0
$\begingroup$

If it were a string connected to an spinning object, I would have a second thought, because for string there doesn't exist any normal force.

But for roller coaster placed on a rail, the rail acts as a surface so normal force does exist to cancel out the weight of the roller coaster. So I think given there is no other force besides the weight, the roller coaster won't be able to complete the loop at the very top.

I got confused about this when I was doing my homework. The question asks for the minimum velocity of the roller coaster at the top of the loop. The answer key says that the centripetal force is equal to the weight, mg, plus the normal force, even though the two forces are in opposite directions; to get the minimum force, I have to assume the normal force to be zero since no change can be made to the weight. I think the book gives the wrong term for something else than the true normal force.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ This is basically equivalent to this question, asked today: physics.stackexchange.com/q/201357 $\endgroup$
    – Rations
    Aug 19, 2015 at 19:14
  • $\begingroup$ The minimum requirement is simply that $\frac{mv^2}{r} \geq mg$, for the roller coaster not to 'come off the track'. From that, calculate the minimum $v$. Your textbook is likely entirely correct on this issue. $\endgroup$
    – Gert
    Aug 19, 2015 at 19:15
  • $\begingroup$ Also, I think the roller coaster is on the inside of the loop rather than on top of it, which is why you add $mg$ and $N$. $\endgroup$
    – Rations
    Aug 19, 2015 at 19:17
  • $\begingroup$ For the minimum velocity, the normal force will be zero so it doesn't matter whether you add or subtract. If the coaster is on the inside of the loop, you would add; if it's on the outside, you subtract. Most likely, if you are talking about "minimum velocity", you are assuming that it would fall if the velocity was less: so that it's on the inside (without the usual safety wheels gripping the rail from the other side). Making a diagram would clear all this up for you. $\endgroup$
    – Floris
    Aug 19, 2015 at 20:34
  • $\begingroup$ Note that this classic question assumes that the cart simply sits on the rails, and will fall off if it is not going fast enough. This is not true of any realistic roller coaster, which is anchored to the track. $\endgroup$
    – Kyle Oman
    Aug 19, 2015 at 22:18

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.