A point mass $m = 1$ is attached to the free end of a mass-less rigid rod of length $L = 1$ . The other end of the rod is attached to the origin with a ball joint, allowing the free end to be positioned at an arbitrary point in $3D$ space (constrained by the length of the rod, or course). The gravitational acceleration is $10$ in the negative $z$ direction.

If the rod rotates in such a way that its free end is constantly at $z = -0.5$ , what is the linear speed of the free end?

The answer is 3.873.

**
This section requires Javascript.
**

You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.

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:

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