How Does a Plumb Bob Deviate at 34.5° North Latitude Due to Earth's Rotation?

[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!

 

[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.

 

[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?

 

[Tide]

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?

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

 

[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.

 

[Tide]

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

 

[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...

 

[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}

 

[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?

 

[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.

 

[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.

 

[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!