# The left bank of the river!

1. Dec 1, 2003

### hhegab

Hi all,
I have the following problem which I have tried to solve for a very long time, very very long time indeed- 2 or more years. I am studying here all by myself and I got no one to help me.

I shall not participate in any talk. I have had enough with it.
this problem is in the Schaum's Book "Theoretical Mechanics" by Moray Speigel, Chapter 6, Moving Coordinate Systems.
Here it is:

A river of width D flows northward with a speed vo at colatitude \lambda. Prove that the left bank of the river will be higher than the right bank by an amount equal to:
(2D\omega vo cos \lambda)(g^2 + 4 \omega^2 vo^2 cos^2 \lambda)^-1/2
where \omega is the angular speed of theearth about its axis.

Last edited: Dec 5, 2003
2. Dec 1, 2003

### gnome

Just to be sure we have the question accurately, are you saying:

$$\Delta h = \frac{2D\omega v_o cos\lambda}{\sqrt{g^2 + 4\omega^2 v_o^2cos^2\lambda}}$$

where$$\Delta h$$ is the difference in height of the surface of the water (with the left side higher)?

3. Dec 2, 2003

### hhegab

Yes!
I shall not ask you how could you make it. I am sure it is written here :)

hhegab

4. Dec 5, 2003

### hhegab

No one cared about it and I have been struggling to do it. Should any one know where can I find the answer will be appreciated.

hhegab

5. Dec 5, 2003

### joc

this is an interesting question and i'd like to see someone shed some light on the solution as well.

6. Dec 5, 2003

### hhegab

Well,
Any one can ask his Classical Mechanics Prof. to solve it then.

hhegab

7. Dec 5, 2003

### gnome

Well, maybe I could, but I'd rather try to do it myself. But it's going to take more time than I can spare now.

I haven't forgotten about it, though.

8. Dec 13, 2003

### hhegab

Well, I thought any one could ask his professor.
I will not lose hope.

hhegab

9. Jan 2, 2004

### gnome

Well, as promised I have tried to solve this river problem, but I don't seem to be getting the same result. If anyone can find an error in my work, please correct me.

Sorry my diagrams are so crude. Best I could do with "KPaint". By the way, does anybody know of a free 'nix program that's good for drawing diagrams?

Anyway, as I try to show in FIG 1, the river is flowing north with velocity v0, and at the same time it is rotating with the surface of the earth with a tangential velocity V (eastward). Let's take a "piece" of water of mass m. It is at latitude &lambda;, so if R is the earth's radius, the radius of the path of this water is Rcos&lambda;.
It has angular momentum of
$$L = mVr$$
or
$$L = m{\omega}r^2$$

But as the water moves toward the north, the radius r of its circular path decreases. &omega; is constant, so the decrease in its angular momentum is given by:
$$\frac{dL}{dt} = 2m{\omega}r\frac{dr}{dt}$$

And since
$$r = Rcos\lambda$$
and
$$\frac{dr}{dt} = -Rsin\lambda\frac{d\lambda}{dt}$$
therefore
$$\frac{dL}{dt} = -2m{\omega}R^2sin{\lambda}cos{\lambda}\frac{d\lambda}{dt}$$

L = R x P (where P is the linear momentum) so L is tangential to the surface, directed towards the north pole. To produce this decrease in angular momentum, there must be a torque T, equal in magnitude to dL/dt but in the opposite direction: tangential to the surface, towards the south.

But this torque T = F x R, so there must be a force F, tangential to the surface and perpendicular to the direction of the linear momentum. Therefore, by the right-hand rule, this F is directed towards the east.

So,
$$F = \frac{\tau}{R} = \frac{1}{R}\frac{dL}{dt}$$
$$F = 2m{\omega}Rsin{\lambda}cos{\lambda}\frac{d\lambda}{dt}$$

Now, note that, if s is the distance along the surface of the earth from the equator to latitude &lambda;, s = R&lambda;, and v0 = ds/dt = Rd&lambda;/dt
Therefore,
$$F = 2m{\omega}{v_0}sin{\lambda}cos{\lambda}$$

Now, FIG 2 attempts to show a cross-section of the river, where H is the (greatly exaggerated) difference in height (or depth) of the water along the left bank as compared to the right bank, and D is the width across the surface. Each "piece" of water feels two forces acting upon it: the normal force equal in magnitude to mg acting vertically upward, and the mysterious (coriolus) force F which is required by the change in angular momentum. I tried to depict these by the large vertical arrow (next to the H) and the short horizontal arrow in FIG 2. These two forces sum to a resultant force &sum;F depicted by the diagonal arrow in the figure.

$${\sum}F = \sqrt{(mg)^2 + (2m{\omega}{v_0}sin{\lambda}cos{\lambda})^2}$$

I believe that the water must align itself so that its surface will be perpendicular to the resultant force. (I can't tell you why; it just seems right.) Therefore, &beta; is the angle that the surface of the water makes with the horizontal.

The angles &alpha; and &beta; are complementary, so
$$sin\beta = \frac{H}{D} = \frac{2m{\omega}{v_0}sin{\lambda}cos{\lambda}}{\sqrt{(mg)^2 + (2m{\omega}{v_0}sin{\lambda}cos{\lambda})^2}}$$
$$H = \frac{2D{\omega}{v_0}sin{\lambda}cos{\lambda}}{\sqrt{g^2 + (4{\omega}^2{v_0}^2sin^2{\lambda}cos^2{\lambda}}}$$

So I believe that this is an expression for the difference in depth of the water. I don't see why your book has it without the sin&lambda; factor.

#### Attached Files:

• ###### river.png
File size:
3.6 KB
Views:
580
10. Jan 7, 2004

### hhegab

Solved!

There's a sideways force (Coriolis force) acting on anything moving
towards or away from the axis of rotation in a rotating frame such as
the earth. The Coriolis force per unit mass (acceleration) is:

F1 = 2 Omega x V

where V is the velocity in the rotating frame. The magnitude of a
cross product is the product of the magnitudes of the two vectors
times the sine of the angle between them which in this case is cos
(lambda), so the magnitude of the Coriolis force/mass is:

F1 = 2 Omega v0 cos(lambda)

and this force acts in the east/west direction (West for a river
flowing north in the northern hemisphere and East in the southern
hemisphere). The other force/mass acting on the river is gravity
which acts in the downward direction:

F2 = g

The sine of the angle between the total force (sqrt(F1^2+F2^2)) and
the vertical is:

F1/sqrt(F1^2+F2^2)

This has to be the same as the sine of the angle between the surface
of the river and the horizontal (because the surface is perpendicular
to the force) which is:

Delta h/D

so:

Delta h = D F1/sqrt(F1^2+F2^2)

= he formula required.

It was solved by a fellow from anoher forum. I dont know his real name.

11. Jan 7, 2004

### gnome

Can you add a little more detail to your explanation? It doesn't really show how you (or he) arrived at that solution. For example:
Why?

And,
Why? Which two vectors? If you're talking about the coriolus force directed East, and gravity directed Down, the angle between them is 90o and has nothing to do with &lambda;, so you must be talking about some other pair of vectors. What are they?

Presumably, the result found in your book is correct, but, having spent all that time trying to work it out, I'd like to know where I went wrong.

12. Feb 7, 2010

### dr_k

Hi Gnome!

I know this problem was posted back in 2004, but I see that your last post was in 09, so you're probably still around!

This problem caught my eye since I've been thinking about these non-inertial frame problems recently. Here's how I "think" the author found the given result...or, this is the only way I could get their result.

In the non-inertial frame, in the N hemisphere, where +x is South, +y is East, and +z is up (radially outwards), one can write down Newton's modified 2nd postulate, for an isolated particle in the river, as

$\rm F_{net_x} \approx 2 m \omega \cos\theta \dot{y} + C_x = C_x$
$\rm F_{net_y} \approx -2 m \omega (\cos\theta \dot{x} +\sin\theta \dot{z}) + C_y = 2m\omega \cos\theta v_0 + C_y$
$\rm F_{net_z} \approx -mg + 2 m \omega \sin\theta \dot{y} + C_z = -mg + C_z$

where $\rm {\bf C} = (C_x,C_y,C_z)$ is the contact force, the velocity, relative to the non-inertial frame, is $\rm {\bf v} = (\dot{x}, \dot{y}, \dot{z}) = (-v_0, 0, 0)$ , and $\rm \theta$ is the colatitude angle. These equations of motion are simplified versions, where terms proportional to $\rm \omega^2$ have been dropped off, i.e. the $\rm \overrightarrow{\omega}~x~ (\overrightarrow{\omega}~x~(\overrightarrow{r}+\overrightarrow{R}))$ terms.

Then one must assume equilibrium, which implies

$\rm C_x = 0$
$\rm C_y = -2m\omega \cos\theta v_0$
$\rm C_z = mg$

http://img190.imageshack.us/img190/5676/riverh.jpg [Broken]

And the result follows...

$\rm \tan\beta = \frac{2m\omega\cos\theta v_0}{mg} = \frac{H}{\sqrt{D^2-H^2}} \Rightarrow H = \frac{2 D\omega\cos\theta v_0}{\sqrt{4\omega^2\cos^2\theta v_0^2 + g^2}}$

Last edited by a moderator: May 4, 2017