Rotation-Energy Question

1. Feb 12, 2017

Arman777

1. The problem statement, all variables and given/known data
A tall,cylindrical chimney falls over when its base is ruptured.Treat the chimney as a thin rod of lenght $55.0m$.At the instant its makes an angle of $35.0^°$ with the vertical as it falls,what are the (a) $a_r$ of the top ? (b) $a_t$ of the top ? (Hint:Use energy considerations,not a torque )
2. Relevant equations
$mgh=\frac 1 2Iw^2$
3. The attempt at a solution
$L=55m$
For tangential veloctiy I found $\frac 1 2Mg(L-Lsin(35))=\frac 1 2(\frac 1 3ML^2)(\frac {(V_{tan})^2} {L^2})$

from that I get $V_{tan}=26.27 \frac m s$

I stucked here... I thought $\frac {V_{tan}} {rt}=∝$

$t$ is the time that end point comes that angle.
but it didnt worked out.

$a_r=\frac {(v_r)^2} {r}$ I dont know how to find radial velocity...

Thanks

2. Feb 12, 2017

TSny

Check the trigonometry that you used to get the left side. (The 35o angle is measuresd from the vertical.)
I don't understand your equation here. There is no need to worry about time in this problem.

If a particle moves in a circle at constant speed, does it have any radial acceleration? Does it have any radial velocity? Thus, does your equation $a_r=\frac {(v_r)^2} {r}$ hold?

3. Feb 12, 2017

Arman777

sorry it will be cos35.
its $w-w_0=∝t$
I think velocity is a vector and as the rod moves , the velocity vector will change,thus, there will be some acceleration

4. Feb 12, 2017

TSny

Yes.
Yes. If you review the basic formula for radial (or centripetal) acceleration of a particle moving in a circle, you will see that it does not involve radial velocity, it involves tangential velocity.

5. Feb 12, 2017

Arman777

so its $a_r=\frac {(v_t)^2} {r}$ ??

6. Feb 12, 2017

TSny

Yes, except be careful with signs (i.e., directions).

7. Feb 12, 2017

Arman777

Ok I found $a_r$ correctly,what about $a_t$ ?

8. Feb 12, 2017

TSny

Despite the wording of the problem, I think using torque is the way to go here.

9. Feb 12, 2017

Arman777

yep I found thanks again