Centripetal Force and Earth's Rotation

[hoodie]

"A plumb bob does not hang exactly along a line directed to the centre of the Earth's rotation. How much does the plumb bob deviate from a radial line at 35 degrees north latitude? Assume the Earth is spherical."

I found the radius of the Earth and rotational velocity at the equator via Wikipedia and plugged them into centripetal acceleration = v^2/r. That's pretty much as far as I got and I'm not even sure if that is right.

I'm having a lot of trouble visualizing this problem. I assume that it has something to do with dividing the tension in the string into components but I can't for the life of me figure out what goes where. Any suggestions?

By the by, I have no idea how to work with radians or frames. Please explain in basic terms.

 

[ideasrule]

It's much easier to solve this problem if you consider it from the perspective of someone on the Earth. The guy at 35 degrees latitude feels as if he was on a spinning circle with the same radius as his circle of latitude. That is, he feels gravity acting towards the center of the Earth as well as a centrifugal force pushing away from the center of the latitude circle. If you can find the (vectorial) sum of these two forces and determine the angle between the resultant and gravity, that's the answer to the question.

 

[hoodie]

And is there an alternate way to solve the problem, without using centrifugal force?

 

[Stonebridge]

Yes. It's possible to use centripetal force.
The bob is moving in a circle of radius r.
You need to calculate this first from the radius of the Earth and the 35 degree latitude.
Draw a diagram showing the pendulum bob hanging to one side (right, say) at an angle x. (This is your unknown)
In order for the bob to move in a circle, it must have a resultant centripetal force on it (mv squared divided by r).
The only two "real" forces acting to produce the centripetal force are
a) the weight of the bob, mg, vertically downwards
b) the tension in the string, T, upwards at the angle x to the vertical.
Thus, the centripetal force equals the vector sum of T and mg.
The direction of this resultant force is downwards at 35 deg to the vertical.
The vector diagram is tricky at first but helps a lot with solving this.