
#1
Jul2913, 12:24 AM

P: 24

As you may know, Foucault's pendulum is an easy way to verify that the earth is rotating about its axis. Its a pendulum that is free to swing in any direction. Since earth rotates under it, the position of the pendulum with respect to the ground changes after some time. So you could put pins on the ground to show that the pendulum knocks down pins at different locations as time goes by.
But my question is this, isn't the pendulum also rotating at the same rate and in the same direction as the earth? Because the pendulum is attached to a frame in some building, and so if earth rotates, so does the building and so the pendulum does too? Wouldn't that mean that with respect to other objects in the room, the pendulum won't really move and so the pins shouldn't fall? 



#2
Jul2913, 12:26 AM

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P: 21,998

What force would make the pendulum rotate? It is held by a wire, not a rod.




#3
Jul2913, 12:39 AM

P: 24

Oh I see, when the pendulum is first given a velocity, that velocity + velocity due to earths rotation  velocity due to earth's rotation means its initial velocity is unchanged and so the effects can be seen. Is this true? 



#4
Jul2913, 01:11 AM

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P: 21,998

How does Foucault's pendulum show the earth's rotation? 



#5
Jul2913, 01:16 AM

P: 24





#6
Jul2913, 01:36 AM

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P: 21,998

Behind what? And the building is built far enough away that the pendulum can't hit it. I'm not sure what you are getting at.




#7
Jul2913, 01:45 AM

P: 24





#8
Jul2913, 05:42 AM

P: 3,538





#9
Jul2913, 05:54 AM

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P: 21,998





#10
Jul2913, 05:59 AM

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P: 21,998

You can't release it in such a way that the tangential velocity component is big enough to matter; you can't make that tangential component cause circular rotation because it is way too slow. 



#11
Jul2913, 07:14 AM

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PF Gold
P: 11,352

I could point out that the mass on the end of the pendulum is rotating, initially, around a vertical axis with a period of 24 hours nothing about that will change over time. There is not even a net torque on the system to affect its net, 'independent' motion about the CM of the Earth..




#12
Jul2913, 07:19 AM

P: 3,538





#13
Jul3013, 06:15 AM

P: 1

Assume that the pendulum oscillates in one fixed plane as it would on equator. Now when the pendulam is at the poles, the forces on the bob when it reaches one extreme position are (i) one due to gravity downwards and (ii) one perpendicular to the first force and in the horizontal direction which tries to push the bob out of the plane,due to the spin of the earth.
When the bob reaches the other extreme same two forces act in the same way. So now we have two pairs of forces two downward forces due to gravity and two oppositely acting forces due to spin. The pair of two oppositely acting spin forces constitute a couple which gradually turn the plane of rotation of the pendulum. ie. The pendulum always oscillates in a plane, but the plane slowly spins about a vertical centre pole. 



#14
Jul3013, 11:31 AM

Engineering
Sci Advisor
HW Helper
Thanks
P: 6,342

Suppose the pendulum is on the equator and swinging northsouth. OK, I know that situation will NOT show the rotation of the earth, but ... after 12 hours, the support at the top of the pendulum has moved round 180 degrees relative to the swinging mass, because if not the pendulum would be swinging upside down! Obviously the reason for this has to do with the fact that the weight of the pendulum always acts towards the center of the earth, but I don't criticize the OP for feeling that a textbook explanation might have skipped over some of the details. If you just think about it without doing the math, it's not obvious (at least to me) why the pendulum would be still swinging in a "straight line" northsouth after 6 hours, and not swinging around in a circle instead.... 


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