M3: banked tracks Watch

Maths&physics
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in the first example.

the car is traveling in a circle. I guess that the speed at which its traveling is whats acting away from the centre, but because its not slipping, friction is acting in the opposite directing to any of the angles the car is traveling in? if we take one position: the cars speed is acting towards the left <- , so friction is acting towards the right, but the reaction force is also acting towards the left (the same as the speed of the car)?

the second example, both the speed and the reaction force are acting towards the centre? so, whats stoping it from traveling in that direction?
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it would help me to know how this acting.

its acting in a circle on the surface, like 1, or is it traveling around it like 2?
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K-Man_PhysCheM
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(Original post by Maths&physics)
it would help me to know how this acting.

is acting in a circle on the surface, like 1, or is it traveling around it like 2?
It's travelling round a bend on a banked track: imagine cyclists going round a velodrome. So 2 is the better representation.
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(Original post by K-Man_PhysCheM)
It's travelling round a bend on a banked track: imagine cyclists going round a velodrome. So 2 is the better representation.
ah, ok.

so, in the second example: the 2 forces are acting inwards, and theres no friction, so whats stopping it from going to the centre?
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K-Man_PhysCheM
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(Original post by Maths&physics)
ah, ok.

so, in the second example: the 2 forces are acting inwards, and theres no friction, so whats stopping it from going to the centre?
The normal contact force has an upwards component.

In this case, the ball it is travelling at such a speed along a particular circle on the hemispherical bowl such that it stays in its circular path, without falling towards the centre. Therefore you can conclude that the upwards component of the normal contact force must exactly equal the weight force, but in opposite direction. This is the only way that the ball could not fall in towards the centre.

Resolving vertically: R\sin (\theta ) = mg

Resolving horizontally:  R \cos (\theta ) = m \omega^2 r

The rest is some simple geometry.
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Notnek
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(Original post by Maths&physics)
in the first example.

the car is traveling in a circle. I guess that the speed at which its traveling is whats acting away from the centre, but because its not slipping, friction is acting in the opposite directing to any of the angles the car is traveling in? if we take one position: the cars speed is acting towards the left <- , so friction is acting towards the right, but the reaction force is also acting towards the left (the same as the speed of the car)?

the second example, both the speed and the reaction force are acting towards the centre? so, whats stoping it from traveling in that direction?
Your commentary on the first diagram seems to lack a bit of understanding so I'll post an explanation below, plus I've never explained this topic on here before!

Speed or velocity is not a force so you can't compare it with friction to determine what will happen to the car. The tangential speed is constant but it is continually changing direction which means the velocity is changing and a change of velocity means that there is acceleration. This acceleration acts towards the centre of the circle and so there must be force acting towards the centre of the circle which is known as the centripedal force.

The centripedal force required to keep an object moving in a circle is

F_c=\frac{mv^2}{r}

So this force is proportional to the square of the tangential speed. The faster the speed, the bigger force is required to maintain circular motion. This will make sense if you imagine yourself in the car.

In your first question since you're considering maximum speed, this is going to require a big centripedal force to maintain circular motion so friction will need to act down the slope to maintain this big force. So you have this situation:


The first diagram shows the three forces acting on the particle and the second shows the horizontal/vertical components.

The centripedal force must satisfy F_c = Fr\cos(15)+R\sin(15).

So the friction needs to be big enough to maintain this big force directed left otherwise the car will slip. There is a maximum friction which means that eventually as you increase the speed there will be a point where the centripedal force cannot be maintained anymore and the car slips.
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old_engineer
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(Original post by Maths&physics)
ah, ok.

so, in the second example: the 2 forces are acting inwards, and theres no friction, so whats stopping it from going to the centre?
The second example is like a ball bearing rolling round the inside of a hemispherical bowl, such that it's motion forms a horizontal circle. Now consider what would happen if the bowl suddenly disappeared. Answer: the particle would become a projectile. The horizontal component of it's motion would remain constant and it would travel in a straight line when viewed from above or below. At the same time it would fall vertically under the influence of gravity.

Now put the bowl back. The bowl is the only thing that is stopping the particle from (a) travelling in a straight line as opposed to a circle and (b) falling vertically under the influence of gravity. To do that, the bowl must be exerting a force on the particle that can be resolved into a vertical component equal to mg and a horizontal component directed towards the centre of rotation equal to mrw^2.
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(Original post by K-Man_PhysCheM)
The normal contact force has an upwards component.

In this case, the ball it is travelling at such a speed along a particular circle on the hemispherical bowl such that it stays in its circular path, without falling towards the centre. Therefore you can conclude that the upwards component of the normal contact force must exactly equal the weight force, but in opposite direction. This is the only way that the ball could not fall in towards the centre.

Resolving vertically: R\sin (\theta ) = mg

Resolving horizontally:  R \cos (\theta ) = m \omega^2 r

The rest is some simple geometry.
thank you so much! brilliantly explained.

and r cos (theta) is acting towards the centre, whilst the speed m \omega^2 r is pushing the particle away, hence why its in equilibrium?
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(Original post by Notnek)
Your commentary on the first diagram seems to lack a bit of understanding so I'll post an explanation below, plus I've never explained this topic on here before!

Speed or velocity is not a force so you can't compare it with friction to determine what will happen to the car. The tangential speed is constant but it is continually changing direction which means the velocity is changing and a change of velocity means that there is acceleration. This acceleration acts towards the centre of the circle and so there must be force acting towards the centre of the circle which is known as the centripedal force.

The centripedal force required to keep an object moving in a circle is

F_c=\frac{mv^2}{r}

So this force is proportional to the square of the tangential speed. The faster the speed, the bigger force is required to maintain circular motion. This will make sense if you imagine yourself in the car.

In your first question since you're considering maximum speed, this is going to require a big centripedal force to maintain circular motion so friction will need to act down the slope to maintain this big force. So you have this situation:


The first diagram shows the three forces acting on the particle and the second shows the horizontal/vertical components.

The centripedal force must satisfy F_c = Fr\cos(15)+R\sin(15).

So the friction needs to be big enough to maintain this big force directed left otherwise the car will slip. There is a maximum friction which means that eventually as you increase the speed there will be a point where the centripedal force cannot be maintained anymore and the car slips.
thank you notnek, this was a great explanation but im going to have to go over it a couple of times to make sure I completely understand it.
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(Original post by old_engineer)
The second example is like a ball bearing rolling round the inside of a hemispherical bowl, such that it's motion forms a horizontal circle. Now consider what would happen if the bowl suddenly disappeared. Answer: the particle would become a projectile. The horizontal component of it's motion would remain constant and it would travel in a straight line when viewed from above or below. At the same time it would fall vertically under the influence of gravity.

Now put the bowl back. The bowl is the only thing that is stopping the particle from (a) travelling in a straight line as opposed to a circle and (b) falling vertically under the influence of gravity. To do that, the bowl must be exerting a force on the particle that can be resolved into a vertical component equal to mg and a horizontal component directed towards the centre of rotation equal to mrw^2.
again, a super explanation! thank you
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Notnek
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(Original post by Maths&physics)
thank you notnek, this was a great explanation but im going to have to go over it a couple of times to make sure I completely understand it.
If you have any questions let me know.

The key concept in my post is that the faster the speed, the bigger the force required to maintain circular motion. Then think about how the forces e.g. friction will affect the force directed towards the centre and how they change for low/high speed.
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(Original post by Notnek)
If you have any questions let me know.

The key concept in my post is that the faster the speed, the bigger the force required to maintain circular motion. Then think about how the forces e.g. friction will affect the force directed towards the centre and how they change for low/high speed.
so, as I understand it.

\frac{mv^2}{r} = Fr\cos(15)+R\sin(15)

 Fr\cos(15)+R\sin(15) is acting towards the centre, whilst is \frac{mv^2}{r} is acting in the opposite direction: therefore, theres equilibrium.
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Notnek
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(Original post by Maths&physics)
so, as I understand it.

\frac{mv^2}{r} = Fr\cos(15)+R\sin(15)

 Fr\cos(15)+R\sin(15) is acting towards the centre, whilst is \frac{mv^2}{r} is acting in the opposite direction: therefore, theres equilibrium.
No there’s no equilibrium directed towards the centre of the circle. The car is actually accelerating towards the centre of the circle. This can seem a bit strange because the car is not moving towards the centre. But it is continually changing direction in a circle which means there must be acceleration.

This resultant force is equal to mv^2/r which is equal to the combined reaction and friction force components. There is no force pointing in the opposite direction in this case.
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Marie12223
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guys could i get some help with chemistry
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guys could i get some help with chemistry
Not here. Please create a thread in the Chemistry forum.
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sure why not 😐
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(Original post by Notnek)
No there’s no equilibrium directed towards the centre of the circle. The car is actually accelerating towards the centre of the circle. This can seem a bit strange because the car is not moving towards the centre. But it is continually changing direction in a circle which means there must be acceleration.

sorry, equilibrium was the wrong word.

This resultant force is equal to mv^2/r which is equal to the combined reaction and friction force components. There is no force pointing in the opposite direction in this case.
this is where I dont understand. if friction and the resultant force are acting in one direction, whats acting in the opposite to stop it?
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Notnek
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(Original post by Maths&physics)
this is where I dont understand. if friction and the resultant force are acting in one direction, whats acting in the opposite to stop it?
There isn't anything. The friction/reaction produce a resultant force (and therefore acceleration) directed towards the centre of the circle which is the centripetal force.

It's like a satellite in orbit around Earth : there is a force pulling it to the centre (Earth's gravity) but no force opposing that. But it doesn't fall towards the Earth because it has a constantly changing velocity.
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(Original post by Notnek)
There isn't anything. The friction/reaction produce a resultant force (and therefore acceleration) directed towards the centre of the circle which is the centripetal force.

It's like a satellite in orbit around Earth : there is a force pulling it to the centre (Earth's gravity) but no force opposing that. But it doesn't fall towards the Earth because it has a constantly changing velocity.
are you able to draw a small diagram illustrating this please? because im having trouble. thanks
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Notnek
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(Original post by Maths&physics)
are you able to draw a small diagram illustrating this please? because im having trouble. thanks
What would you like the diagram to show? This diagram is a basic illustration of circular motion:



The centripetal force is directed towards the centre of the circle and there are no other forces. This produces an acceleration towards the centre of the circle.

On a banked track it is a combination of weight (and reaction force) and friction that produces this force. Depending on the speed, friction may act in the opposite direction but there will be a resultant force directed towards the centre if the particle is moving in a circle.
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