# uniform circular motion

• September 28th 2008, 11:59 AM
sarabolha
uniform circular motion
A mass m = 8.5 kg is suspended from a string of length L = 1.37 m. It revolves in a horizontal circle. The tangential speed of the mass is 2.75 m/s. What is the angle q between the string and the vertical (in degrees)?

help?
i know F= m v^2 /R...
so..
Tsinx = m v^2/Lsinx
and
Tcosx - mg = 0

i'm having trouble solving for x though...
• September 28th 2008, 12:20 PM
skeeter
$T\cos(q) = mg$

$T = \frac{mg}{\cos(q)}$

$T\sin(q) = \frac{mv^2}{L\sin(q)}$

$\frac{mg\sin(q)}{\cos(q)} = \frac{mv^2}{L\sin(q)}$

$\frac{\sin^2(q)}{\cos(q)} = \frac{v^2}{gL}$

$\sin^2(q) = \frac{v^2}{gL}\cos(q)
$

$1 - \cos^2(q) = \frac{v^2}{gL}\cos(q)$

$0 = \cos^2(q) + \frac{v^2}{gL}\cos(q) - 1$

$a = 1$, $b = \frac{v^2}{gL}$ , $c = -1$
$\cos(q) \approx 0.75726...
$q \approx 41^{\circ}$
$T \approx 110 \, N$