Trains along equator
Two trains of equal masses are running on a track along the equator of the Earth. Train A runs from east to west, and train B from west to east. The speedometers of the trains show that they are running at equal and constant speeds.
Which train exerts a greater force on the track?
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4 solutions
Really helpful. Easily understood thelogic. Thanks a lot.
Excellent and easy explanation.
This doesn't sound quite right, the reason a exerts more force it that it is rotating slower around earth and so will be pushed less by centripetal force, so it will appear to weigh more.
No. B has a greater acceleration than A. In both cases the acceleration is towards the centre of the earth, causing the train's velocity to change direction rather than change speed. But (in layman's terms) B is being 'whirled around' faster than A and hence requires more acceleration to achieve its trajectory.
What about the force exerted on the track? The acceleration is produced by a force downwards (weight due to gravity) but with the upward force (between track and train) subtracted. So if B has a greater acceleration, then A exerts more force.
This thought experiment might help: Imagine tying a magnet to a piece of string and attaching a metal object to the magnet. Now whirl the string around faster and faster until the metal object breaks loose. The magnetic force in this thought experiment is equivalent to the gravitational force in the puzzle. Although the magnetic force remains constant, the net force between magnet and object is reduced as the string is swung faster until the point where it drops to zero and the two break free.
Relevant wiki: Uniform Circular Motion
Let the angular velocity for train A be ω A , and that for train B be ω B (relative to the Earth's centre).
Because of the rotation of the Earth, ω B > ω A .
The centripetal force of each train is m ω 2 r = m g − R , where R is the reaction, which implies that R = m g − m ω 2 r .
For train A , R A = m g − m ω A 2 r , and for train B , R B = m g − m ω B 2 r .
Since ω B 2 > ω A 2 , R A > R B , so the reaction force of the track on train A is greater than that of train B .
By Newton's Third Law, the contact force exerted on the track by train A is greater than that of train B .
How much do you think ω would need to be so that the force makes some measurable difference?
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Yeah I did consider this. Massively high!
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Any point on the equator moves at a great speed of around 4 6 5 m/s . To have a significant effect, the trains must move at comparable speeds.
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@Rohit Gupta – Yes, although ω would still only need to be very small, as the angular velocity of the Earth is very small. Due to the massive radius, this translates to a massive linear velocity, like you said.
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@Dan Ley – Yes, by the same reasoning, the angular velocities of the trains with respect to the center of the Earth will be even smaller and they have to travel at high speeds to create a significant impact.
@Dan Ley – If the trains are moving in a circle closer to the poles, then they can travel at smaller linear velocity to get the same angular velocity.
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@Pranshu Gaba – But the centres of their motion changes so the set up doesn't apply surely.
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@Dan Ley – If we keep the axis of rotation same and shifts the plane of rotation of the trains closer to the pole, then, the same setup will apply.
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@Rohit Gupta – But then their weights would not be acting towards the centre of the earth? Would we just take components?
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@Dan Ley – Ahh, yes, you are correct. We will have to talk in terms of the components. The trains at moving at greater speed with respect to the center of the Earth will exert lesser radial/normal force on the track.
and how would this force differential compare to the force differential from to actual speed differences due to the kinematic time dilation of the clocks in their respective speedometers?
Nice explanation! Not a big deal, but I believe you're missing an "m" in your R[A] equation.
assuming the problem has "one" solution, I took the speed to be zero. Then the force exerted is equal. Obviously the problem was not stated specifically enough to guarantee only one solution.
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Well, yes! this is a loophole in the problem. But, just to give credit to the author, he mentioned the directions in which the trains are running, so, it won't be fair to take the speeds as zero. Otherwise, there will be no sense of mentioning the directions.
One has to assume level tracks at equal heights above sea level. I understand the target was to illuminate angular velocity and centripetal force, but technically speaking, there is insufficient information to conclude. Perhaps a simple, "all else being equal"?
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The question mentions about a single track along the equator. It should go along the curvature of the Earth, shouldn't it?
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I am imagining a slightly more complex topography. The Earth is not smooth. If the part of the track on which train A appears is significantly higher (e.g. on a plateau), and the part of the track on which train B appears is significantly lower (e.g. in a valley), their respective distances from the center of the Earth could change the answer to the question. Given a practical speed of the trains of 60 MPH, a sea level above the center of the Earth at the equator of roughly 4,000 miles, the velocity of the Earth's rotation at sea level around 1000 MPH, then if train A was...better than 1,000 miles higher than train B then it would be B exerting the greater force. OK, bad example. But my point is made?
Relevant wiki: Formulas in Uniform Circular Motion
We can assume the surface of the Earth as a convex bridge.
Now, for Train A, m g x - m w r e l 2 r = R (normal force)
w r e l is angular velocity/frequency ( RELATIVE to earth) and m w 2 r is cenripetal force which is resultant of Normal force and weight in the x component.
Similarly for Train B.
BUT,
The trick lies in w r e l .A rotates east to west and Earth west to east so w r e l for A is w e − w A .
Except for Train B it is w r e l = w e + w B .
So R is greater for Train A than Train B .
Thus F ( A ) > F ( B ) ( because R is nothing but F.)
Can you elaborate more on what do you mean by "weight in the x component"? In which direction x points?
i think the speed on speedometer is already the relative speed between the train and the earth, so the relative angular must be the same for both train
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The speedometer gives the speed of the trains relative to the ground which will be the same. However, their velocities with respect to the center of the Earth will be different.
Let us consider the velocities of trains A and B to be v1 and v2 respectively and their velocities according to speedometers be 'v' each. Then we calculate the magnitudes of v1 and v2 by analysing their motions on rotating earth. Thus, v1 = v - Rw where w = angular velocity of earth, R = Radius of earth and v2 = v + Rw. This their respective centrifugal forces are of the order CF(A) < CF(B), where CF(A) represents centrifugal force on A. Since centrifugal forces operate away from the centre of the circular motion of each of the two trains, The normal contact force exerted by the train having less centrifugal force, will be more in magnitude. Hence Train A exerts greater Normal force than that of Train B.
Yes, the train moving at a faster speed with respect to the center of the Earth will experience a greater centrifugal force and will exert less force on the track. This makes me wonder that if the trains move at great enough speeds, can they actually leave the ground and fly in the sky? If yes, what would that minimum speed be?
As the speed of the train increases, it experiences a greater centrifugal force and a lesser normal force as time passes. At one particular stage, it achieves a speed which balances the force due to gravity with centrifugal force exactly. At this moment, Normal force becomes zero. And, the velocity of the train, will be tangential to its orbit of revolution around the earth. But, gravitational force from now will be responsible for the circular motion of the train irrespective of its contact or normal force on earth, rather just by the virtue of its velocity, viz, Now the motion of the train around the earth is in a way similar to that of a planet around sun. That minimum speed would be calculated by using a simple equation that : (1/2)m(v^2) = mg ,i.e, v = squareroot(rg) = 7.9km/s.
In layman language :- The earth rotates from west to east therefore it opposes the motion of A whereas it supports B now since both travel with equal speed therefore A must have a greater acceleration than B and thus exerts more force on th track than B.