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Old Jun 18th 2012, 03:32 PM   #1
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A ball lies on the Earth surface, undergoing circular motion together with the Earth.

Hi:
Because of the rotation of the Earth, a plumb bob does not hang exactly in the direction of its weight. If I make a free-body diagram I can see this is the case. Assume the latitude is, say 40 degrees. The tension in the string added to the weight gives a force F lying on the 40 deg parallel plane and pointing towards the Earth's axis. F provides the acceleration the bob is subjected to, in his motion about that axis.

A different case is this: a ball lies on the surface of the Earth, assumed to be perfectly spherical. Here again, the total force on the ball is F. And the string above can be replaced by a contact force N, normal to the surface and a force T tangent to it. So, the pull of gravity, N and T add up to F (stated otherwise, as the string no longer exists, I must replace it by N', a contact force equal in magnitude and direction to the string tension). Where does T come from? Am I missing something?
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Old Jun 19th 2012, 06:03 AM   #2
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Why would you say that there is a tension T if there is just a ball lying on the surface of earth without a string?
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Old Jun 19th 2012, 10:06 AM   #3
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If you draw a free-body diagram of the string and ball you have three components:

1. Tension in the string,
2. Force due to gravity,
3. Acceleration of the body due to the rotation of the earth.

The acceleration due to the earth's rotation is , where R=radius of the earth and is the latitude of the position on earth. The tension in the string is the vector sum:



To put this in perspective: the magnitude of 'a' at the earth's equator is about 0.35% of g, and at 45% latitude about 0.25% of g. At the north or south poles it's 0. The angle of the resulting tension vector compared to vertical varies from 0 at the equator to about 1 degree at 45 degrees latitude then back to zero at the poles.

Last edited by ChipB; Jun 19th 2012 at 10:08 AM.
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Old Jun 19th 2012, 04:02 PM   #4
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Originally Posted by werehk View Post
Why would you say that there is a tension T if there is just a ball lying on the surface of earth without a string?
Thanks for your kind reply. Well, forget about the string. I'll make the following question: if the ball is initially at rest with respect to the geographic coordinate system, will it begin to move along its meridian line?

And thank you ChipB. The real problem, which I proposed to myself, is that of the ball alone. The first one was intended to get some insight into the latter.
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Old Jun 19th 2012, 04:30 PM   #5
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In the case of a ball on the earth's surface the forces acting on it are the normal force, the force of gravity, and a sideways inertial force due to centripedal acceleration caused by the earth's rotation. The magnitude of the inertial force is its mass times the acceleration that I wrote about earlier: , and the component of that force acting horizontally to the earth's surface is . There is no string here, so nothing to oppose that force, and consequently the ball begins to roll according to , where T = torque acting on the ball (which is the force described above times the radius of the ball:), I is the moment of inertia of the ball, and alpha is the resulting rotational acceleration of the ball.

Of course in real life no ball is absolutely smooth and the surface of the earth is not absolutely smooth, so the small amount of torque created on the ball by the earth's rotation is insignificant.
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Old Jun 20th 2012, 05:18 PM   #6
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My doubt has now vanished. The ball will be making a trip towards the equator. Now that if it rolls, there must be friction. Otherwise the ball would slide without turning. And so the torque is less than F_h * r, according to

F_h - F_f = m * A
r * F_h = I * alpha
A = r * alpha

where r * F_h is the torque, F_f force of friction, A, horizontal acceleration of the center of the ball. From which, assuming I = (2/5) * m * r^2, I get

F_f = (2/5) m * A
F_h = (7/5) m * A

That is, F_f = (2/7) F_h. Kind regards.
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