Polar Co-ordinate proofs

Question

The expression for acceleration in spherical polars is

$$ \ddot{\mathbf r} =( \ddot r -r\dot\theta^2-r\dot\phi^2\sin^2\theta) \mathbf e_r + (r\ddot\theta+2\dot r \dot\theta-r\dot\phi^2\sin\theta\cos\theta ) \mathbf e_\theta + \frac {1}{r\sin\theta} \frac {d}{dt}(r^2\dot \phi\sin^2\theta) \mathbf e_\phi $$

Hence show there can be a solution to the equation of motion where the particle orbits at constant latitude $\theta_0 $, constant radius $ r_0$ , and constant angular velocity $ \dot \phi= \omega $. Write down the constraints on the values of $ \theta, r_0, $ and $ \omega $ .

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Im really lost with this question, not sure how to begin, all help is greatly appreciated!

Answer 1

If both $r=r_0$ and $\theta=\theta_0$ are constants, then their time derivatives are zero.

That is to say, $\dot r=\ddot r=0$ and $\dot \theta=\ddot \theta=0$.

We also are given $\dot \phi =\omega$ is a constant.

Therefore,

$$\begin{align} \frac{d^2\vec r}{dt^2}&=\hat r(\theta_0,\phi) (-r_0\omega^2 \sin^2\theta_0)+\hat \theta(\theta_0,\phi)(-r_0 \omega^2 \sin \theta_0 \cos \theta_0) \\\\ &=-\omega^2 (\hat x x+\hat yy) \tag 1 \end{align}$$

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NOTE: In arriving at $(1)$, we made use of the following relationships:$$x=r\sin \theta \cos \phi$$ $$y=r\sin \theta \sin \phi$$ $$\hat x=\hat r \sin \theta \cos \phi+\hat \theta \cos \theta \cos \phi-\hat \phi \sin \phi$$ $$\hat y=\hat r \sin \theta \sin \phi+\hat \theta \cos \theta \sin \phi+\hat \phi \cos \phi$$ Putting that all together, we find that $$\hat xx+\hat yy=r\sin \theta \left(\hat r \sin \theta+\hat \theta \cos \theta\right)$$

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Thus, the equations of motion reduce to

$$\begin{align} &\ddot x+\omega^2 x=0\\\\ &\ddot y+\omega^2 y=0\\\\ &\ddot z=0 \end{align}$$

from which we obtain

$$\begin{align} &x(t)=\frac{\dot x(0)}{\omega}\sin \omega t+x_0\cos \omega t\\\\ &y(t) =\frac{\dot y(0)}{\omega}\sin \omega t+y_0\cos \omega t\\\\ &z(t)=r_0\cos \theta_0 \end{align}$$

Answer 2

If latitude and radius are constant, then then all derivatives of them are 0. If the derivative of phi is the constant, omega, then its second derivative is 0. so the right side of your equation reduces to $$(0- r(0)- \omega r \sin^2(\theta)e_r+ (r(0)+ (0)(0)- (r \sin(\theta)\cos(\theta) \omega)e_{\theta}+ \omega' e_{\phi}$$

Answer 3

$ \theta$ is constant, $ \dot \theta, \ddot\theta $ vanish

$$ \ddot{\mathbf r} =( \ddot{r} -r\omega^2\sin^2\theta) \,\mathbf e_r + (-r \omega^2\sin\theta\cos\theta )\, \mathbf e_\theta + \frac {1}{r\sin\theta} \frac {d}{dt}(r^2\omega\sin^2\theta) \,\mathbf e_\phi $$

ODE is between $ r,t.$

EDIT1:

It is also known as the Conical Pendulum problem in steady state for pendulum fixed at cone vertex,mass circling on base circle $r_0$. An even solution is:

$ r = r_0 \cos (\omega t ) $ , $ z = z_0, \theta $ are constants.