Equations of motion of a block on a spinning wedge

[Like Tony Stark]

Homework Statement

The wedge spins with angular velocity $\omega$. There's a block of mass $m$ on it which is restricted by a slot to slide down. Determine the acceleration of the block. Is there any force acting on the walls of the slot? How can you calculate it?

Relevant Equations

$\Sigma F_{e\theta}=m(2\dot r \dot {\theta})$
$\Sigma F_{er}=m(\ddot r -r\dot \theta^2)$

I think that the only force acting on the wall is the normal force caused by Coriolis force, so it can be calculated this way:
$N=m2\dot r \dot \theta$

But $\dot r$ is not constant, so how can I calculate it?

Then, I can't calculate the acceleration either since I don't have the value of $\ddot r$ to plug in the equation
$\Sigma F_{er}=m(\ddot r -r\dot \theta^2)$

 

[haruspex]

Are you not told the angle of the wedge?

 

[wrobel]

Yes the angle must be given. As I understand there is a vertical plane that contains the block, the wedge and the axis z and this vertical plane rotates with constant angular velocity $\omega.$ The block is compelled to stay in this plane by the ideal constraint. The system has one degree of freedom.
If it is so then it is sufficient to write the law of energy conservation relative the frame that rotates together with the plane. The centrifugal force is a potential force

 

[Like Tony Stark]

Are you not told the angle of the wedge?

It just says $\theta$ so I think I can assume that it is known

 

[Like Tony Stark]

As I understand there is a vertical plane that contains the block, the wedge and the axis z and this vertical plane rotates with constant angular velocity $\omega.$

There's a rotating platform and there's a wedge fixed to it. The block is on the wedge and it is compelled to stay in it.

If it is so then it is sufficient to write the law of energy conservation relative the frame that rotates together with the plane. The centrifugal force is a potential force

Is mechanical energy conserved? How should I consider the pseudo-forces?

 

[wrobel]

It just says θ\theta so I think I can assume that it is known

and this is the angle you differentiate in the initial post

 

[haruspex]

It just says $\theta$ so I think I can assume that it is known

Sure, but you omitted that.
You might confuse yourself by also using θ(t) for the angle of rotation of the wedge

 

[haruspex]

Is mechanical energy conserved? How should I consider the pseudo-forces?

It is not conserved in the lab frame. But @wrobel suggests that in the rotating frame centrifugal force represents a potential gradient. It's not an approach I'm comfortable with, so I'll leave it to him to assist further.

 

[wrobel]

I think that this approach will not be the main one in this thread, so I bring it completely as an alternative version of some main solution which will appear here by somebody's efforts

Let $s$ be a distance from the bottom angle of the slope to the block. Then the kinetic energy of the block (relative the rotating frame) is $T=m\dot s^2/2$. And the potential energy of the centrifugal force is (think why)
$$W=-\frac{1}{2}m\omega^2s^2\cos^2\alpha,$$ here $\alpha$ is slope's incline angle.
So the energy is conserved:
$$T+W+mgs\sin\alpha=const$$
Differentiating this equality in time we get the equation of motion: $\ddot s=\ldots$.
I also feel that this problem is actually about relative equilibrium of the block. The equilibrium can be found as a critical point of the potential energy $V(s)=W+mgs\sin\alpha$
This equilibrium is unstable as a maximum of the potential energy

UPD: there must be square on cosine. Corrected

 

[Like Tony Stark]

I've made a mistake with my notation... when I wrote $\dot \theta$ I meant "the $\theta$ direction from polar coordinates". So the angle of the wedge has nothing to do with it. Let's call it $\alpha$

 

[Like Tony Stark]

It is not conserved in the lab frame.

Why? There are no non-conservative forces acting on the system, I think.

 

[haruspex]

Why? There are no non-conservative forces acting on the system, I think.

Something is keeping the wedge rotating at a constant rate. If the block changes radius its KE will change. This must be from a normal force from the side of the slot.

 

[wrobel]

This must be from a normal force from the side of the slot.

and this force does not work relative to the rotating frame but it does work relative to the lab frame

 

[Like Tony Stark]

and this force does not work relative to the rotating frame but it does work relative to the lab frame

That force would be caused by Coriolis force, wouldn't it?

 

[wrobel]

That force would be caused by Coriolis force, wouldn't it?

for the observer sitting inside the rotating frame the side reaction from the wedge is equalized by the Coriolis force ; in the lab frame the Coriolis force does not exist and the reaction causes acceleration of the block