Centripetal force direction confusion

[tellmesomething]

Homework Statement

A motorcyclist wants to drive on the vertical surface of a wooden well of radius 5m. In horizontal plane with speed of 5 root 5 m/s. find the min value of coefficient of friction between tyres and the wall of the well

Relevant Equations

Ah

So far: I an having trouble in the FBD. I drew one completely opposite to one I found on google . in this image NORmal force is pointed away from the cyclist and centripetal force is pointed away from the centre . mine was the complete opposite am i wrong?

 

[BvU]

I drew one

Don't see it !?

in this image NORmal force is pointed away from the cyclist

In your image or in the image that you did post ?

[edit] ah, you elaborated :​

​

mine was the complete opposite​

​

Doesn't seem right. Whereas

is correct.​

Do post your picture, to avoid confusion and ambiguity...​

Anyway, the

isn't a real force on the cyclist.

 

[Chestermiller]

I agree with your assessment.

 

[tellmesomething]

Don't see it !?In your image or in the image that you did post ?

[edit] ah, you elaborated :​

​

Doesn't seem right. Whereas View attachment 333711 is correct.​

Do post your picture, to avoid confusion and ambiguity...​

Anyway, the View attachment 333710 isn't a real force on the cyclist.

As far as I know normal force acts away from the body when two bodies are pressed as some sort of repulsion. Therefore the normal force on the cyclist from the well should also be away from the body? Also why is centrifugal force coming into play?

 

[tellmesomething]

I agree with your assessment.

So this is correct?

Sorry I am new here im not sure if you replied to my post or to @BvU 's reply.

 

[Chestermiller]

So this is correct?

Yes, the normal force on the cycle is also equal to the net force, which is the mass times the centripetal acceleration.

 

[haruspex]

NORmal force is pointed away from the cyclist

The normal force comes from contact with the wall, so points towards the centre, as in the diagram. Whether it is shown as from the cyclist towards the centre or from the wall towards the cyclist is irrelevant.

centripetal force is pointed away from the centre

The problem is that the text says "centripetal" but the diagram is for centrifugal.

The centripetal force would not be shown in an FBD because it is not an applied force; it is just a component of the resultant of all the applied forces.
In this case the normal force supplies the centripetal force: $\vec N=\vec F_{centripetal}=-\frac{mv^2\hat r}r$. That's in an inertial frame.

In the frame of reference of the cyclist, there is no acceleration. The centrifugal force is effectively an applied force, so would be in the FBD. It balances a component of the net of all the other forces, namely, the component normal to the velocity: $\vec N=-\vec F_{centrifugal}=-\frac{mv^2\hat r}r$.

 

[gmax137]

A free body diagram should show the forces acting on one body. I think your googled drawing shows more, leading to confusion.

 

[BvU]

is definitely NOT correct. Such a force would not act on the cylist,

$\$

 

[tellmesomething]

View attachment 333716 is definitely NOT correct. Such a force would not act on the cylist,

$\$

Im stumped! Can you elaborate?

 

[tellmesomething]

The normal force comes from contact with the wall, so points towards the centre, as in the diagram. Whether it is shown as from the cyclist towards the centre or from the wall towards the cyclist is irrelevant.

The problem is that the text says "centripetal" but the diagram is for centrifugal.

The centripetal force would not be shown in an FBD because it is not an applied force; it is just a component of the resultant of all the applied forces.
In this case the normal force supplies the centripetal force: $\vec N=\vec F_{centripetal}=-\frac{mv^2\hat r}r$. That's in an inertial frame.

In the frame of reference of the cyclist, there is no acceleration. The centrifugal force is effectively an applied force, so would be in the FBD. It balances a component of the net of all the other forces, namely, the component normal to the velocity: $\vec N=-\vec F_{centrifugal}=-\frac{mv^2\hat r}r$.

I dont understand why centrifugal force comes into play here, isn't the cyclist moving in that circle because theres a force pulling it towards the centre, which should be centripetal?

 

[Lnewqban]

I dont understand why centrifugal force comes into play here, isn't the cyclist moving in that circle because theres a force pulling it towards the centre, which should be centripetal?

From the point of view of an observer standing on the ground (inertial frame of reference):
There is velocity and a radius or circular trajectory that the bike is influenced by; therefore, there is centripetal acceleration.

From the point of view of the motorcycle (non-inertial rotating frame of reference):
The tires of the bike are pushing against the wall with centrifugal force, which makes possible the existence of the vertical friction keeping the bike from sliding down the wall.

Comparing magnitude of both forces:
From the first (inertial) point of view: The wall is pushing against the tires of the bike with centripetal force, which induces the horizontal circular trajectory.
That centripetal or normal force coming from the wall is the reaction par (third Newton’s law) that equals in magnitud the centrifugal force experienced from the second (non-inertial) point of view.

Edited confusing post (see posts 14 and 15 below).

 

[haruspex]

I dont understand why centrifugal force comes into play here, isn't the cyclist moving in that circle because theres a force pulling it towards the centre, which should be centripetal?

It depends on the reference frame. In the present case, there are two main options. You can use the 'ground' frame (or any inertial frame) or you can use the frame of reference of the cyclist.

If you choose an inertial frame then there are three forces acting: the normal force, the frictional force and the gravitational force. The result of these forces is the acceleration.
We can choose to resolve the acceleration into the component parallel to the velocity (the tangential component) and that normal to the velocity (the radial or centripetal component).
In the present case there is no tangential component, so all the acceleration is centripetal.

If you choose the cyclist's reference frame then, by definition, the cyclist is stationary, and certainly not accelerating. So all forces must be in balance. To make this work, in a rotating reference frame we have to add in a virtual (or "fictitious" force) acting outwards from the centre of rotation. This is what is called the centrifugal force.

 

[haruspex]

There is velocity and a radius or circular trajectory that the bike is influenced by; therefore, there is centripetal acceleration.
The tires of the bike are pushing against the wall with centrifugal force, which induce the vertical friction.
The wall is pushing against the tires of the bike with centripetal force, which is the reaction par that equals in magnitud the centrifugal force, which induce the horizontal circular trajectory.

No! Please do not confuse the OP.

 

[PeterDonis]

There is velocity and a radius or circular trajectory that the bike is influenced by; therefore, there is centripetal acceleration.
The tires of the bike are pushing against the wall with centrifugal force, which induce the vertical friction.
The wall is pushing against the tires of the bike with centripetal force, which is the reaction par that equals in magnitud the centrifugal force, which induce the horizontal circular trajectory.

You're mixing up two different frames here, which is not helpful to the OP.

Your first sentence is true in an inertial frame; but in an inertial frame there is no centrifugal force.

Your second sentence is true in a non-inertial rotating frame in which the cyclist is at rest; in this frame the cyclist has zero acceleration and the "velocity" $v$ is the velocity of the frame, relative to an inertial frame. Also there is no "circular trajectory" of the cyclist in this frame; he's at rest, staying at the same place.

Your third sentence would need to be restated in either frame since it mixes up things from both frames, which doesn't work.

 

[Lnewqban]

No! Please do not confuse the OP.

Not trying to do so on purpose!
So far, it seems that none of the previous posts has been able to make the OP less confused.

I have just edited my previous post, please see if you find anything to be still incorrect.
Thank you both, @haruspex and @PeterDonis.

 

[PeterDonis]

Commenting on edited version of post:

From the point of view of an observer standing on the ground (inertial frame of reference):
There is velocity and a radius or circular trajectory that the bike is influenced by; therefore, there is centripetal acceleration.

Yes.

From the point of view of the motorcycle (non-inertial rotating frame of reference):
The tires of the bike are pushing against the wall with centrifugal force, which makes possible the existence of the vertical friction keeping the bike from sliding down the wall.

Yes.

Comparing magnitude of both forces:

This is a misnomer in the inertial frame, since in that frame there is only one force in the horizontal direction.

From the first (inertial) point of view: The wall is pushing against the tires of the bike with centripetal force, which induces the horizontal circular trajectory.

And there is no other horizontal force in this frame.

That centripetal or normal force coming from the wall is the reaction par (third Newton’s law) that equals in magnitud the centrifugal force experienced from the second (non-inertial) point of view.

Here you are describing things in the non-inertial frame in which the bike is at rest. The fact that the magnitudes of both forces (equal and opposite in direction) in this frame are equal to the magnitude of the centripetal force in the inertial frame is a consequence of how the two frames are chosen.

(Also, I'm not sure there is agreement in the literature about whether "fictitious" forces like centrifugal force can participate in Newton's Third Law pairs. But that's a relatively minor point here: in the non-inertial frame the bike is at rest so the forces on it must all balance. Whether the horizontal balance can be considered a Newton's Third Law pair or not does not affect the fact that the forces are in balance.)