Metal Pole Falls Homework: Find Speed of Upper End

[Arman777]

Homework Statement

A uniform metal pole of height $30.0m$ and a mass $100kg$ is initially standing upright but then falls over one side without its lower end sliding or losing contact with the ground.What is the linear speed of the pole's upper end just before impact ?

Homework Equations

$τ=Frsinθ$
$τ=I∝$

The Attempt at a Solution

I think there must be some $F$ force that the pole will start to fall.And there's also $F'$ force between pole and ground.
so total torque of the object will be ($I$ take the touching point between pole and ground rotation axis).
So If total length is $L$ then I can write for this instance $(t=0)$
$FL=τ(t=0)$
In between hitting the ground (lets call that time $T$).The torque will be
$FL+mgsinθ\frac L 2=τ(0,T)$, but here sinθ will change every moment.
And in the impact it will be $FL+mg\frac L 2=τ(T)$
then from here I tried to take a some time interval like when $sinθ=\frac {\sqrt 2} {2}$ and substract these values so that $Fr$ will cancel out but I don't know I stucked.
Then I thought I can just use the center of mass of pole's motion.It will make a parabola.And If I can calculate its speed when hits the ground I can easily calculate the end poing of pole.
Here I used normal projectile motion equations to find speed.but came out wrong.Maybe I am forgetting the centripetal force.Or Writing some equation wrong.

Thanks

 

[BvU]

It will make a parabola

As a function of what ? Why ?

 

[Arman777]

Cause its the center of mass and the acceleration in that point will be g.A function of time ?

 

[haruspex]

Homework Equations

Any conservation laws that might be relevant?

 

[Arman777]

Any conservation laws that might be relevant?

Yeah I can apply that maybe or I apllied $mgL=\frac 1 2mv^2$ but I get wrong result

 

[kuruman]

Yeah I can apply that maybe or I apllied $mgL=\frac 1 2mv^2$ but I get wrong result

That's because in writing the above equation, you are assuming that the entire mass of the rod is concentrated at the tip of the rod at distance L from the pivot. Is it?

 

[Arman777]

That's because in writing the above equation, you are assuming that the entire mass of the rod is concentrated at the tip of the rod at distance L from the pivot. Is it?

yeah that's not true..then I find the energy change of the center of mass.which its $mg\frac L 2=\frac 1 2mv^2$
from that the velocity of the top end should be $2v$ ,which is not correct answer

 

[kuruman]

You cannot assume that the entire mass is concentrated at the center of mass either. The rod is not in free fall. It is rotating about its end. You need to consider the rotational kinetic energy of the rod just before it hits the ground. It is not $\frac{1}{2}mv_{cm}^2$. What is it?

 

[Arman777]

You cannot assume that the entire mass is concentrated at the center of mass either.

It says uniform but ok

$E=\frac 1 2Iω^2$ and $I=\frac 1 6mL^2$

 

[kuruman]

The moment of inertia of a uniform rod about its end is $I = \frac{1}{3}mL^2$, otherwise it's OK. Now conserve mechanical energy remembering that the change in potential energy of the CM is what's relevant.

 

[Arman777]

The moment of inertia of a uniform rod about its end is I=13mL2

how can we calcualte it?

 

[kuruman]

Calculate what?

 

[Arman777]

Calculate what?

Moment of inertia

 

[kuruman]

I just gave you the correct formula for it. Doesn't the statement of the problem give you the numbers that go in it?

 

[Arman777]

I just gave you the correct formula for it. Doesn't the statement of the problem give you the numbers that go in it?

I am asking the formula the equation..

 

[Arman777]

OK I found it never mind

 

[Arman777]

Ok I solved thanks a lot