed of the tree?

Yes, air resistance does play a role in the speed of the tree. However, for objects with large mass and short distances, the effects of air resistance are minimal and can be neglected.

5. How can the speed of the tree be calculated in different units?

The speed of the tree can be calculated in different units by converting the height of the tree to the desired unit and using the formula v = √(2gh) where v is the speed in the desired unit, g is the acceleration due to gravity in that unit, and h is the height of the tree in that unit.

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