3D Pendulum (Part 2)

Pendulum parameterised in spherical coordinates:

$x = \sin{\theta} \cos{\phi}$ $y = \sin{\theta} \sin{\phi}$ $z = \cos{\theta}$

The equations of motion were derived in the previous version of this problem. They read:

$\ddot{\theta} = \dot{\phi}^2 \ \sin{\theta} \cos{\theta} + 10\sin{\theta}$ $\ddot{\phi} = -2 \dot{\theta} \ \dot{\phi} \cot{\theta}$

It is given that $z= -0.5$ .

$\implies \theta = \arccos(-0.5) = \frac{2\pi}{3}$

This implies that $\theta$ is always constant and its higher derivatives are always zero. Using this fact in the equations of motion:

$\ddot{\phi} = 0 \ \because \dot{\theta}=0$ $\ddot{\theta}=0 \ \implies \dot{\phi}^2 = \frac{-10}{\cos{\theta}} \implies \dot{\phi}^2 = 20$

Finally taking the time derivatives of the coordinates of the pendulum gives:

$\dot{x} = -\sin{\theta} \sin{\phi} \ \dot{\phi}$ $\dot{y} = \sin{\theta} \cos{\phi} \ \dot{\phi}$ $\dot{z} = 0 \ \because \theta(t) = \frac{2\pi}{3}$

FInally, the speed is:

$V = \sqrt{\dot{x}^2 + \dot{y}^2} = \sin{\theta}\dot{\phi} = \frac{\sqrt{3}}{2}\sqrt{20} = \sqrt{\frac{3 \times 20}{4}} = \sqrt{15}$

Answer is:

$\boxed{V = \sqrt{15}}$

Taking the assumption that it is constrained at $x=-0.5$ , we can infer three things:

(1) vertical force is 0

(2) Horizontal force is $\frac{mv^{2}}{r}$

(3) The angle subtended by the rod with z axis is 60 degrees

Based on these, we can draw a F.B.D and get the following equations by letting $F$ , the force exerted by the rod on point mass:

(1) $F\sin (30) =mg$

$F=20$

(2) $F\cos (30)= \frac{mv^{2}}{r}$

$10\sqrt{3}=\frac{1\times v^{2}}{1\times \cos(30)}$

$15=v^{2}$

$v=\sqrt{15}$

$v\approx 3.873$

This problem actually describes a conical pendulum . Let the rod subtends an angle of $α$ with vertical. Then the linear velocity $v$ is related to the angular velocity $\omega$ through $v=\omega \sin α$ . Now, $\cos α=0.5=\dfrac{g}{\omega^2}$ . So $\omega=\sqrt {2g}\implies v=\omega\sin α=\sqrt {2g}\times \dfrac{\sqrt 3}{2}=\sqrt {\frac{3g}{2}}=\sqrt {15}\approx \boxed {3.873}$ .