B How can gravity be greater than the centripetal force?

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Naser Tay

So I saw this in my physics textbook and according to the highlighted sentence, it is possible for the weight to be greater than the centripetal force when the car is at the top of the circle. But how is this possible when the weight itself is PART of the centripetal force when the car is at the top?

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russ_watters

Mentor
But how is this possible when the weight itself is PART of the centripetal force when the car is at the top?
The weight is not part of the centripetal force. I don't really like the diagram because it shows normal force instead of centripetal force. Normal force is the sum of centripetal force and weight, so why not just show the two components separately? It looks confusing to me.

Bandersnatch

So I saw this in my physics textbook and according to the highlighted sentence, it is possible for the weight to be greater than the centripetal force when the car is at the top of the circle. But how is this possible when the weight itself is PART of the centripetal force when the car is at the top?
Centripetal force is just some value the sum of all actual forces must have in order to make something go in circles.
These forces here are the weight and the reaction (normal) force. They can be lower or higher than the centripetal force, but only if they together are exactly equal to it will the car follow a circular path.
So, if weight at the top is larger than the required value of centripetal force, then so is the sum of weight and reaction force, and the car will follow a different trajectory than circular (=it will detach from the track and fall).

Once again, the sum of all forces being equal to centripetal force is the condition for circular motion. If these forces don't meet this condition, then there's no circular motion.

Normal force is the sum of centripetal force and weight
Is it? At the top you'd have $W+F_c=R_t$. At the Bottom you'd have $W-F_c=R_b$ And yet, $R_t<R_b$.

"How can gravity be greater than the centripetal force?"

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