Time period of oscillations of a particle revolving in a paraboloid

Clearly, if initially the angular velocity was $\omega_{0}$ , then there would have been no vertical motion, as the components of normal reaction would have balanced both weight and centrietal force.

When , $\omega = \omega_{0}(1+ \eta)$ , the particle will initially go up, and then after a certain height it will come down.

Consider the instant when the particle is at a height $y$ has angular velocity $\omega$ , and a velocity along the surface $v$

By conservation of angular momentum,

$\omega r^2 = \omega_{0}(1+\eta) r_{0}^2$

$\Rightarrow \omega y = \omega_{0}(1+\eta)y_{0}$

$\Rightarrow \omega = \omega_{0}(1+\eta)\dfrac{y_{0}}{y}$

By using conservation of mechanical energy,

$\dfrac{1}{2}m\omega_{0}^2(1+\eta)^2r_{0}^2 - \dfrac{1}{2} m (\omega^2r^2 + v^2) = mg(y-y_{0})$

Using $(1+\eta)^2 \approx 1+2\eta$

$\omega_{0}^2(1+2\eta) y_{0} - (\omega^2y + v^2) = 2g(y-y_{0})$

Eliminating $\omega$ from the two equations, we get

$\displaystyle v = \sqrt{\dfrac{(\omega_{0}^2(1+2 \eta)y_{0} - 2gy)(y-y_{0})}{y}}$

Using $\omega_{0}^2 = 2g$ , $y_{0} = 1$

$v = \displaystyle \sqrt{\dfrac{2g(1+ 2 \eta - y)(y-y_{0})}{y}}$

At the highest point, $v = 0$ , hence

$y = 1+2\eta$

Slope, $tan \theta = \dfrac{dy}{dr} = 2r = 2\sqrt{y}$

Hence, $v_{y} = v \sin \theta = v \times \dfrac{2 \sqrt{y}}{\sqrt{1+4y}} = \displaystyle \displaystyle \sqrt{\dfrac{8g(1+ 2 \eta - y)(y-1)}{1+4y}}$

Hence, $\dfrac{dy}{dt} = \displaystyle \displaystyle \sqrt{\dfrac{8g(1+ 2 \eta - y)(y-1)}{1+4y}}$

Hence, $\displaystyle \dfrac{1}{2 \sqrt{2g}} \int_{1}^{1+ 2\eta} \sqrt{\dfrac{1+4y}{(1+2\eta-y)(y-1)}} { \mathrm dy} = \int_{0}^{T/2} dt$

Putting $y-1 = z$ ,

$\dfrac{T}{2} = \dfrac{1}{2 \sqrt{2g} }\displaystyle \int_{0}^{2 \eta} \sqrt{\dfrac{5+4z}{2 \eta z - z^2}} dz$

Now, as $z$ is very small, $\sqrt{5+4z} \approx \sqrt{5}$

Hence, $\dfrac{T}{2} = \dfrac{\sqrt{5}}{2 \sqrt{2g}} \displaystyle \int_{0}^{2 \eta} \dfrac{1}{ \eta^2 - (z -\eta)^2} dz$

Hence, $\boxed{T = \pi \sqrt{\frac{5}{2g}}}$

Instead of having a small perturbation in the angular velocity at a known radius, we could instead have a small perturbation in the radius at a known angular velocity and get the same answer. Therefore, let the initial angular velocity be exactly $\omega_0$

First, we determine that the potential energy is $mgr^2$ , and the kinetic energy is $\frac { 1 }{ 2 } mv^{ 2 }=\frac { 1 }{ 2 } m(\dot { r } ^2 +r^2\dot{\theta}^2+4r^2\dot{r}^2)$ , so the Lagrangian is $L=\frac { 1 }{ 2 } m(\dot { r } ^2 +r^2\dot{\theta}^2+4r^2\dot{r}^2) -mg^2$ , where $\theta$ is the angular position.

Using the Euler-Lagrange Equations, $\frac { \partial L }{ \partial \theta } = 0 = \frac{d}{dt}\frac{\partial L}{\partial\dot{\theta}}= \frac{d}{dt}mr^2\dot{\theta} \Rightarrow mr^2\dot{\theta}=mr^2\omega_0=l$ and $\frac { \partial L }{ \partial r } =mr\dot { \theta } ^{ 2 }+4mr\dot { r } ^{ 2 }-2mgr=\frac { d }{ dt } \frac { \partial L }{ \partial \dot { r } } =\frac { d }{ dt } (m\dot { r } +4mr^{ 2 }\dot { r } )\Rightarrow \frac { l }{ mr^{ 3 } } -2mgr=m\ddot { r } +4mr\dot { r } ^{ 2 }+4mr^{ 2 }\ddot { r }$

We substitute $r$ with $r_0(1+a)$ , where $r_0 = 1$ and $a \ll 1$ function of time. (Note that, based on the way we changed the problem statement, $a\neq 0$ initially). Also, $\frac{l^2}{mr_0^3} = 2mgr_0$

The differential equation then becomes $\frac { l }{ mr_0^{ 3 } }(1+a)^{-3} -2mgr_0(1+a)=mr_0\ddot {a} +4mr_0^3\dot { a } ^{ 2 }+4mr_0^{3 }\ddot { a }$

Since $a \ll 1$ , for a first-order-approximation simple harmonic oscillation, $\dot{a}^2 \ll \ddot{a}$ , so we simplify to $-3\frac { l }{ mr_0^{ 3 } }a -2mgr_0a=mr_0\ddot {a} +4mr^{3 }\ddot { a }$ .

Therefore, if $\omega_y$ is the angular frequency of vertical oscillations, $\omega_y^2 = -\frac{\ddot{a}}{a} = \frac{6mgr_0+2mgr_0}{mr_0+4mr_0^3} = \frac{8g}{5}$ . The period in seconds is then calculated as $\frac{2\pi}{\omega_y}$ .