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In the bucket experiment when the bucket reaches the top of the circle why will it have a normal force acting on the water downwards? Doesn't normal force oppose any other force? There is no force acting upwards... (As we are an observer in an inertial frame so we wont consider the centrifugal force because it is a pseudo force)

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  • $\begingroup$ In fact there is a reaction to the normal force and it is given by the contact force the liquid applies on the bucket. This happens even when the bucket is at the top of the trajectory, unless the water falls off. $\endgroup$ – Diracology May 26 '16 at 12:53
  • $\begingroup$ And what would be the magnitude of this contact force? $\endgroup$ – oshhh May 26 '16 at 13:01
  • $\begingroup$ Why are you removing the homework tag? $\endgroup$ – user36790 May 27 '16 at 11:10
  • $\begingroup$ because it isn't my homework and the discussion is more about a concept than the solution of this question in particular $\endgroup$ – oshhh May 27 '16 at 11:14
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    $\begingroup$ Hi Osheen Sachdev. Welcome to Phys.SE. Echoing @MAFIA36790's above comments, if you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic May 27 '16 at 13:19
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I want to focus on one thing, here. The nature of the normal force.

You write

Doesn't normal force oppose any other force?

which is a easy impression to get when you are introduced to the normal force in the context of things sitting on other things in a gravitational field, but that's not the best way to think about it.

The normal force keeps things from occupying the same space.

So consider a book sitting on the lab bench. It's subject to gravity, and in the absence of other forces would fall. But to fall it would have to occupy the same space as the solid top of the bench. The normal force is the interaction between to two objects that resists their inter-penetrating. In this case it has to provide a force that is equal and opposite to that of gravity to make that happen.

In the case of the water in the bucket, it's inertia would take it in a straight line, gravity modifies that into a parabola, but the sides and bottom of the bucket are moving in a circle for both those things to be true (the bucket goes in a circle and the water goes in a parabola) the water would have to move through the bottom of the bucket. The normal force serves to prevent the water from penetrating the solid material of the bucket and must supply whatever forces (beyond that of gravity) are necessary to cause the motion to be in a circle.

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  • $\begingroup$ But shouldn't normal force point sideways then...because water wants to go tangentially but the sides of the bucket prevent that? $\endgroup$ – oshhh May 26 '16 at 16:37
  • $\begingroup$ The normal from the sides of the bucket points toward the inside of the bucket all the way around. The normal from the bottom of the bucket points (roughly) toward the center of the circle (my shoulder joint when I do this demo in class). Remember that for circular motion $\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}$ points to the middle of the circle and that is the direction of the force that must be imposed on the water to keep it moving in a circle. $\endgroup$ – dmckee May 26 '16 at 16:43
  • $\begingroup$ As there is normal force pushing water from the bottom of the bucket, there must be a force pushing the water towards the bucket. I don't understand when the tension is inwards and water wants to go tangentially due to inertia then what force is pushing it outwards? considering the observer to be in an inertial frame there is no centrifugal force too... $\endgroup$ – oshhh May 27 '16 at 10:58
  • $\begingroup$ Remember that the water is not in equilibrium (because it is moving in a circle, which means that it is accelerating). The forces not only don't need to cancel out they need to not cancel out. $\endgroup$ – dmckee May 27 '16 at 15:06
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Imagine a scenario where the bucket is rotated at just the right speed so that the centripetal acceleration required to keep the water on a circular path is exactly 9.81 $ms^{-1}$. Then at the top of the rotation, all the centripetal acceleration is supplied by gravity.

However the bucket might be rotating faster in any given scenario but it still rotates on a fixed radius. Since the force required to maintain circular motion is given by

$F=\frac{mv^2}{r}$

the force needs to be greater at higher velocities if the radius is constant. This means that gravity is no longer able to supply all the acceleration required to keep the water inside the bucket on its circular path.

The water wants to move to a larger radius but it cannot because the bucket is in the way. Hence the water feels a reaction force from the bucket. This contact force is simply the force required to hold the water at a given radius minus the force of gravity:

$F_N=\frac{mv^2}{r}-F_g$

Another way to think about it is if you desperately tried to move through a wall the wall would exert a force on you and the force would increase the harder you tried.

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  • $\begingroup$ I don't understand when the tension is inwards and water wants to go tangentially due to inertia then what force is pushing it outwards? considering the observer to be in an inertial frame there is no centrifugal force too... $\endgroup$ – oshhh May 27 '16 at 10:55
  • $\begingroup$ You have to remember that the bucket is also following the circular path that the water takes $\endgroup$ – Jaywalker May 27 '16 at 10:56
  • $\begingroup$ There isn't actually a force acting outwards, it is only perceived that way. As explained in dmckee's answer the normal force is just what is keeping the water from going through the bucket. $\endgroup$ – Jaywalker May 27 '16 at 10:58
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    $\begingroup$ Yes but think about it. When the water moves tangentially, the bucket has also moved a bit so it is actually hitting the bottom of the bucket. $\endgroup$ – Jaywalker May 27 '16 at 11:25
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    $\begingroup$ Well that is the explanation. I'll try one more time with a different analogy. Its like you keep catching the water with the bucket by moving it such that it can never go off on its tangent. The whole thing is a constant game of "catch" between bucket and water. $\endgroup$ – Jaywalker May 27 '16 at 11:49
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In the bucket experiment when the bucket reaches the top of the circle why will it have a normal force acting on the water downwards?

The normal contact force is exerted by the bottom of the bucket as explicitly mentioned by the author. So, it is acting downward when the bucket is inverted.

Doesn't normal force oppose any other force? There is no force acting upwards

hmm... as said by Dirocology the bucket is forced by the contact force from the liquid; the normal force is reactive to that force.

Normal force plays the role of centripetal force when the velocity of the liquid is greater than $\sqrt{rg}\;.$

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Wanted to give a different angle to thinking about this.

Since you mentioned you wanted to consider this from the inertial frame (non rotating), then there is indeed no (fictitious) force. But in that frame, we can consider the water is indeed falling; however, the rate at which the bucket is also "falling" is such that the two stay together - in other words at sufficient speed of rotation the bucket follows the water and the water stays in the bucket.

It is sufficient to demonstrate that the vertical acceleration of the bucket is $\ge g$; for this you can consider the curvature and resulting change in velocity vector. Which looks a lot like the derivation of centripetal force, of course...

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