Extreme Confusion of centripetal acceleration force and normal force

In summary, the conversation discusses a question about the maximum speed of a car traveling over a humpback bridge and the concept of "just in contact with the bridge." The conversation also brings up a question about the net centripetal force in a Ferris wheel model. The answer is provided that the net centripetal force must be equal to the mass times acceleration and that at the point where the car leaves the surface, N = 0. This means that Fc=mg = mv^2/r where r>45m.
  • #1
SecretSnow
66
0
Hi guys, first off, thanks for answering my previous question...

Now here's another one. I got confused by my school's homework a bit...so there's a car that travels over a humpback bridge with a radius of 45m, so basically it's a semicircle bridge, and so I'm supposed to calculate the maximum speed of the car if the wheels are to just stay in contact with the bridge. I don't get what the question mean when it says "just in contact with the bridge" does it mean that normal contact force will be zero? I imagine it to be right on top of this bridge, and checked the answer. I know the net force is the one that gives the centripetal force, and so

mg - N (where N is just the magnitude of normal force alone) must be greater or equal to mv^2/r. Is this correct? By the way, when I say greater than, I'm already assuming the car doesn't have constant speed along the bridge as it travels right?

Next, the answer told me that for the wheels to stay "just in contact with the bridge", the N at the top of the bridge will be zero. This, I don't get it at all, why would there even be N=0? The car is always in contact with the bridge. Does "just in contact" mean N=0? If that is the case, is N= 0 at other parts of the bridge?

Next, there's this ferry wheel model type of question, so imagine a object moving in loops around the Ferris wheel circumference (inner rim of the Ferris wheel of course, not on it). I understand that at the bottom of the loop, the N force must be greater than mg such that there will be a net centripetal force upwards turning the object. Then, at this bottom point, N must be 2mg such that 2mg-mg of the object will give a centripetal mg force on the object. THEN, at the top of the loop, it will make sense for N to be zero since the weight can entirely provide the centripetal force, but this is different from the car question above, so why is the car enjoying N=0 too? And there is something really funny about this Ferris wheel model as well, why must the net centripetal force be mg (at the top and bottom of the loop)?? There is just no explanation, I searched everywhere but there's nothing. What about at the sides of the loop? Must the centripetal force in this Ferris wheel model be mg too? These are extremely confusing qns to me..

Any help would be greatly appreciated, thanks a lot!
 
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  • #2
SecretSnow said:
Hi guys, first off, thanks for answering my previous question...

Now here's another one. I got confused by my school's homework a bit...so there's a car that travels over a humpback bridge with a radius of 45m, so basically it's a semicircle bridge, and so I'm supposed to calculate the maximum speed of the car if the wheels are to just stay in contact with the bridge. I don't get what the question mean when it says "just in contact with the bridge" does it mean that normal contact force will be zero? I imagine it to be right on top of this bridge, and checked the answer. I know the net force is the one that gives the centripetal force, and so

mg - N (where N is just the magnitude of normal force alone) must be greater or equal to mv^2/r. Is this correct? By the way, when I say greater than, I'm already assuming the car doesn't have constant speed along the bridge as it travels right?

Next, the answer told me that for the wheels to stay "just in contact with the bridge", the N at the top of the bridge will be zero. This, I don't get it at all, why would there even be N=0? The car is always in contact with the bridge. Does "just in contact" mean N=0? If that is the case, is N= 0 at other parts of the bridge?

Next, there's this ferry wheel model type of question, so imagine a object moving in loops around the Ferris wheel circumference (inner rim of the Ferris wheel of course, not on it). I understand that at the bottom of the loop, the N force must be greater than mg such that there will be a net centripetal force upwards turning the object. Then, at this bottom point, N must be 2mg such that 2mg-mg of the object will give a centripetal mg force on the object. THEN, at the top of the loop, it will make sense for N to be zero since the weight can entirely provide the centripetal force, but this is different from the car question above, so why is the car enjoying N=0 too? And there is something really funny about this Ferris wheel model as well, why must the net centripetal force be mg (at the top and bottom of the loop)?? There is just no explanation, I searched everywhere but there's nothing. What about at the sides of the loop? Must the centripetal force in this Ferris wheel model be mg too? These are extremely confusing qns to me..

Any help would be greatly appreciated, thanks a lot!

Just stick to the free-body diagram. There are two force vectors (2 forces. mg and N, and the vector representing the sum of those two forces, which is equal to the mass x acceleration. The only acceleration is centripetal.

If the car were to go over at too great a speed and leave the surface it would be because the centripetal force required to keep it following the road cannot be supplied by mg - N. But at that point N = 0 since it is not touching. So Fc=mg = mv^2/r where r>45m. If it is just touching the road but not putting any force on the road (N = 0) gravity is able to supply the needed centripetal force ie. Fc = mg = mv^2/r where r=45m.

AM
 
Last edited:

1. What is centripetal acceleration force and how does it differ from normal force?

Centripetal acceleration force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle and is proportional to the square of the object's speed. Normal force, on the other hand, is the perpendicular force exerted by a surface on an object in contact with it. While centripetal acceleration force is a type of force that causes circular motion, normal force is a reaction force that prevents objects from passing through each other.

2. Why do people often confuse centripetal acceleration force with normal force?

People often confuse centripetal acceleration force with normal force because both forces have a similar direction, towards the center of the circular motion. Additionally, both forces are often present in circular motion problems and are necessary for an object to maintain its circular path.

3. How can one differentiate between centripetal acceleration force and normal force in a problem?

The easiest way to differentiate between centripetal acceleration force and normal force is to remember that centripetal acceleration force is a type of force that causes circular motion, while normal force is a reaction force that prevents objects from passing through each other. Additionally, centripetal acceleration force is usually calculated using the equation F=ma, while normal force is calculated using Newton's third law of motion.

4. Can centripetal acceleration force ever be equal to normal force?

No, centripetal acceleration force and normal force can never be equal. Centripetal acceleration force is a type of force that causes objects to move in a circular path, while normal force is a reaction force exerted by a surface on an object in contact with it. These forces serve different purposes and are never equal in magnitude.

5. How does understanding the difference between centripetal acceleration force and normal force help in solving problems?

Understanding the difference between centripetal acceleration force and normal force is crucial in solving problems involving circular motion. It helps in correctly identifying the forces acting on an object and using the appropriate equations to solve for unknown variables. It also allows for a deeper understanding of the physics behind circular motion and how different forces interact with each other.

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