# Thread: Fly, little birdie!!

1. ## Fly, little birdie!!

OK, I may just be stupid, having just taken a long break from physics, but I'm having trouble with a problem about--you'll never guess--a bird.

It's flying--sorry, soaring--in a horizontal circular path. It's bank angle (relative to the horizontal) is estimated to be $\displaystyle 25^o$ and the bird takes 13 s to complete one circle. How fast is the bird flying and what is the radius of the circle?

I would start by taking the normal force exerted on the bird by the air, at an angle of [MATH65^o[/tex] to the horizontal (90-25), and then finding the percentage of that force pushing the bird toward the center of the circle, but I don't know what to do after that.

2. let $\displaystyle L$ = lift force, $\displaystyle m$ = bird mass

three equations ... three unknowns $\displaystyle v$, $\displaystyle r$, and $\displaystyle L$

$\displaystyle \displaystyle v = \frac{2\pi r}{t}$

$\displaystyle \displaystyle L\sin(25) = F_c = \frac{mv^2}{r}$

$\displaystyle L\cos(25) = mg$

solve the system

3. OK, I did this:

$\displaystyle v=\frac{2\pi R}{t}$

$\displaystyle L\sin(25)=\frac{mv^2}{R}$

$\displaystyle L\cos(25)=mg$

$\displaystyle L=\frac{mv^2}{R\sin(25)}=\frac{mg}{\cos(25)}$

$\displaystyle R=\frac{\sin(25)g}{v^2\cos(25)}=\frac{4.57}{v^2}$

$\displaystyle v=\frac{2\pi4.57}{tv^2}$

So, solving this, I get about 1.3 m/s for V. My book gives a different answer, though. What am i doing wrong?

4. The problem is in your equation for R - check your units and you'll see that you got a couple of terms upside down. It should be:

$\displaystyle R=\frac{\cos(25)v^2}{g \sin(25)}=\frac{v^2}{14.67 m /s^2}$

$\displaystyle v=\frac {2 \pi R} t = \frac{2\pi v^2}{14.67 t}$

Solve for v, and I get 30.3 m/s