I'm looking at a problem involving a simplified rollercoaster with only 3 cars (connected with weightless rods) for which the friction and distance from the carts to the rails can be ignored. The part I was thinking about is the following: at a certain moment cart 2 is situated on top of an arc as I have drawn here. rollercoaster

At a certain point in the set of questions, we seem to rely on the fact that all carts share the same acceleration $a = \frac{{v_2}^2}{R}$. Now I understand that they all share the same normal acceleration since they are performing a circular motion, but what is stopping them from having a tangential acceleration? In other words: why do we know $a_t = 0$ for each cart?

  • $\begingroup$ if they are being acted on only by gravity and the force of constraint, then the tangential acceleration is certainly NOT zero (and the normal acceleration isn't constant, either) $\endgroup$ Jan 23 '13 at 18:45
  • $\begingroup$ Are we to also assume that there is no gravity in this situation? If gravity is taken into effect, then the only point at which there actually isn't a tangential component of acceleration is at the very top of the loop. $\endgroup$
    – Mik Cox
    Jan 23 '13 at 18:45
  • $\begingroup$ @Jerry Isn't normal acceleration constant in every point of a circular motion? Mik Yes, there is definitely gravity. $\endgroup$ Jan 23 '13 at 19:00
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    $\begingroup$ To clarify, if you are interested in only the exact instant shown above, then indeed each cart's tangential acceleration is the same (and is 0). This is because the tangential component of the acceleration due to gravity on cart 3 is in the opposite direction (tangentially) as cart 1 at this instant, and thus the net tangential acceleration due to gravity on the train at this instant is zero (therefore tangential acceleration on each cart is also zero since they are connected). At this instant, thus, the only acceleration is normal and equals v^2/r for each cart, as suggested. $\endgroup$
    – Mik Cox
    Jan 23 '13 at 19:04
  • $\begingroup$ @MikCox Yes, it only needs to be true for this exact instance. Come to think about it, ofcourse the total tangential acceleration must be zero, because this is the exact moment the acceleration passes through zero to switch from left to right. $\endgroup$ Jan 23 '13 at 19:16

At the exact instance above, we can indeed prove that the tangential acceleration on each of the 3 individual carts is zero:

Cart 2 obviously experiences no tangential acceleration as gravity pushes it in the same direction as the normal force due to the track, which is perpendicular to the cart's direction of motion.

The force due to gravity on cart 3 does have a tangential component though, as gravity acts straight down with a component along the direction of the track. This causes the third card to "want" to speed up due to gravity, as would be expected.

However, since the same exact force is also exerted by gravity on cart 1 (but is in the opposite direction tangentially to that on cart 3), the force of gravity causing cart 1 to "want" to slow down will exactly cancel, and thus we see that the net tangential acceleration across the entire train at this exact instant is zero. Since each cart is connected, this proves that the net tangential acceleration on each individual cart is also zero, and the normal acceleration on each cart is the same and as suggested above, $a=\frac{{v}^2}{r}$.

Another perhaps simpler way to think of this is that the instant above describes when the train, as a whole, passes the top of the loop. Thinking of the train as a smaller mass (a ball or point mass, whichever) it then becomes clear that at the top of the loop is the moment when the train moves from slowing down to speeding up, which is to say in this case that the tangential acceleration is zero.


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