Deflection of a Falling Mass

[Pushoam]

Homework Statement

Because of the Coriolis force, falling objects on the Earth are deflected
horizontally. For instance, a mass dropped from a tower lands to the
east of a plumb line from the release point. In this example we shall
calculate the deflection of a mass m dropped from a tower of height h
at the Equator.

I have attached the example for your convenience, But I want to solve it on my own with your help.

Homework Equations

3. The Attempt at a Solution

Does it matter whether I consider any point on the Earth or equator?

[/B]
The physical force acting on the system is the gravitational force : $\vec F_{ph} = m \vec g$
Assuming that h is small in comparison with the radius of the earth.

Wrt. the Earth frame( which is non - inertial ),
$\vec F_{n-in} = \vec F _{ph} + \vec F_{pseudo}$
$\vec a_{n-in} = \vec g -\omega ^2 r \hat r - \omega v_{n-in} \hat n$
where I don't know the direction of Coriolis force i.e. $\hat n$ as I don't know the direction of $\vec v_{in}$.
$\hat r$ is the direction from the center of the Earth towards the system

Now, what to do further?

Qualitatively, what I understand is , if the Earth is not rotating, then the path of the mass will be straight as observed from the Earth's frame. Let's mark the point on the Earth where the mass falls by B in this case.
In case of the rotating Earth, before the ball reaches to the Earth's surface, the point marked by B will have moved ahead and so it will fall on another point of the Earth's surface.To a person on the earth, it will appear as a deflection in the motion of mass.

 

[scottdave]

It is not that the ball is still and the Earth moves under it. They are both moving at approximately the same tangential speed. The top of the tower is actually moving slightly faster, because of the longer radius to the center.

 

[haruspex]

Does it matter whether I consider any point on the Earth or equator?

At a pole, there would be no deflection.
At a general point, the released object would temporarily be in an elliptical orbit focussed on the Earth's centre, so would be deflected towards the equator as well as towards the East.
The equator is the simplest case.

 

[Pushoam]

Wrt. inertial frame,
Taking cylindrical coordinate system whose origin coincides with the center of the Earth and the +ve z- axis is along the axis of the rotation of the Earth,
Let's say that the equator is rotating with angular speed $\omega \hat z$. As long as the mass m is in contact with the tower, it's also rotating with angular velocity $\omega\hat z$.
Let's say that when the mass leaves the tower its velocity is $\vec v = \omega \left (R + h \right )\hat x$.
Taking h<< R ,where R is the radius of the earth, the net physical force acting on the mass is $\vec F_{ph} = m \vec g$

So, the information I have is:
Net external physical force acting on the system i.e. $\vec F_{ph} = m \vec g$
Initial position of the system i.e.$\vec r_i = \left ( R+ h \right )\hat y$
Magnitude of the final position of the system i.e. $r_f = R$
Initial velocity of the system i.e. $\vec v_i = \omega \left ( R+h \right )\hat x$
I have to solve$\vec r$(t). This will tell me the time when the mass reaches the Earth's surface.
Then, I will have to calculate the final position of the point A i.e. $\vec r_{Af} = \left (R, \omega t , 0 \right )$

By comparing the final positions of A and the mass, I have to show the deflection.

For calculating r(t),

According to Newton's 2nd law,
$m g\left (\hat r\right )= m \left [\{\ddot r - r{ \dot \theta} ^2 \}~\hat r + \{2 \dot r \dot \theta + r \ddot \theta\} ~\hat \theta\right ]$
This gives the two following differential equations,
$\ddot r -r {\dot \theta}^2 +g = 0$
$2 \dot r \dot \theta + r \ddot \theta = 0$

Is this correct so far?
If yes, any hint on how to solve these differential equations.

 

[haruspex]

You are overcomplicating it. Just treat it as a trajectory in a uniform gravitational field. Good old SUVAT.

 

[Pushoam]

Just treat it as a trajectory in a uniform gravitational field.

I didn't get what you mean by this.
What I understood from your remark is:

Approximating the Earth to an inertial frame, the mass will reach the Earth's surface in time t = $\sqrt {\frac {2h} g}$
By this time the Earth will moved by $\theta = \omega t = \omega \sqrt {\frac {2h} g}$
So, the point A will be moved by $R \omega \sqrt {\frac {2h} g}$
So, the deflection = $R \omega \sqrt {\frac {2h} g}$

 

[Pushoam]

SUVAT

Please tell me its meaning.

 

[scottdave]

It does seem too complicated, for me, too. I had to look up what SUVAT meant @haruspex . I don't recall it being called that (just equations of motion), when I learned it.

What I was trying to say is: the tangential velocity v = ωr, where ω is the angular velocity, and r is the radius. For the Earth, let radius = R, so at height h above R you have radius = (R + h).
So v_ground = ωR, and v_ball = ω(R + h).

So v_ball - v_ground is the velocity of the ball relative to a spot on the ground. When you do this calculation, you will find that you don't even need to know the value of R {radius of the earth}.
The value for ω is 2π radians per day.

 

[Pushoam]

When the mass leaves the tower, it has velocity $\vec v_i = \omega \left ( R+h \right )\hat x$.
And the gravitational force acting on it is along the $- \hat r$ direction. Earlier, I thought that as the mass moves horizontally as well as toward the Earth,too, the $\hat r$ goes on changing. Now, I think what haruspex meant is that I should take this change in the $\hat r$ to be negligible.

Then, the mass reaches the Earth's surface in time t = $\sqrt \frac {2h} g$
By this time, the mass travels in the horizontal direction by a distance $s = \omega \left ( R+h \right )\sqrt \frac {2h} g$ , while the point A on the Earth's surface travels a distance $S = \omega R\sqrt \frac {2h} g$ (actually it is an arc, but , we can approximate it to horizontal distance as it is an small arc (h<< R) of a very big circle).
Then, the deflection is d = $\omega h \sqrt \frac {2h} g$
Is this correct?

 

[TSny]

When the mass leaves the tower, it has velocity $\vec v_i = \omega \left ( R+h \right )\hat x$.
And the gravitational force acting on it is along the $- \hat r$ direction. Earlier, I thought that as the mass moves horizontally as well as toward the Earth,too, the $\hat r$ goes on changing.

Yes. This must be taken into account.

Then, the deflection is d = $\omega h \sqrt \frac {2h} g$
Is this correct?

This is not the correct answer. As you suspected, you need to include the effect of the changing direction of $\hat{\bf r}$.

 

[haruspex]

I didn't get what you mean by this.
What I understood from your remark is:

Approximating the Earth to an inertial frame, the mass will reach the Earth's surface in time t = $\sqrt {\frac {2h} g}$
By this time the Earth will moved by $\theta = \omega t = \omega \sqrt {\frac {2h} g}$
So, the point A will be moved by $R \omega \sqrt {\frac {2h} g}$
So, the deflection = $R \omega \sqrt {\frac {2h} g}$

No, you are again ignoring the initial horizontal velocity of the falling object. Other than that this approach would work.
SUVAT is an acronym formed by s for distance, u for initial speed, v for final speed, a for acceleration, t for time. It is the system of five equations relating those (four in each equation) for the case of uniform acceleration in one dimension.

 

[Pushoam]

No, you are again ignoring the initial horizontal velocity of the falling object. Other than that this approach would work.
SUVAT is an acronym formed by s for distance, u for initial speed, v for final speed, a for acceleration, t for time. It is the system of five equations relating those (four in each equation) for the case of uniform acceleration in one dimension.

Please see the post #9.

 

[haruspex]

Please see the post #9.

I had assumed that your answer in post #9 would be correct to a first order approximation. But having looked at it in more detail I believe TSny is right, that it is off by a factor.
A completely accurate solution would involve treating it as being in a long thin orbit.
Conservation of angular momentum looks helpful, but the answer I get smells wrong to me, so I maybe made a mistake.

Edit: no, maybe my answer is right. I get an extra factor of 3/2.
I took the radial motion from the SUVAT s=-gt²/2, but used conservation of angular momentum to find the angular velocity as a function of time.

 

[TSny]

Edit: no, maybe my answer is right. I get an extra factor of 3/2.
I took the radial motion from the SUVAT s=-gt²/2, but used conservation of angular momentum to find the angular velocity as a function of time.

I believe the correct answer is 2/3 of the result that Pushoam got for d in post #9. She introduced an inertial frame with origin at the center of the Earth as shown below

As the object falls, there is a non-constant, negative x-component of acceleration, $g_x$, of the object relative to this inertial frame. (The axes are non-rotating.)

The 2/3 correction factor is due to $g_x$.

 

[ehild]

Using rotating frame of reference, there is the Coriolis force, perpendicular to the velocity of the falling object. We can use a local Cartesian frame of reference, as the high of the tower is very small with respect to the size of Earth. There are then two inertial forces acting on the ball: the vertical centrifugal force, but it can be ignored with respect to mg, and the Coriolis force perpendicular to the velocity: $\vec Fc=-2m\vec v\times \vec Ω$. As this is a small force, the horizontal component of velocity is small, so we take the direction of Fc horizontal. Let be the vertical velocity v=gt, then the horizontal acceleration is $\ddot x = 2vΩ =2Ωgt$. Integrating twice and substituting $\sqrt{2h/g}$ for the time, we get the deflection. The factor 2/3 is there.

 

[Pushoam]

The physical force acting on the mass is $\vec F_{ph} = m\vec g = mg \left ( - \hat r = - \sin\theta \hat x - \cos \theta \hat y \right )$
Since h<<R, I approximate $\theta$ to $\omega t$, $\cos \theta$ to 1 and $\sin\theta$ to $\theta$.
Considering motion in y- direction,
-g = $\ddot y$
$\dot y = -gt$ as at t =0, v_yi = 0
$y = -\frac 1 2 g t^2$
$-h = -\frac 1 2 g t^2$
$t = \sqrt {\frac {2h} g }$

Considering motion in x - direction,

$\ddot x = -g\omega t$
$\dot x = \frac {-g\omega t^2} 2 + \{\dot x_i = \omega \left( R+h\right) \}$
$x= \frac {-g\omega t^3} 6 + \omega \left( R+h\right) t ~as~ x_i = 0$

So, horizontal distance traveled by the mass $x_m = \left (\frac {-\omega h } 3 + \omega \left( R+h\right)\right ) t$
horizontal distance traveled by the Earth $x_E = \omega Rt$
deflection $D = x_m - x_E = \left (\{\frac {-\omega h } 3 + \omega h \right ) t = \frac {2\omega h } 3 t = \frac {2\omega h } 3 \sqrt {\frac {2h} g }$

Is this correct?

What simplify this calculation from that done in post # 4 is proper choice of coordinate system. Right?

 

[TSny]

View attachment 207755
The physical force acting on the mass is $\vec F_{ph} = m\vec g = mg \left ( - \hat r = - \sin\theta \hat x - \cos \theta \hat y \right )$
Since h<<R, I approximate $\theta$ to $\omega t$, $\cos \theta$ to 1 and $\sin\theta$ to $\theta$.
Considering motion in y- direction,
-g = $\ddot y$
$\dot y = -gt$ as at t =0, v_yi = 0
$y = -\frac 1 2 g t^2$
$-h = -\frac 1 2 g t^2$
$t = \sqrt {\frac {2h} g }$

Considering motion in x - direction,

$\ddot x = -g\omega t$
$\dot x = \frac {-g\omega t^2} 2 + \{\dot x_i = \omega \left( R+h\right) \}$
$x= \frac {-g\omega t^3} 6 + \omega \left( R+h\right) t ~as~ x_i = 0$

So, horizontal distance traveled by the mass $x_m = \left (\frac {-\omega h } 3 + \omega \left( R+h\right)\right ) t$
horizontal distance traveled by the Earth $x_E = \omega Rt$
deflection $D = x_m - x_E = \left (\{\frac {-\omega h } 3 + \omega h \right ) t = \frac {2\omega h } 3 t = \frac {2\omega h } 3 \sqrt {\frac {2h} g }$

Is this correct?

Yes. Nice. Agrees with the standard Coriolis force method as outlined by ehild.

 

[Pushoam]

Yes. Nice. Agrees with the standard Coriolis force method as outlined by ehild.

Will you please tell me name of the software which you used for drawing the diagram?

 

[TSny]

Will you please tell me name of the software which you used for drawing the diagram?

Microsoft Paint.

 

[Pushoam]

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,

$\vec F_n = \vec F_{ph} + \vec F_{pseudo} = m\vec g + m \omega ^2 r~\hat r - 2m \vec \omega \times\vec v_n$
$\vec a_n = \{g+ \omega ^2 r\} \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n$

And $\vec a_n = g\left( -\hat r \right )$ , $\vec v_n = \omega h \hat x + v_r\left (-\hat r\right )$


$g\left( -\hat r \right ) = \{g+ \omega ^2 r\} \left (- \hat r \right ) - 2 \omega \hat z\times\{ \omega h \hat x + v_r\left (-\hat r\right )\}$
$\omega ^2 r\hat r = -2 \omega ^2 h ~\hat y + 2\omega v_r~\hat \phi$ where $\phi = \frac {\pi} 2 - \omega t$ is the angle made by the position vector of the mass m with x- axis.

$\hat y = \sin \phi \hat r +\cos \phi \hat \phi$

$\omega ^2 r\hat r = -2 \omega ^2 h ~ \sin \phi \hat r -2 \omega ^2 h ~\cos \phi \hat \phi+ 2\omega v_r~\hat \phi$
So, we get the following two equations,
$\omega ^2 r\hat r = -2 \omega ^2 h ~ \sin \phi \hat r$

$\omega ^2 h ~\cos \phi \hat \phi = \omega v_r~\hat \phi$

$\omega ^2 r= -2 \omega ^2 h ~ \sin \phi$

I am going on the wrong track.
Am I somewhere conceptually mistaken?

 

[TSny]

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

$\vec F_n = \vec F_{ph} + \vec F_{pseudo} = m\vec g + m \omega ^2 r~\hat r - 2m \vec \omega \times\vec v_n$
$\vec a_n = \{g+ \omega ^2 r\} \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n$

There is a sign error in the first term on the right side of the second equation.

And $\vec a_n = g\left( -\hat r \right )$ , $\vec v_n = \omega h \hat x + v_r\left (-\hat r\right )$

I don't understand either of these two equations.

$\vec a_n \neq g\left( -\hat r \right )$ in the rotating frame of the earth. $\vec a_n$ is given by your previous equation (after correcting the sign error).

$\vec v_n \neq \omega h \hat x + v_r\left (-\hat r\right )$. In the rotating frame of the earth, the object has zero initial velocity. It might help to make a rough sketch of the shape of the trajectory of the object in the rotating frame.

 

[TSny]

What simplify this calculation from that done in post # 4 is proper choice of coordinate system. Right?

Yes. But it is not too hard to solve your equations of #4. You have

$\ddot r -r {\dot \theta}^2 +g = 0$
$2 \dot r \dot \theta + r \ddot \theta = 0$

In the first equation you can argue that the term $r {\dot \theta}^2$ can be neglected. Then you can solve for $r$ as a function of time, $r(t)$ (to a good approximation).

The second equation is equivalent to $\frac{d}{dt}(r^2 \dot \theta) = 0$. So, $r^2 \dot \theta$ remains constant during the fall. This is an expression of conservation of angular momentum in your frame of reference. You can use this equation with the result for $r(t)$ to determine the horizontal deflection during the fall.

 

[Pushoam]

There is a sign error in the first term on the right side of the second equation.

$\vec a_n = \{g- \omega ^2 r\} \left (- \hat r \right ) - 2 \vec \omega \times\vec v_n$

$\vec a_n ≠ g\left( -\hat r \right )$ in the rotating frame of the earth.

Why is it so? Won,t a person sitting on the Earth's surface will feel that the mass is coming down with acc. g?

In the rotating frame of the earth, the object has zero initial velocity

The object is moving with velocity $\omega\left(R+h \right)\hat x$ and A is moving with $\omega R\hat x$ wrt inertial frame. So, won't the object has an initial velocity $\omega h \hat x$ wrt the Earth's frame?

 

[TSny]

Why is it so? Won,t a person sitting on the Earth's surface will feel that the mass is coming down with acc. g?

No. If that were true, then the object would not have any horizontal deflection relative to the Earth as it falls.

The object is moving with velocity $\omega\left(R+h \right)\hat x$ and A is moving with $\omega R\hat x$ wrt inertial frame.

Yes.

So, won't the object has an initial velocity $\omega h \hat x$ wrt the Earth's frame?

No. The xyz axes of the Earth's frame rotate with the earth. So, before the object is released, the object maintains constant values of x, y, and z. Therefore, the rate of change of these coordinates is zero at the instant the object is released.

 

[Pushoam]

In the first equation you can argue that the term $r {\dot \theta}^2$can be neglected. Then you can solve for r as a function of time, r(t) (to a good approximation).

$\ddot r = -g~ gives~ r = h+R - \frac 1 2 g t^2 \\ \text{This gives}~ t= \sqrt{\frac {2h} g } \\ r^2 \dot \theta = ~ constant ~= ~ \left (h+R\right ) ^2\omega ≡ C\\ \frac{d\theta}{dt} =\frac C { \{h+R - \frac 1 2 g t^2 \}^2}\\ \int^{\theta }_0 \, d\theta = \int ^ t_0 \frac C { \{h+R - \frac 1 2 g t^2 \}^2} \, dt$

Is this correct so far?

Solving the above integration will give me $\theta$ traveled by the mass. Then deflection D = R$\left(\theta - \{\theta_{Earth} = \omega t\} \right)$

 

[ehild]

The object is moving with velocity $\omega\left(R+h \right)\hat x$ and A is moving with $\omega R\hat x$ wrt inertial frame. So, won't the object has an initial velocity $\omega h \hat x$ wrt the Earth's frame?

When you stand on the ground, do you see the top of a tower moving? and waiting for long enough time will the top of the tower moving away?

 

[Pushoam]

When you stand on the ground, do you see the top of a tower moving? and waiting for long enough time will the top of the tower moving away?

Yes, I got it.
I understood the following post with the help of the above example.

The xyz axes of the Earth's frame rotate with the earth. So, before the object is released, the object maintains constant values of x, y, and z. Therefore, the rate of change of these coordinates is zero at the instant the object is released.

 

[TSny]

$\ddot r = -g~ gives~ r = h+R - \frac 1 2 g t^2 \\ \text{This gives}~ t= \sqrt{\frac {2h} g } \\ r^2 \dot \theta = ~ constant ~= ~ \left (h+R\right ) ^2\omega ≡ C\\ \frac{d\theta}{dt} =\frac C { \{h+R - \frac 1 2 g t^2 \}^2}\\ \int^{\theta }_0 \, d\theta = \int ^ t_0 \frac C { \{h+R - \frac 1 2 g t^2 \}^2} \, dt$

Is this correct so far?

Solving the above integration will give me $\theta$ traveled by the mass. Then deflection D = R$\left(\theta - \{\theta_{Earth} = \omega t\} \right)$

Looks good. You may use the approximation $\frac 1 {(1-x)^2} \approx 1 + 2x$ for small $x$ to simplify the integration.

 

[Pushoam]

You may use the approximation $\frac 1 {(1-x)^2} \approx 1 + 2x$ for small x to simplify the integration.

$\int^{\theta }_0 \, d\theta = \int ^ t_0 \frac C { \{h+R - \frac 1 2 g t^2 \}^2} \, dt$
$\\ = \int ^ t_0 \frac { \frac C {\left ( h + R)^2\right )}} {1- \frac 1 2 \frac g {\left ( h+R\right) } t^2 }$
$\\ \frac g {\left ( h+R\right) } << 1, \text{so taking the above approximation,}$
$\\ = \frac C {\left ( h + R)^2\right )} \int ^ t_0 \left({1+ \frac g {\left ( h+R\right) } t^2}\right) \, dt \\ \text {so,}~ \theta =\frac C {\left ( h + R)^2\right )}\{ t + \frac g{3\left ( h+R\right) } t^3\} \\ \text { putting the expression for C which is C= }~ \left ( h + R\right )^2 \omega ~, \\\theta = \omega \{ t + \frac g{3\left ( h+R\right) } t^3\} \\ \text { Hence , the deflection D} ~ = \left(\theta - \{\theta_{Earth} = \omega t\} \right) \\ = \omega \frac g{3\left ( h+R\right) } t^3 = \omega \frac{2 h }{3\left ( h+R\right) } \sqrt {\frac {2h}g}$
$\\ \text{ taking }~ h+R ~≅ R ,$
$D = \frac{2 h }{3 R } \omega \sqrt {\frac {2h}g}$

 

[TSny]

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

 

[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.