# I Am STUCk! pls enlighten me

1. Sep 10, 2004

### xcutexboax

Hi Guys,

I tried drawing a free body diagram, But the term 34.5 north latitude is getting on my nerves... the closest feeling i got wat it meant was the angle made between the Weight of the bob(line of action drawn Diagonally) and the horizontal.

Q. A plumb bob does not hang exactly along a line directed to the center of the Earth, because of the Earth's rotation. How much does the plumb bob deviate from a radial line at 34.5° north latitude? Assume that the Earth is perfectly spherical and of radius 6.40 x 106m.

i understand there is a formula for the deviation but wifout the formula is there any other way to solve it? Shd i try resolving the F component using the angle 34.5.. Wat equations?? PLS help!

2. Sep 10, 2004

### Tide

If $\hat i$ is up and $\hat j$ is horizontal (pointed toward local south) then the effective acceleration due to gravity accounting for the rotation is $\vec a = (-g + \Omega ^2 R \cos \lambda) \hat i + \Omega ^2 R \sin \lambda \hat j$ where $\lambda$ is the latitude and $\Omega$ is the angular rotation rate.

3. Sep 10, 2004

### xcutexboax

I understand that u have resolved the components into the horizontal and vertical axis, but still how do u apply this equation to get the deviation with tension unknown... although it would be canceled eventually... Any more takers?

4. Sep 10, 2004

### Tide

All you need to do is determine by what angle the total vector I gave you deviates from the ordinary grav acceleration vector.

5. Sep 11, 2004

### xcutexboax

Hmmm sorry but how do u do that... i jus found out that the 2 equations to resolve the three vectors into their vertical and horizontal components are:

$\T \cos \phi + m \Omega ^2 R \cos^2 \theta = mg$ ---> vertical forces

$\T \sin \phi = m \Omega ^2 R \cos \theta \sin \theta$ ----> horizontally

where phi is the deviation to be found....
BUT i DUn understand why it is cos^2 and not cos, and for the bottom equation, there is an additional cos theta.

6. Sep 11, 2004

### Tide

You can find the angle between two vectors $\vec A$ and $\vec B$ from $\vec A \cdot \vec B = A B \cos \phi$.

7. Sep 11, 2004

### xcutexboax

I am totally confused by you now that u have introduced the dot product and u oni started with an acceleration vector... Oh mAn... Sorry to bother u... can u Show me at least half of the full working towards getting the deviation...

8. Sep 11, 2004

### Tide

Sorry - didn't mean to confuse you. Here's a simpler way:

Look at the vector I gave you in #2. Divide the $\hat j$ component by the $\hat i$ component. The result is the (negative) tangent of the deflection angle so:

$$\tan \phi = \frac {\Omega ^2 R \sin \lambda}{g - \Omega ^2 R \cos \lambda}$$

9. Sep 11, 2004

### xcutexboax

Ok based on ur equation, when i subs in all the values of the question:

$$\tan \phi = 0.001959$$
$$\phi = 6.43 \deg$$ ---> which is still considerably bigger

Actual ans is 0.0925 deg..
Is there any link u can establish with ur equation and the 2 equations i gave you?

10. Sep 11, 2004

### Tide

I think you made an error in your calculations. I get 0.14 degrees from my equation but I realize that in my original equations R should really be the distance from the axis of your location which, in your example is the radius of the Earth divided by $\sqrt 2$ in which case I get 0.1 deg. We may have used slightly different numbers for the radius of the Earth and for g.

11. Sep 12, 2004

### xcutexboax

Ehh yupz...i agree tat it is 0.11196 degrees, but in fact the ans shd be 0.0925 degrees, my mistake for that 6.43.... i found out that the angle of deviation is given by a formula:

$$\phi= \frac {\Omega ^2 R \sin \lambda \cos \lambda}{g - \Omega ^2 R \cos^2 \lambda}$$ , where $\lambda$ is the latitude and $\Omega$
is the angular rotation rate. g=9.81, and Earth radius=6.4 x 10^6. and $/phi$ is in radian form.

SomeHow the oni thing that puzzles me is the additional $\cos \lambda$ on the top and bottom.

12. Sep 12, 2004

### Tide

One $\cos \lambda$ is from breaking the centripetal acceleration into horizontal and vertical components and the other is from the fact that the point in question travels about a circle of radius $R \cos \lambda$ as the Earth rotates.

I'm only guessing here but the number we got and "the answer" are sufficiently close that I suspect "the answer" was obtained using slightly different numbers than we used - e.g. someone might have used g = 10 m/s^2 instead of 9.81 m/s^2 or they might have ignored the small term in the denominator. If you find out do let us know!