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omegasquared
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Homework Statement
A ring of mass m slides on a smooth circular hoop with radius r in the vertical plane. The ring is connected to the top of the hoop by a spring with natural length r and spring constant k.
By resolving in one direction only show that in static equilibrium the angle the spring makes to the vertical is θ where:
[tex]cos\theta = \frac{1}{2}\cdot \frac{1}{1-\frac{mg}{kr}}[/tex]
Homework Equations
[tex]T=kx[/tex]
The Attempt at a Solution
To find the equation for tension:
[tex]L=2rcos\theta\\
T = kr(2cos\theta - 1)[/tex]
Resolving vertically I got:
[tex]Tcos\theta = Rcos\theta + mg[/tex]
and horizontally:
[tex]Rsin\theta = Tsin\theta[/tex]
The above can't be correct and also I've resolved in two directions where the question has only asked for one. The forces I have drawn on my diagram are tension and reaction parallel but in opposite directions and weight straight down.