Mechanics - Gravity at 55 Degrees North

[ronny45]

[SOLVED] Mechanics - gravity

Homework Statement

A pendulum bob is suspended from a long pole at a latititude of 55 degrees north on the Earth. When the pendulum is at rest, the combined action of gravitation and Earth's rotation makes the bob deviate towards the south. By how much does the bob deviate?
(Radius of Earth = 6378km)

The Attempt at a Solution

Okay... firstly, I'm assuming that the tension is divided into components so that the vertical component equals the weight of the bob. Therefore, the horizontal force is indeed acting towards the south. Now, the rotational speed of the Earth is obviously 1 day = 86400s. The angle of deflection (theta) is what I'm looking for. Any ideas?

 

[Kurdt]

What will be the centrifugal force acting on the pendulum?

P.S. The rotational speed of the Earth is not one day.

 

[ronny45]

Centrifugal force = m (Omega)^2 r

Sorry, the time it takes the Earth to rotate is one day... how would I get the rotational speed then? Is it the angular velocity?

 

[Kurdt]

Omega is just the angular velocity and is given by $\omega = \frac{2\pi}{T}$. Be careful with r.

 

[ronny45]

2pi/86400 = 7.27*10^-5m/s
(7.27*10^-5)^2 *6378000*5.9742 × 10^24 kilograms (mass of earth)
=2.01388*10^23 N (not sure if Newtons are the correct unit of centrifugal force)
Does this seem a reasonable figure?

 

[Kurdt]

I said to be careful with r. The radius will not be that of the Earth. There is no need to work out the speed anyway since you have an equation for force with angular velocity and r in anyway.

Drawing a diagram may help you visualise this.

 

[ronny45]

Of course... it's the radius at 55 degrees north, which is 962,461m.
Centrifugal force = m (Omega)^2 r = 5.9742 × 10^24(7.27*10^-5)^2(962461)
= 3.04*10^22

 

[Kurdt]

The mass will be that of the pendulum bob and I don't know how you've calculated the radius at 55 degrees north but that's not correct.