# Homework Help: 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