Understanding Pseudo-Forces in Non-Inertial Frames

[Like Tony Stark]

Homework Statement

The picture shows a block on an arm. This arm is rotating with constant angular velocity. There is friction between the block and the arm, and the block is about to slide down.

Relevant Equations

Newton's equations

I know that, if my $X$ axis is aligned with the arm, there is friction pointing to $O$ and the $x$ component of the weight. Then, on the $Y$ axis, there is the normal force and the $y$ component of the weight.
Supposing that I am situated on the arm (non inertial frame), where is the centrifugal force pointing to? Because, if I think the problem as a spinning ball, the centrifugal force is aligned with the tension, so in this case this pseudo-force would be in the $x$ axis. But if I think as if I were on the arm, I think that I would feel pushed against the arm, so the centrifugal force would be in the $y$ axis.
What's the correct option?

 

[SammyS]

Homework Statement: The picture shows a block on an arm. This arm is rotating with constant angular velocity. There is friction between the block and the arm, and the block is about to slide down.
Homework Equations: Newton's equations

I know that, if my $X$ axis is aligned with the arm, there is friction pointing to $O$ and the $x$ component of the weight. Then, on the $Y$ axis, there is the normal force and the $y$ component of the weight.
Supposing that I am situated on the arm (non inertial frame), where is the centrifugal force pointing to? Because, if I think the problem as a spinning ball, the centrifugal force is aligned with the tension, so in this case this pseudo-force would be in the $x$ axis. But if I think as if I were on the arm, I think that I would feel pushed against the arm, so the centrifugal force would be in the $y$ axis.
What's the correct option?

What is the axis of rotation?

 

[Andrew Mason]

If I understand the configuration correctly, the rotation is about the vertical axis through the central point. Is that correct? If so the centipetal force is horizontal toward the axis of rotation. So there is a normal component perpendicular to the arm and a component along the arm.

AM

 

[Like Tony Stark]

If I understand the configuration correctly, the rotation is about the vertical axis through the central point. Is that correct? If so the centipetal force is horizontal toward the axis of rotation. So there is a normal component perpendicular to the arm and a component along the arm.

AM

It says that the arm rotates about a horizontal axis through $O$

 

[Like Tony Stark]

What is the axis of rotation?

It says that the arm rotates about a horizontal axis through $O$

 

[haruspex]

friction pointing to O

To O?

I think that I would feel pushed against the arm, so the centrifugal force would be in the y axis.

The only reason you might feel that is because of gravity. If the rotation were in the horizontal plane the normal force would be vertical.

 

[jbriggs444]

As I understand the setup, it is like a mouse-trap or a catapult (classic onager style like Wile E Coyote uses), but without angular acceleration. Just smooth rotation.

Adopting the rotating reference frame, one can identify four forces on the object: The downward force of gravity, the outward centrifugal force, the normal component of the contact force and an unknown frictional force.

Given that "the block is about to slide down", one can easily identify the direction in which the frictional force must be acting.

 

[Like Tony Stark]

As I understand the setup, it is like a mouse-trap or a catapult (classic onager style like Wile E Coyote uses), but without angular acceleration. Just smooth rotation.

Adopting the rotating reference frame, one can identify four forces on the object: The downward force of gravity, the outward centrifugal force, the normal component of the contact force and an unknown frictional force.

Given that "the block is about to slide down", one can easily identify the direction in which the frictional force must be acting.

That's my problem. If I think it as a catapult, what would be the force that doesn't allow the block to slide? Because while the catapult is moving, the thing that is in the catapult won't be thrown away until the catapult stop. And if that's the case, the centrifugal force would be in the direction of the weight, like case 2 ($f*$ is the centrifugal force)

 

[jbriggs444]

If I think it as a catapult, what would be the force that doesn't allow the block to slide?

Friction. Same as any other problem involving an inclined plane. If it is about to slide down, the rotation rate cannot be high enough to fling it outward.

Edit: Looking at your drawing, you have the weight force drawn normal to the plane. Why is that?

 

[Like Tony Stark]

To O?

The only reason you might feel that is because of gravity. If the rotation were in the horizontal plane the normal force would be vertical.

I see... so the pseudo-force is the centrifugal force. But if the block starts sliding, would it feel also Coriolis force? Because it would have velocity relative to the non inertial frame

 

[haruspex]

I see... so the pseudo-force is the centrifugal force. But if the block starts sliding, would it feel also Coriolis force? Because it would have velocity relative to the non inertial frame

Yes, but more specifically, because its relative velocity has a component normal to the axis of rotation of the frame.