Forces applied by a slot on a slider

[Like Tony Stark]

Homework Statement

A disk with a slot rotates on a horizontal plane about a tree that passes through its centre, while a slider $P$ of mass $450 g$ moves on the slot.
The disk is at rest and the slider is in the position depicted in the picture when an angular acceleration of $40 \frac{rad}{s^2}$ is applied to the disk with a clockwise direction.
Find the horizontal force exerted on $P$ by the slot when the motion is started and the inicial acceleration of $P$ with respect to the slot.

Relevant Equations

Newton's equations

What forces are exerted by the slot on the slider? You have a normal force since it's inside it, and that's the only force that I can imagine. But then? I think about the pseudo-forces since it's on a rotating reference frame (centrifugal force, Coriolis force, Euler force), but those forces are not exerted by the slot itself.

 

[tnich]

You are asked to find the forces on the slider at a moment in time. What is the angular rate of the disk at that time? What does that imply about the rotating reference frame forces at that time?

 

[Like Tony Stark]

You are asked to find the forces on the slider at a moment in time. What is the angular rate of the disk at that time? What does that imply about the rotating reference frame forces at that time?

The only forces that I've identified are normal force and Euler force.
Then, for Newton's equation you have two options: you choose tilted axis, or you choose the axis depicted in the picture.

If we choose the second option, as Euler points to the negative direction of X, the normal force will point to the positive direction. So you'll have
$x) N-F_E=m.a_{rel}$
And for $y$ you don't have any forces, altough you have acceleration. What force is causing it??Then, if you choose the first option, you have
$x) N.cos(\alpha) -F_E cos(\alpha)=m.a_{rel_x}$
$y) F_E sin(\alpha)-N.sin(\alpha)=m.a_{rel_y}$
$a_{rel_y}=0$ so you have two unknowns (the normal force and the acceleration) but just one equation that contains them (the $x)$ equation contains both but $y)$ equation just contains $N$)

 

[tnich]

You can simplify this problem by looking at it in the inertial frame. At the point in time we are interested in, the angular rate of the disk is zero and the speed of the slider is zero.
As you pointed out in your original post, the fictional forces of the rotating reference frame are not exerted by the slot itself, so they are irrelevant.

 

[Like Tony Stark]

You can simplify this problem by looking at it in the inertial frame. At the point in time we are interested in, the angular rate of the disk is zero and the speed of the slider is zero.
As you pointed out in your original post, the fictional forces of the rotating reference frame are not exerted by the slot itself, so they are irrelevant.

Like this?

 

[tnich]

Your equations are very difficult to read. Are they still in the rotating frame? In post #1 you said that the normal force is the only force imposed on the slider by the slot. Do you now believe that is incorrect?

 

[Like Tony Stark]

Your equations are very difficult to read. Are they still in the rotating frame? In post #1 you said that the normal force is the only force imposed on the slider by the slot. Do you now believe that is incorrect?

Sorry, I'll write them down
$\hat e_{r}) N.cos(\theta)=m.(\ddot r -r.\dot \theta^2)$
$\hat e_{r}) N.cos(\theta)=m.\ddot r$

Also
$\hat e_{\theta}) N.sin(\theta)=m.(2\dot r \dot \theta -r.\ddot \theta)$
$\hat e_{\theta}) N.sin(\theta)=-m.r.\ddot \theta$

 

[tnich]

I do find it a bit confusing. You seem to be using $\theta$ both for the angle of the slot and for the angle of rotation. So when you say $N\sin\theta = -mr\ddot\theta$, you seem to imply that $N$ is not perpendicular to the surface of the slot. I would have thought that $N$ represented the normal force, but if it is not perpendicular to the slot, it is not a normal force by definition. In your diagram you show $\hat e_{\theta}$ as normal to the slot, and $\vec N$ in the direction of the acceleration of the disc at the location of the slider, but your equations do not match.
How about drawing a free body diagram for the slider and labeling the forces involved?

 

[Like Tony Stark]

I do find it a bit confusing. You seem to be using $\theta$ both for the angle of the slot and for the angle of rotation. So when you say $N\sin\theta = -mr\ddot\theta$, you seem to imply that $N$ is not perpendicular to the surface of the slot. I would have thought that $N$ represented the normal force, but if it is not perpendicular to the slot, it is not a normal force by definition. In your diagram you show $\hat e_{\theta}$ as normal to the slot, and $\vec N$ in the direction of the acceleration of the disc at the location of the slider, but your equations do not match.
How about drawing a free body diagram for the slider and labeling the forces involved?

Because I think that if the disk wasn't accelerating, there wouldn't be any force (there wouldn't be normal force because the slider wouldn't being pushed against the wall). But if we think in a non inertial frame (as I thought earlier) we see that the Euler force pushes the slider against the slot, so the normal force must be in the direction of Euler force, which is not perpendicular to the slot.

Then, can you tell me why my equations don't match?

 

[tnich]

It seems to me that $mr\ddot\theta$ must be in the direction of $\vec N$ in your diagram, but you show it in your equations as being in the direction of $-\hat e_{theta}$.
The problem with trying to solve this problem in a non-inertial frame is that the fictional forces are not forces, so none of them are actually applied by the slot. Besides, there are no fictional forces at time 0 (the time you need a solution for) because $\dot\theta=0$.
Assuming the slot is frictionless, there is only one force acting on the slider, the normal force of the slot, which is perpendicular to the slot by definition.

 

[Like Tony Stark]

It seems to me that $mr\ddot\theta$ must be in the direction of $\vec N$ in your diagram, but you show it in your equations as being in the direction of $-\hat e_{theta}$.
The problem with trying to solve this problem in a non-inertial frame is that the fictional forces are not forces, so none of them are actually applied by the slot. Besides, there are no fictional forces at time 0 (the time you need a solution for) because $\dot\theta=0$.
Assuming the slot is frictionless, there is only one force acting on the slider, the normal force of the slot, which is perpendicular to the slot by definition.

So the equation is
$\hat e_{\theta}) N=m.r.\ddot \theta$?

Where $r=\frac{x}{sin \theta}$

 

[tnich]

So the equation is
$\hat e_{\theta}) N=m.r.\ddot \theta$?

Where $r=\frac{x}{sin \theta}$

No, not quite. If the acceleration of the disk where it touches the slider is $mr\ddot\theta\hat x$, where $r$ is the distance from the center to the slider, and $\hat x$ is a unit vector in the horizontal direction in your diagram, then what would be the component of $mr\ddot\theta\hat x$ that is normal to the slot?

 

[Like Tony Stark]

then what would be the component of $mr\ddot\theta\hat x$ that is normal to the slot?

Are you saying that I shouldn't write the acceleration in the transversal coordinate but a component of it?
If that's what you mean I don't get the idea. The normal force is in $\hat e_{\theta}$, so I should write the acceleracion in $\hat e_{\theta}$, that's the component

 

[tnich]

The normal force is in $\hat e_{\theta}$, so I should write the acceleracion in $\hat e_{\theta}$, that's the component

I think I mostly agree with that. If you take the acceleration in the $\hat x$ direction and find the component of it in the $-\hat e_{\theta}$ direction, you would have the normal force. That would be:
$$mr\ddot \theta \hat x \cdot (-\hat e_{\theta})$$

 

[Like Tony Stark]

I think I mostly agree with that. If you take the acceleration in the $\hat x$ direction and find the component of it in the $-\hat e_{\theta}$ direction, you would have the normal force. That would be:
$$mr\ddot \theta \hat x \cdot (-\hat e_{\theta})$$

Sorry but I don't understand why $\hat x$. I mean, why do you use another direction ($\hat x$) if you are interested just in $\hat e_{\theta}$?

 

[tnich]

$\hat x$ is the horizontal direction in your diagram and $mr\ddot \theta \hat x \cdot (-\hat e_{\theta})$ is the component of the disk acceleration normal to the slot at the point of contact with the slider. I used a dot product to project the disk acceleration on $-\hat e_{\theta}$.
I've urged you to do a free body diagram of the slider. I think that if you drew the vector acceleration of the disk at the point of contact with the slider and the component of that vector in the $-\hat e_{\theta}$ direction, you could work out the answer from there.