Conical pendulum in rotating frame

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A conical pendulum at the North Pole exhibits different periods when moving east (T_E) versus west (T_W) due to the Earth's rotation. The pendulum moving east has a longer period because it must "catch up" with the rotating ground. The equation of motion includes the Coriolis force and a modified gravitational acceleration, which accounts for the Earth's rotation. To analyze the problem, one can start with a free body diagram in the rotating frame or switch to a non-rotating frame for clarity. Understanding the modified gravitational acceleration and the role of centrifugal force is crucial for solving the problem accurately.
StrangelyQuarky
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Homework Statement


A pendulum of length l at the north pole is moving in a circle to the east at an angle \theta to the vertical. It has some period T_E as measured in the rotating Earth frame. The experiment is then repeated except now the pendulum is moving to the west with period T_W. The question asks which period is longer, and to calculate the relative time difference in the periods.

Homework Equations



In the rotating frame, the equation of motion of the pendulum involves the Coriolis force and a modified gravitational acceleration due to the rotation of the Earth: m\mathbf{a'} = m\mathbf{g'} + \mathbf{T} -2m(\mathbf{\omega '} \times \mathbf{v'})

where T is the tension in the wire.

The Attempt at a Solution


Since the Earth rotates counterclockwise as viewed looking down on the north pole, the pendulum that moves to the east has a longer period because the Earth's rotation means it has to "catch up" with the rotating ground (I think?). However, I am at a loss as to how to calculate the period from equation of motion.
 
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StrangelyQuarky said:
In the rotating frame, the equation of motion of the pendulum involves the Coriolis force and a modified gravitational acceleration due to the rotation of the Earth: m\mathbf{a'} = m\mathbf{g'} + \mathbf{T} -2m(\mathbf{\omega '} \times \mathbf{v'})
Can you describe in more detail the "modified gravitational acceleration" ##\mathbf{g'}##? What is its magnitude and direction?

What does the symbol ##\mathbf{\omega '}## stand for?

What about the "centrifugal force" in the rotating frame? Or does your ##\mathbf{g'}## include that?

Since the Earth rotates counterclockwise as viewed looking down on the north pole, the pendulum that moves to the east has a longer period because the Earth's rotation means it has to "catch up" with the rotating ground (I think?).
Yes, the "catching up" is from the point of view of an observer in a nonrotating. In fact, you can work the problem by analyzing it from the nonrotating frame. But, if you want to work it out in the rotating frame, then follow the usual recipe of starting with a free body diagram, etc.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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