- #1
Benzoate
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Homework Statement
A particle is suspended from a fixed point by a light inextensible string of length a. Investigate 'conical motions' of this pendulum in which a string maintains a constant angle [tex]\theta[/tex]with the downward vertical. Show that , for any acute angle theta
ex], a conical motion exists and that the particle speed u is given by u^2=a*g*sin([tex]\theta[/tex])*tan([tex]\theta[/tex])
Homework Equations
dv/dt= (r''-r([tex]\theta[/tex])^2)r-hat + (r([tex]\theta[/tex])''+2r'([tex]\theta[/tex])')[tex]\theta[/tex]-hat
The Attempt at a Solution
since the motion of the particle goes around in a circle, I know that r''=r'=0; therefore :
dv/dt='-r([tex]\theta[/tex])^2)r-hat + (r([tex]\theta[/tex]'[tex]\theta[/tex]-hat
let \rho be the radius of the cricle; therefore \rho = L*sin([tex]\theta[/tex], L is the length of the string. the pendulum makes a circle in they xy plane. In the z direction, from the circle that lies on the xy plane to the top of the pendulum , k=L*cos(\theta)
I think there are three forces that act on the string: centripetal force, the gravitational force, the force of the inextensible string, and I think air resistance is small , so I can neglect air resistance.
-m*L*([tex]\theta[/tex])''=mg*cos([tex]\theta[/tex]) - T+ mv^2/(L*cos([tex]\theta[/tex])
-m*L*([tex]\theta[/tex])''=mg*sin([tex]\theta[/tex])
I not sure if I need to integrate dv/dt to calculate the equation of motion and then square v after I have integrated dv/dt