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A train is moving with a constant speed on a straight portion of track. Then, it encounters a curve in the tracks, and goes along the curve. However, it does not change the magnitude of acceleration during its time on the curve.

When the train comes back to a straight portion of track (and factoring in the resistance faced by the wheels because of the track), it will have a reduced speed, correct?

Thank you in advance.

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    $\begingroup$ Pretty sure it does accelerate when it goes around the curve. $\endgroup$ – imallett Apr 19 '15 at 23:16
  • $\begingroup$ I think the OP means the train does not change its forward velocity during the curve. $\endgroup$ – Jiminion Apr 20 '15 at 3:00
  • $\begingroup$ There will be a change in acceleration because of the direction, but I am talking about the change in the magnitude of the acceleration. $\endgroup$ – SR1 Apr 20 '15 at 6:23
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The train does accelerate as it goes around a curve.

Velocity is a vector, with magnitude and direction. Speed is the magnitude. The train changes direction.

Acceleration is caused by a force. If the force causes a change in direction with no change in speed, it must be perpendicular to the direction of velocity. For example, A planet in a circular orbit travels at a constant speed because the force of gravity is toward the sun and velocity is along the orbit.

Likewise a frictionless spinning top spins at a constant speed because internal inter atomic forces hold each atom in place. Velocity is along each atom's circular path. The net force is toward the center of rotation. (If the inter atomic forces suddenly vanished, each atom would travel in a straight line tangent to its circular path. The forces deflect atoms away from a straight line towards the center of rotation. This is centripetal acceleration.)

If the train has a reduced speed, it is not because the track deflects it sideways. It is because the force of friction is opposed to the direction of velocity. A train will coast to a stop just as quickly on a straight track. If it keeps a constant speed, it is because the engine supplies just as much forward force as friction supplies backward.

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    $\begingroup$ "An electron in a uniform magnetic field travels at constant speed in a circle" is not correct. The electron loses energy (emit photons) and travels on a spiral path. This time it loses speed too. $\endgroup$ – HolgerFiedler Apr 19 '15 at 19:12
  • $\begingroup$ @HolgerFiedler - What was I thinking? You are absolutely correct. I will remove that. $\endgroup$ – mmesser314 Apr 19 '15 at 19:17
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    $\begingroup$ "A train will coast to a stop just as quickly on a straight track" - not exactly. The rolling resistance of rail-wheel systems is higher in curves :-) $\endgroup$ – Bergi Apr 20 '15 at 1:00
  • $\begingroup$ @Bergi - I did not know that. So I should have said "Given the same friction force, ..." On the other hand, let's assume the engineer knows this, and adjusts the force of the engine to compensate. So the engine and friction still cancel all along the track. The net force is perpendicular to the train's velocity. In that case, the speed of the train will stay constant. $\endgroup$ – mmesser314 Apr 20 '15 at 1:05
  • $\begingroup$ He says speed, not velocity. Speed is a scalar. $\endgroup$ – Jiminion Apr 20 '15 at 3:01
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First of all, you need to be careful when saying the train does not accelerate. In Physics, acceleration is the rate of change of velocity. As velocity is vector, if the velocity of the train changes direction, as it does on the curved portion of the track, then the velocity is changing over time (changing in direction). Therefore, there must be an acceleration to allow the velocity to change direction.

In circular motion at a constant speed (not velocity), this acceleration is called the centripetal acceleration, and its direction is always perpendicular to the velocity. As a result, the velocity only changes direction, but not magnitude (constant speed).

However, I assume that what you mean is that the train's engine is not contributing any change of speed.

If this is the case, we have two key forces to worry about: the centripetal force, which the rails exert on the wheels, and friction, which includes air resistance. The centripetal force causes the centripetal acceleration which, as said earlier, causes no change in speed.

However, the friction forces will generally act in the opposite direction as the train's velocity (its more complicated in reality). As a result the friction forces cause an acceleration that decreases the magnitude of velocity over time.

So, yes, the train will be slower after the curve due to friction. However, since there is always frictional forces, the train would slow down regardless of the presence of the curve anyways. Whether the curve causes any significant increase in frictional forces is more difficult to say, as that depends on the mechanisms of the wheels, and for simplified analysis, it could be ignored.

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  • $\begingroup$ So it wouldn't matter if the train was on a curve or a straight path? Does this mean that they can maintain constant speeds (not velocities) in either case, without a greater magnitude of acceleration on the curve? $\endgroup$ – SR1 Apr 20 '15 at 6:19
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No its SPEED will remain same, velocity has a direction, it will change, both while entering and leaving. On the curve, frictional force will balance the centrifugal force. Assuming constant coefficient of friction throughout the track and constant force provided to engine.

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