How Does the Coriolis Effect Influence Falling Objects at the Equator?

[Pushoam]

Looks good except for a missing factor of R in D=(θ−{θEarth=ωt})D=(θ−{θEarth=ωt})D = \left(\theta - \{\theta_{Earth} = \omega t\} \right). (Note that your final answer for D does not have the dimensions of distance).

Oh, yes.
$\\ \text { Hence , the deflection D} ~ = R \left(\theta - \{\theta_{Earth} = \omega t\} \right) \\ = R\omega \frac g{3\left ( h+R\right) } t^3 = R\omega \frac{2 h }{3\left ( h+R\right) } \sqrt {\frac {2h}g} D = \frac{2 }{3 } h \omega \sqrt {\frac {2h}g}$

 

[ehild]

$\ \\ \text { Hence , the deflection D} ~ = \left(\theta - \{\theta_{Earth} = \omega t\} \right)$

You wanted the solution in the Earth's frame of reference. With respect to the rotating Earth, theta_Earth =0. The Earth does not rotate with respect to itself.

 

[Pushoam]

You wanted the solution in the Earth's frame of reference. With respect to the rotating Earth, thetaEarth =0. The Earth does not rotate with respect to itself.

No, it is wrt an inertial frame. Please look at post #{4,22,25,28,29,30,31}

 

[TSny]

You wanted the solution in the Earth's frame of reference. With respect to the rotating Earth, theta_Earth =0. The Earth does not rotate with respect to itself.

Pushoam wanted to work the problem two ways: one way using an inertial frame and one way using the Earth frame. The various posts for the two methods got interlaced. Sorry for the confusion.

 

[TSny]

Oh, yes.
$\\ \text { Hence , the deflection D} ~ = R \left(\theta - \{\theta_{Earth} = \omega t\} \right) \\ = R\omega \frac g{3\left ( h+R\right) } t^3 = R\omega \frac{2 h }{3\left ( h+R\right) } \sqrt {\frac {2h}g} D = \frac{2 }{3 } h \omega \sqrt {\frac {2h}g}$

That's it.

 

[ehild]

No, it is wrt an inertial frame. Please look at post #{4,22,25,28,29,30,31}

See your post #20.

Now I am trying to solve it wrt the Earth's frame.

. You have the solution in the inertial frame of reference in Post #16. How would you solve in the Earth's frame of reference, using the pseudo forces?

 

[TSny]

See your post #20. . You have the solution in the inertial frame of reference in Post #16. How would you solve in the Earth's frame of reference, using the pseudo forces?

Pushoam's solution in #16 is for the inertial frame using Cartesian coordinates. At the end of that post, she asks if using Cartesian coordinates simplified the problem compared to using cylindrical coordinates (as first attempted in post #4). So, posts {4,22,25,28,29,30,31} develop the solution in the inertial frame using cylindrical coordinates.

 

[Pushoam]

Continuing from the post #{20 ,21,23,24}

Now I am trying to solve it wrt the Earth's frame.

The sub-script n implies that the corresponding quantity is measured wrt the Earth's frame.

Now, Considering cylindrical co. system with its origin coinciding with the center of the Earth,

in which I started solving the problem wrt. the Earth's frame,
$\vec F_n = \vec F_{ph} + \vec F_{pseudo} = m\vec g + m \omega ^2 r~\hat r - 2m \vec \omega \times\vec v_n \\\vec a_n = \{g- \omega ^2 r\} \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n \\ \text {ignoring }~\omega^2 r, \\ \vec a_n = g \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n \\ \text{Now , how to relate } \vec a_n~ and ~\vec v_n \text{ with their unit vectors? }$

$\\ \text {Initially, the mass has no angular velocity.} \\ \text{ I assume that the mass gets angular velocity as time of motion increases. Then,} \\ \vec v_n = \dot r \hat r + r \dot \Theta \hat \Theta \\ \text{ I am using }\Theta \text { for the angular displacement wrt the Earth's frame } \\ \vec a_n =\{ \ddot r - r {\dot \Theta}^2 \} \left( - \hat r \right) + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta \\ \text { Is this correct so far?}$

 

[TSny]

$\\ \text {ignoring }~\omega^2 r, \\ \vec a_n = g \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n$
$\\ \vec v_n = \dot r \hat r + r \dot \Theta \hat \Theta \\ \vec a_n =\{ \ddot r - r {\dot \Theta}^2 \} \left( - \hat r \right) + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta \\ \text { Is this correct so far?}$

Yes. ehild has outlined the solution in #15 using Cartesian coordinates in the Earth frame.

 

[Pushoam]

$\\ \vec a_n =\{ \ddot r - r {\dot \Theta}^2 \} \left( - \hat r \right) + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta \\ \text { Is this correct so far? }$

There is a sign mistake there $- \hat r$. The correct one is
$\\ \vec a_n =\{ \ddot r - r {\dot \Theta}^2 \} \hat r + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta$

$\{ \ddot r - r {\dot \Theta}^2 \} \hat r + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta = g\left(-\hat r\right ) - 2\omega\dot r \hat \Theta + 2\omega r \dot \Theta \hat r$

$\{ \ddot r - r {\dot \Theta}^2 \} = -g + 2\omega r \dot \Theta ~~~~~~~~~~~~~ \text{ eqn 1} \\ \{2\dot r \dot \Theta + r \ddot \Theta\}= - 2\omega\dot r ~~~~~~~~~~~~~ \text{ eqn 2} \\ \text { considering eqn. 1,}$
$\\ \\ \text{ ignoring } r {\dot \Theta}^2 ~and~ 2\omega r \dot \Theta ,$
$\\ \text { Here, I understand that since ω is very small,we can ignore } 2\omega r \dot \Theta \text{ but what does allow us to ignore } r {\dot \Theta}^2$
$\\ \\ \ddot r =-g ~gives ~r = C - ½ gt^2, \text{where C is an appropriate constant }$
$\\ \\ \text { considering eqn. 2,} \\ \\ \text{ ignoring } - 2\omega\dot r , ~ we ~have, \\ 2\dot r \dot \Theta =- r \ddot \Theta \\ \frac {gt} { C-½ gt^2 } dt = \frac {d \dot \Theta}{\dot \Theta} \\ \text{ Is this correct so far?}$

 

[TSny]

There is a sign mistake there $- \hat r$. The correct one is
$\\ \vec a_n =\{ \ddot r - r {\dot \Theta}^2 \} \hat r + \{2\dot r \dot \Theta + r \ddot \Theta\}\hat \Theta$

Yes. Good.