Why is the speed of a particle in a conical pendulum related to the angle theta?

[Benzoate]

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 \thetawith 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(\theta)*tan(\theta)

Homework Equations

dv/dt= (r''-r(\theta)^2)r-hat + (r(\theta)''+2r'(\theta)')\theta-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(\theta)^2)r-hat + (r(\theta'\theta-hat

let \rho be the radius of the cricle; therefore \rho = L*sin(\theta, 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*(\theta)''=mg*cos(\theta) - T+ mv^2/(L*cos(\theta)
-m*L*(\theta)''=mg*sin(\theta)

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

 

[tiny-tim]

I think there are three forces that act on the string: centripetal force, the gravitational force, the force of the inextensible string …

-m*L*(\theta)''=mg*cos(\theta) - T+ mv^2/(L*cos(\theta)
-m*L*(\theta)''=mg*sin(\theta)

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

Hi Benzoate!

I'm not really following your equations …

There are two forces … tension, and gravitation … they have to match the acceleration …

and the acceleration is purely radial (zero tangential) …

and you need to find T anyway …

try again!

 

[Benzoate]

Hi Benzoate!

I'm not really following your equations …

There are two forces … tension, and gravitation … they have to match the acceleration …

and the acceleration is purely radial (zero tangential) …

and you need to find T anyway …

try again!

I would used the equation of acceleration in polar for right where:

a=(-r*(\theta)^2')r-hat+(r*\theta'')theta-hat

Would the sum of the gravitational and tension forces be the sum of the centripetal force? If so then ma_y=T*cos(\theta)-mg=0==> T=mg/cos(\theta), and ma_x=T*sin(\theta)==>a_x= mg/cos(\theta)*sin(\theta)=mg*tan(\theta) right?

r=L*sin(\theta)

 

[tiny-tim]

Hi Benzoate!

(have a theta: θ and a squared: ² )

I would used the equation of acceleration in polar for right where:

a=(-r*(\theta)^2')r-hat+(r*\theta'')theta-hat

Would the sum of the gravitational and tension forces be the sum of the centripetal force? If so then ma_y=T*cos(\theta)-mg=0==> T=mg/cos(\theta), and ma_x=T*sin(\theta)==>a_x= mg/cos(\theta)*sin(\theta)=mg*tan(\theta) right?

r=L*sin(\theta)

Wow! That's perfect!

Now (since θ'' = 0 and rθ' = v), just use v²/r instead of rθ'² and you have the answer!

 

[Benzoate]

Hi Benzoate!

(have a theta: θ and a squared: ² )


Wow! That's perfect!

Now (since θ'' = 0 and rθ' = v), just use v²/r instead of rθ'² and you have the answer!

Don't quite understand why \theta'' is zero. Here are my calculations :


mv²/L*sin(\theta)=mg*tan(\theta) , m cancel out and therefore I am left with: v²=g*tan(\theta)*L*sin(\theta)

 

[tiny-tim]

Hi Benzoate!

(what happened to that θ I gave you?)

v²=g*tan(\theta)*L*sin(\theta)

That's right! That's the given answer (with L instead of a).

(and θ'' = 0 because the string is always radial, and there's never any tangential acceleration )