Deviation of free falling objects (Coriolis effect) using conservation of angular momentum

Question

I read this pdf on non inertial frame, in particular I have a question on the deviation of free falling object due to Coriolis effect.

Consider a ball let go from a tower at height $h$. The displacement due to Coriolis effect, calculated with formulas in Earth system, is $(4.19)$, after it there is explanation of the effect that uses the conservation of the angular momentum of the ball in a inertial frame.

$$x =\frac{2\sqrt{2}ωh^{3/2}}{3g^{1/2}} \tag{4.19}$$ Just before being dropped, the particle is at radius $(R+h)$ and co-rotating, so it has speed $(R+h)ω$ and angular momentum per unit mass $(R+h)^2ω$. As it falls, its angular momentum is conserved (the only force is central), so its final speed v in the (Eastward) direction of rotation satisfies $Rv = (R+h)^2ω$, and $v= (R+h)^2ω/R$. Since this is larger than the speed $Rω$ of the foot of the tower, the particle gets ahead of the tower. The horizontal velocity relative to the tower is approximately $2hω$ (ignoring the $h^2$ term), so the average relative speed over the fall is about $hω$. We now see that the displacement $(4.19)$ can be expressed in the form (time of flight) times (average relative velocity) as might be expected.

But $$v_{average} t_{flight}=h \omega \sqrt{\frac{2h}{g}}$$

Which differs by $\frac{2}{3}$ from $(4.19)$. Is that due to the approximation made?

I also don't understand completely why the average relative velocity $v_{average}$ is taken to be half the relative velocity found. Isn't this valid only for constant accelerated linear motions?

Answer 1

The error is just to consider an average speed $h\omega$.

When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$, ignoring higher order terms in $z$. During the time interval $dt$ the particle falls $dz$ with vertical speed $u(z)$. Hence $$dt=\frac{dz}{u(z)}=\frac{dz}{\sqrt{2gz}},$$ where $u(z)=\sqrt{2gz}$ can be obtained from Torricelli's formula. The horizontal distance traveled during this $dt$ is $$dx=vdt=2z\omega\frac{dz}{\sqrt{2gz}}.$$ Integrating from $0$ to $h$ we obtain the total horizontal displacement $$x=\sqrt{\frac 2g}\omega\int_0^h\sqrt zdz=\frac{2\omega}{3}\sqrt{\frac{2h^3}{g}}.$$

Answer 2

Considering conservation of angular momentum for the dropped ball, $\omega(z)$, the angular velocity of the ball as a function of z, is not constant for the dropped ball. $\omega(z) = {(R + h)^2 \over (R + z)^2} \omega_e$, where $\omega_e$ is the angular velocity of the earth. As the ball falls, $z$ decreases and its angular velocity increases. The answer by @Diracology assumes $\omega(z)$ is constant at $\omega_e$ for the dropped ball; this is good approximation for $h << R$ and hence $z << R$.

Eqn. 4.19 assumes the ball is dropped at a latitude of zero degrees where $\vec v = \vec \omega \times \vec R$ has magnitude $\omega R$. In general, $\vec \omega \times \vec R$ has magnitude $\omega R \enspace cos\lambda$ where $\lambda$ is the latitude. Considering the latitude, eqn. 4.19 should be multiplied by $cos\lambda$.

See the textbook, Fowles Analytical Mechanics, for the derivation of eqn. 4.19 in the non-inertial frame considering latitude and you will find the factor $cos\lambda$ in the result.

Answer 3

You're right about the wrong average velocity. But you can indeed use the conversation of angular momentum to calculate the right displacement. In order to do that let $\omega_0$ be the angular velocity of the earth and $h_0$ be the initial height of the ball. Then it follows from the conservation of angular momentum that $$(R+h_0)^2\omega_0 = (R+h)^2 \omega(h)\Rightarrow \omega(h)=\omega_0(1+\frac{2}{R}(h_0-h))$$ Let $\Delta \omega$ be the difference of the angular velocities of the ball and the earth and $\Delta \phi$ be the difference of their angles. Then it follows that $$\Delta \phi = \int_0^T \Delta \omega \,\mathrm{d}t = \frac{2\omega_0}{R}\int_0^T h_0-h(t) \,\mathrm{d}t$$ Plugging in $h_0-h(t) = \frac{g}{2}t^2$ and $T = \sqrt{\frac{2h_0}{g}}$ one obtains $$\Delta \phi = \frac{\omega_0 g}{R}\int_0^{\sqrt{\frac{2h_0}{g}}}t^2 \, \mathrm{d}t$$ and therefore $$x = R\Delta \phi = \frac{2\omega_0 h_0^{\frac{3}{2}}\sqrt{2}}{3\sqrt{g}}$$