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The question that I'm trying to answer is part b) of the question below:

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My first thought was to prove that the reaction force at R was equal to or greater than the radial force at R, but seeing that there is no way to calculate the reaction force at R I looked at the solution, as shown below:

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I tried to understand the solution and I just couldn't understand why the radial force would have to be greater than or equal the weight of the car. I tried to reason the solution by seeing if the radial force at R is somehow equal to the reaction force at R and I didn't succeed at that either. So, I was wondering why if anyone could explain the premise of the solution.

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The force exerted by gravity on the car may be split into two components at any given time: the component acting tangential to the track, and the component acting perpendicular to it. The former affects the speed of the car, while the latter doesn't do much except cause pressure between the car and the track.

Centripetal force is a measure of the required force to keep the car constrained to its circular path. If the actual force acting on the car toward the center of the circular path is greater, than the car will lift off the track. If it is less than the centripetal force, then it will be pushed into the track. The track is solid of course and the car can't pass through, so this will just result in pressure between the car and the track. (i.e. a non-zero normal force)

So what we should look at to see if the car remains on the track through this journey is whether or not the force acting upon it perpendicular to the track by gravity remains equal or greater than the centripetal force. That's the liftoff requirement: $$mg \sin \theta > \frac{mv(\theta)^2}{a} \tag{$\theta$ is zero when the car is vertical}$$ $$mg \sin \theta - \frac{mv(\theta)^2}{a} > 0 \tag{Liftoff requirement}$$ Then this is when the force applied by gravity to the car toward the center of the circular section is greater than what it required to keep it constrained to the track. This will cause it to lift off the track due to the excess force toward the center.


There's still a question remaining though. Suppose we use conservation of energy to find $v_R$ and from there verify that: $$mg \leq \frac{mv_R^2} a$$ We have found that if the car makes it to the top of the ramp, then yes, it will have sufficient speed to remain pressed into the track and not lift off. But who's to say that it will make it to the apex of the track in the first place? Well, consider that as the car climbs the ramp, $v(\theta)$ is decreasing while $\sin \theta$ is increasing. This tells us that that the quantity $$mg \sin \theta - \frac{mv(\theta)^2} a$$ is increasing as the car climbs the track. If it still hasn't increased sufficiently for liftoff (i.e. still not greater than $0$) by the apex of the track, then it certainly never increased sufficiently for liftoff at any point of the track.

So that's my long-winded way of saying why simply confirming $$mg - \frac{mv_R^2} a \leq 0$$ (where $v_R$ is determined by conservation of energy) is sufficient to conclude that the car does not lifoff the track at any point of its journey around the circular arc.

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I think you could argue the following way:

The radial force $F_r$ is the force needed to maintain the circular movement of the car. It's direction is always perpendicular to the track, i.e. in $R$ it points downwards. No we can distinguish between 3 cases (in the point $R$):

  • $F_r = F_G$: In this case the whole weight of the car (and no other force) plays the role of the radial force. The car touches the track, but there is no contact force.
  • $F_r < F_G$: In this case the weight of the car exceeds the (needed) radial force for the track, therefore it falls down. Another point of view is that the weight forces the car in a path with a smaller radius (Smaller radius = higher radial force)
  • $F_r > F_G$: In this case the weight can't adapt the role of the radial force on its own, the car is forced into a path with a larger radius. However there is the track: Since we assume it to be fixed and solid, its contact force $F_C$ takes the missing part of the needed radial force: $F_R = F_G + F_C$. With $F_C > 0$ one could easily argue that the car must touch the track.
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