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I'm trying to understand the situation where a roller coaster goes around a loop, and we need to find the minimum speed it takes not to fall off. What I'm not understanding is why we would set the centripetal force equal to $mg$. At the top, the normal force is 0. Centripetal force acts towards the center of loop so I have that pointed down. I have weight pointed down as well.

The only way that we get $F_c = \frac{mv^2}{r}=mg$ is if $F_c$ is acting away from the center and I don't understand why this would be the case.

What I am piecing together is that centripetal force in and of itself is not a force, but if something is moving in a circular path it has centripetal force. So if its moving in a circular path or we want it to continue on its path, then the centripetal force must equal the weight. What I would like to know is how to reconcile this idea of centripetal force acting towards the center.

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  • $\begingroup$ "Centripetal force" is a catch-all phrase that describes any center-seeking force. At the top of the roller coaster loop, gravity is providing all of the centripetal force, so normal force with the coaster car's track goes to zero. If you add $mg$ and centripetal force together, you are doubling the force that actually exists. $\endgroup$ Commented Feb 11, 2023 at 20:38
  • $\begingroup$ en.wikipedia.org/wiki/Loop-the-Loop_(disambiguation) $\endgroup$
    – hft
    Commented Feb 11, 2023 at 23:42
  • $\begingroup$ Why was this question downvoted? I think its a perfectly fair question to ask that I didn't see a clear answer to in other questions. $\endgroup$ Commented Feb 12, 2023 at 2:07
  • $\begingroup$ @DavidWhite Your answer makes sense to me. In this case what is keeping the cart at the top of the loop? Is it inertia that keeps it from falling? I could see it being in sort of equilibrium and continuing moving because of Newton's 3rd law $\endgroup$ Commented Feb 12, 2023 at 2:10
  • $\begingroup$ Downvoted because this is a common physics homework problem. Alternatively, downvoted because this problem is commonly described in an introductory physics textbook and so shows little research effort. Alternatively, downvoted because the answers to this question will be of little use to others, and so it is not a good question for this site. This site has specific guidelines for good questions--this site is not just an any-question-goes site. You could try Physics Forums or the homework help room instead. Here are some question guidelines: physics.stackexchange.com/help/dont-ask $\endgroup$
    – hft
    Commented Feb 12, 2023 at 3:10

2 Answers 2

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Consider the case that the roller coaster does make it around the loop, and draw a free-body diagram at the top. Then write down the associated Newton's Second Law equation. It is: $$ \vec{F}_N + \vec{F}_g = m \vec{a} $$ where all three vectors — the normal force, $\vec{F}_N$; the gravitational force, $\vec{F}_g$; and the acceleration of the car, $\vec{a}$ — point downward. The first two are obvious, and the third is true because of kinematics$^1$: for an object traveling in a circle, the acceleration points inward. Therefore, if you choose an upward-pointing $y$ axis, you can write the $y$-component of this equation as: $$ - F_N - F_g = m (-a) $$ where these are now expressed in terms of magnitudes (e.g, $F_N = \left|\vec{F}_N \right|$). We know$^1$ that the magnitude of the acceleration vector is $v^2/R$, where $R$ is the radius of the circular path (the loop), so: $$ F_N + F_g = m v^2 / R $$ Now, imagine what happens as the speed $v$ decreases. The right-hand side of this equation gets smaller, so the left-hand side must also get smaller. The only way for the latter to occur is for $F_N$ to decrease (the weight of the coaster car, $F_g$, is fixed). At some value of the speed, $v_{\rm min}$, $F_N$ reaches zero and we can then write: $$ 0 + mg = m v_{\rm min}^2 / R \quad \rightarrow \quad v_{\rm min} = \sqrt{gR} $$

1 - The story is slightly more complicated than that: the radial component of the acceleration for an object traveling on a circular path is always $a_r = - v^2/R$.

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You wouldn't. You may be thinking of setting the magnitude of centrfugal force $m\omega^2 r \ge mg$ such that at the top of the loop the vector sum of centrifugal force (pointing radially outward, hence straight up at the top of the loop) and gravity (pointing down) is zero (in the case of minimum speed to stay on the track) or points upward (the engineer believes in safety margins).

If you have a normal force between the car and the track of $mg$, that would indeed mean that the car is being accelerated downward by the track at twice the acceleration due to gravity - that is to say, the magnitude of centrifugal force at the top of the loop was twice the magnitude of the force of gravity, and the track is pushing back with the difference to keep the car constrained to the loop.

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