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

• Pushoam
In summary, the conversation discusses the Coriolis force and how it affects falling objects on Earth. The example given is a mass dropped from a tower at the Equator, and the conversation delves into calculating the deflection of the mass due to the Coriolis force. Differential equations are mentioned, but it is suggested to use simpler methods such as SUVAT equations. The final goal is to show the deflection of the mass by comparing its final position with the final position of a fixed point on Earth's surface.
Pushoam said:
No, it is wrt an inertial frame. Please look at post #{4,22,25,28,29,30,31}
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?

ehild said:
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}
Pushoam said:
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?}

##

Pushoam said:
##\\ \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 said:
##\\ \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?}
##

Last edited:
Pushoam said:
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.

• Introductory Physics Homework Help
Replies
3
Views
432
• Introductory Physics Homework Help
Replies
3
Views
808
• Introductory Physics Homework Help
Replies
1
Views
327
• Introductory Physics Homework Help
Replies
1
Views
583
• Introductory Physics Homework Help
Replies
9
Views
930
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
995
• Introductory Physics Homework Help
Replies
3
Views
661
• Introductory Physics Homework Help
Replies
23
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
825