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Imagine a loop, in a rollercoaster. When the car of some mass is at the top of the loop, the occupants would be pointing downwards.

At this point, in an ideal situation, the force of gravity would be mass x gravitational acceleration. What about the normal force? Is it equal to the gravitational force, so the net forces, in an ideal situation, is normal force + gravitational force? Is normal force even applicable?

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  • $\begingroup$ The normal force and the weight add up. The sum is equal to the centripetal force. $\endgroup$
    – sazan
    Jan 8, 2023 at 12:18

1 Answer 1

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At the top of the loop, the normal force can still be present, but it is not necessarily equal in magnitude to the car's weight. If the car is following the track then we know there must be a centripetal force on the car equal to $\frac {mv^2} r$ where $m$ is the car's mass, $v$ is its speed and $r$ is the radius of the loop. Since the normal force $N$ and the car's weight $mg$ both point downwards, we have

$\displaystyle N+mg = \frac {mv^2} r$

which we can re-arrange to get

$\displaystyle N = m \left( \frac {v^2} r - g \right)$

Note, however, that $N$ cannot be negative (unless that car is held onto the track in some way). So if $\frac {v^2} r < g$ then the car will fall of the track.

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  • $\begingroup$ Thanks, this answer helps. Is there a point where the normal force can be 0? $\endgroup$
    – user6076
    Jan 9, 2023 at 9:29
  • $\begingroup$ @user6076 If $\frac {v^2} r = g$ then the normal force will be zero just at the top of the loop. If $\frac {v^2} r < g$ then the normal force becomes zero before the top of the loop and the car falls off the track. $\endgroup$
    – gandalf61
    Jan 9, 2023 at 9:54
  • $\begingroup$ Thank you, that makes perfect sense. $\endgroup$
    – user6076
    Jan 9, 2023 at 11:29

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