# When does the particle fall off the dome

In summary: Yes, it is true that part of the path having the same curvature of a circle implies the radial acceleration within that part of the path is always v^2/r.
Homework Statement
Calculating when does a particle falls off a dome
Relevant Equations
a = v^2 / r

A particle was resting on top of a dome and given a negligible push such that it falls. The question is at what angle will the particle fall off the dome.
The solution is that,
$$m\frac{v^2}{a} = mg\cos\theta - R$$
and by conservation of energy
$$0 + mga = \frac{1}{2}mv^2 + mga\cos\theta$$
then the angle is found to be
$$\theta=\cos^{-1}\frac{2}{3}$$

What I don't understand is, when the particle exerts force on the dome due to gravity, why doesn't the dome follows Newton's third law and exert an equal but opposite force on the particle
$$mg\cos\theta = R$$
but exerts a smaller force back to the particle such that a net acceleration towards the center is resulted in the particle instead?

You forget that the particle is moving and that a net acceleration is necessary to keep it on the circular path.

Orodruin said:
You forget that the particle is moving and that a net acceleration is necessary to keep it on the circular path.
Thank you for your answer. But I still don't quite understand why it is necessary, because if the particle does not fit the condition for circular motion, then it is not in a circular motion. In fact, it does fall off eventually. Nothing is really keeping it in a circular path.

Thank you for your answer. But I still don't quite understand why it is necessary, because if the particle does not fit the condition for circular motion, then it is not in a circular motion. In fact, it does fall off eventually. Nothing is really keeping it in a circular path.

If a particle, under the influence of gravity, is on any surface, then it will try to fall through the surface. But, the surface will prevent it. Gravity and the normal force from the surface combine to move the particle along some trajectory on the surface. The particle may get faster or slower. In this case, it will get faster.

Once the particle is moving, it has a natural path under gravity: some parabola, based on the instantaneous velocity of the particle.

If at some point on its trajectory the natural path under gravity is above the surface, then it will leave the surface.

In this case, a particle is on the surface of a sphere/dome is compelled to move in a circular path (if it is simply given an initial impulse) until it reaches a speed where its natural trajectory is above the surface, at which point it leaves the surface.

This coincides with the point where the normal force reduces to zero; which also coincides with the point where gravity is no longer sufficient to continue the circular motion.

DaveE and anorlunda
PeroK said:
If a particle, under the influence of gravity, is on any surface, then it will try to fall through the surface. But, the surface will prevent it. Gravity and the normal force from the surface combine to move the particle along some trajectory on the surface. The particle may get faster or slower. In this case, it will get faster.

Once the particle is moving, it has a natural path under gravity: some parabola, based on the instantaneous velocity of the particle.

If at some point on its trajectory the natural path under gravity is above the surface, then it will leave the surface.

In this case, a particle is on the surface of a sphere/dome is compelled to move in a circular path (if it is simply given an initial impulse) until it reaches a speed where its natural trajectory is above the surface, at which point it leaves the surface.

This coincides with the point where the normal force reduces to zero; which also coincides with the point where gravity is no longer sufficient to continue the circular motion.

I agree with you that the surface is preventing the particle from falling vertically through the surface and this force together with gravity forms the trajectory of the particle. But I don't understand why the resulting net force isn't tangential.

While it is proven to be true that fixed radial acceleration v^2/r implies an object is moving in circular path, is it true that part of the path having the same curvature of a circle implies the radial acceleration within that part of the path is always v^2/r? If it is true, then it seems to be violating Newton's third law in our case (I explained in the problem). Is Newton's third law violated here? If yes, how do we tackle that problem? If no, could you explain how it is not violated?

Thanks.

While it is proven to be true that fixed radial acceleration v^2/r implies an object is moving in circular path, is it true that part of the path having the same curvature of a circle implies the radial acceleration within that part of the path is always v^2/r? If it is true, then it seems to be violating Newton's third law in our case (I explained in the problem). Is Newton's third law violated here? If yes, how do we tackle that problem? If no, could you explain how it is not violated?

Thanks.

Clearly Newton's third law is not violated. There is a normal force on the particle, which is equal and opposite to the force that the particle exerts on the surface.

Your conceptual problem is that the particle is not exerting its full weight on the surface. Imagine, for example, that you construct a parabolic track. This track is exactly the shape that a projectile will take if fired at a certain speed at the top of the track. Now, in this case, the projectile will move along the track (theorectically it may be touching the track the whole way). But, it exerts no force on the track, because the projectile is actually in free-fall. In other words there is no normal force in this case.

Imagine, then, that the projectile is fired a little more slowly. Now the track is no longer the ideal parabola for the projectile. The projectile still moves along the track, but it's not quite in free fall this time. It's trying to move below the track. In this case there will be a non-zero normal force. But, this force will not be the full weight of the projectile.

If we analyse what is happening: some of the gravitational force is used to accelerate the particle in a curved path, leaving only some of the force to act as a force on the surface.

A related example is where an elevator accelerates downwards. You are still standing on the floor of the elevator, but only some of your weight is acting on the elevator floor. If you stand on a set of scales, they will record a lower weight while the elevator is accelerating. Once the elevator reaches a fixed speed, your weight will return to normal, as recorded on the scales.

The moral is that when an object is accelerating, you have to take that acceleration into account when looking at the force exerted by that object on an adjacent surface.

Orodruin
PeroK said:
Clearly Newton's third law is not violated. There is a normal force on the particle, which is equal and opposite to the force that the particle exerts on the surface.

Your conceptual problem is that the particle is not exerting its full weight on the surface. Imagine, for example, that you construct a parabolic track. This track is exactly the shape that a projectile will take if fired at a certain speed at the top of the track. Now, in this case, the projectile will move along the track (theorectically it may be touching the track the whole way). But, it exerts no force on the track, because the projectile is actually in free-fall. In other words there is no normal force in this case.

Imagine, then, that the projectile is fired a little more slowly. Now the track is no longer the ideal parabola for the projectile. The projectile still moves along the track, but it's not quite in free fall this time. It's trying to move below the track. In this case there will be a non-zero normal force. But, this force will not be the full weight of the projectile.

If we analyse what is happening: some of the gravitational force is used to accelerate the particle in a curved path, leaving only some of the force to act as a force on the surface.

A related example is where an elevator accelerates downwards. You are still standing on the floor of the elevator, but only some of your weight is acting on the elevator floor. If you stand on a set of scales, they will record a lower weight while the elevator is accelerating. Once the elevator reaches a fixed speed, your weight will return to normal, as recorded on the scales.

The moral is that when an object is accelerating, you have to take that acceleration into account when looking at the force exerted by that object on an adjacent surface.

Did you mean that when the velocity of the particle is changed, an impact force is exerted on the particle by the dome and an equal but opposite force is exerted on the dome by the particle, so the normal reaction R is not just the gravitational force component but also includes this impact force?

Thanks.

Did you mean that when the velocity of the particle is changed, an impact force is exerted on the particle by the dome and an equal but opposite force is exerted on the dome by the particle, so the normal reaction R is not just the gravitational force component but also includes this impact force?

Thanks.

No. There are no "impacts" on a dome.

What I don't understand is, when the particle exerts force on the dome due to gravity, why doesn't the dome follows Newton's third law and exert an equal but opposite force on the particle
$$mg\cos\theta = R$$
but exerts a smaller force back to the particle such that a net acceleration towards the center is resulted in the particle instead?
The particle does not exert force on the dome "due to gravity". The gravitational force mg is exerted on the particle by the Earth, and the particle exerts equal and opposite force on the Earth.
mg cos(theta) is the radial component of this force of gravity.
The other force acting on the particle is R, the outward normal force due to the surface of the dome. According to Newton's Third Law, particle exerts equal and opposite force on the dome.
The vector sum of these two forces - gravity and normal force- determines the acceleration of the particle while it moves on the surface of the dome. The radial component of the resultant force is equal to the centripetal force:
##m\frac{v^2}{a} = mg\cos\theta - R##, so
##R=mg\cos\theta-m\frac{v^2}{a}##.
The tangential component determines the acceleration along the circular track.
If the centripetal force needed requires zero or negative (inward) normal force, the particle does not stay on the surface any more, as the dome can only push it and is not able to pull.

PeroK said:
No. There are no "impacts" on a dome.

Sorry, I didn't say clearly. I meant impact on the particle, because there is change in velocity, hence momentum. This force should be exerted by the dome. In return, the particle also exerts an equal but opposite force on the dome, as stated in Newton's third law. Is that right?

Thanks.

Sorry, I didn't say clearly. I meant impact on the particle, because there is change in velocity, hence momentum. This force should be exerted by the dome. In return, the particle also exerts an equal but opposite force on the dome, as stated in Newton's third law. Is that right?

Thanks.
The normal force from the dome is a reaction force. The acceleration is due to gravity: tangential to the dome, which increases the speed of the particle; and normal to the dome, which causes the circular motion. The normal force on the dome is what's left over. This residual force eventually reduces to zero as the particle leaves the dome.

To be honest, I'm not sure any of the posts on this thread have successfully changed your fundamental misconception.

ehild said:
The other force acting on the particle is R, the outward normal force due to the surface of the dome. According to Newton's Third Law, particle exerts equal and opposite force on the dome.

Thanks for your answer. What I said "due to gravity" is the situation you described here.

The particle exerts force on the dome because it has got pulled by the gravitational force by the Earth and the dome is stopping it.

I didn't mean "the gravitational force exerted by the Earth on the particle in return exerts force on the dome".

Sorry I didn't say clearly.

PeroK said:
The normal force from the dome is a reaction force. The acceleration is due to gravity: tangential to the dome, which increases the speed of the particle; and normal to the dome, which causes the circular motion. The normal force on the dome is what's left over. This residual force eventually reduces to zero as the particle leaves the dome.

To be honest, I'm not sure any of the posts on this thread have successfully changed your fundamental misconception.

Would you mind telling me other than mg cos \theta, what force has been exerted on the dome by the particle?

Would you mind telling me other than mg cos \theta, what force has been exerted on the dome by the particle?
As explained in posts #4 and #6, the force is less than ##mg \cos \theta##.

Trying reading and understanding these posts.

Thanks for your answer. What I said "due to gravity" is the situation you described here.

The particle exerts force on the dome because it has got pulled by the gravitational force by the Earth and the dome is stopping it.
That is better...

You wrote in the first post
What I don't understand is, when the particle exerts force on the dome due to gravity, why doesn't the dome follows Newton's third law and exert an equal but opposite force on the particle
mgcosθ=R
You can not say that the particle exerts force on the dome due to gravity. It is the Earth that exerts force due to the gravitational interaction. The particle and the dome exert force on each other because they touch each other and interact. And this force is not equal to mgcosθ. The force depends on the motion of the particle, its speed. If the particle does not move,the force is mgcosθ, otherwise it is less
You can determine this force of interaction from the resulting motion of the particle. ##R=mg\cos\theta-m\frac{v^2}{a} ##

PeroK
ehild said:
The force depends on the motion of the particle, its speed. If the particle does not move,the force is mgcosθ, otherwise it is less
Thanks a lot.

ehild said:
If the particle does not move,the force is mgcosθ,
I don't think that's quite what you mean. Maybe, "if the only other forces and any acceleration are in the tangential direction then..."?

haruspex said:
I don't think that's quite what you mean. Maybe, "if the only other forces and any acceleration are in the tangential direction then..."?

Sorry, I didn't say clearly. I meant impact on the particle, because there is change in velocity, hence momentum. This force should be exerted by the dome. In return, the particle also exerts an equal but opposite force on the dome, as stated in Newton's third law. Is that right?
I think you mean impulse, not impact. I think you were on the right track.

You're correct in thinking that because the particle's momentum changes, there's an impulse imparted to it.

What you referred to as the "impact force," however, is simply the net force on the particle. It's not a separate force, as you seemed to imply.

Other than that, though, your reasoning was okay: momentum changing -> a non-zero impulse -> non-zero net force -> forces don't cancel. In the radial direction in particular, R won't be equal to ##mg\cos\theta##.

vela said:
I think you mean impulse, not impact. I think you were on the right track.

You're correct in thinking that because the particle's momentum changes, there's an impulse imparted to it.

What you referred to as the "impact force," however, is simply the net force on the particle. It's not a separate force, as you seemed to imply.

Other than that, though, your reasoning was okay: momentum changing -> a non-zero impulse -> non-zero net force -> forces don't cancel. In the radial direction in particular, R won't be equal to ##mg\cos\theta##.
Thank you. I looked it up and I think you are right. "impulse" is the right term. I learned this stuff ages ago so I made mistake on their names...

## 1. When does the particle fall off the dome?

The exact time at which a particle falls off a dome depends on various factors such as the initial velocity, the shape and size of the dome, and the force of gravity. It is not a fixed time and will vary in each scenario.

## 2. What is the force that causes a particle to fall off the dome?

The force of gravity is the main force that causes a particle to fall off a dome. As the particle moves closer to the edge of the dome, the force of gravity increases and eventually becomes greater than the force of the dome's surface tension, causing the particle to fall off.

## 3. Can a particle stay on top of the dome indefinitely?

No, a particle cannot stay on top of a dome indefinitely. Eventually, the force of gravity will overcome the surface tension of the dome and the particle will fall off.

## 4. Does the shape of the dome affect when the particle falls off?

Yes, the shape of the dome can affect when the particle falls off. A dome with a steeper curve will have a higher force of gravity, causing the particle to fall off sooner than a dome with a flatter shape.

## 5. How does air resistance affect the falling of a particle off the dome?

Air resistance can slow down the particle's fall, but it will not change the fact that it will eventually fall off the dome. The force of gravity will still overcome the force of surface tension, causing the particle to fall off at some point.

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