Find the equilibrium angle of this unusual driven pendulum geometry

[Alif Yasa]

Homework Statement

In Picture

Relevant Equations

Force Equilibrium and Torque Equilibrium

-I tried to draw the forces on the hoop when it is in the equilibrium state. I know there are friction and normal force on the contact point of the shaft and the hoop
-I also put the weight force to the M object
-But when i used the torque equilibrium, where the pivot is the contact point of the shaft and the hoop, the only torque exist is the weight force that is caused by the M object
-I expect I'm missing other forces, but i don't know what is it

 

[haruspex]

when i used the torque equilibrium, where the pivot is the contact point of the shaft and the hoop, the only torque exist is the weight force that is caused by the M object

So what does that tell you about the location of the mass?

 

[Alif Yasa]

I find that the torque by the mass is
T=Mg sin(x) *(R+L)
where x is the equilibrium angle

 

[haruspex]

I find that the torque by the mass is
T=Mg sin(x) *(R+L)
where x is the equilibrium angle

Your diagram must be wrong. Can you post it?

 

[Alif Yasa]

it is something like this

 

[haruspex]

it is something like this
View attachment 241436

Ok, but then I do not understand how you got what you did for the torque about the point of contact.
You correctly determined that the weight of the mass is the only force that could have such a torque. So in equilibrium, what does that tell you about the position of the mass in relation to the point of contact?

 

[Alif Yasa]

Oh, so is it like
Mg cos(x)R-Mg sin(x)(R+L)=0
and tan(x) = R/(R+L)

Also, from the force equilibrium, i got
tan(x)=1/u
cos(x)=u/sqrt(1+u^2)

The answer from the book is
x=arcsin((Ru)/((R+L)sqrt(1+u^2))

I can get it if i multiply tan(x) from the torque with cos(x) from the force,
but why it isn't just arctan(1/u) or arctan(R/(R+L)) ?
Do it have to be done that way ?

 

[haruspex]

Oh, so is it like
Mg cos(x)R-Mg sin(x)(R+L)=0

No.
Call the centre of the hoop C, the point of contact P and the mass M.
You seem to be assuming PCM is a right angle. It need not be.
E.g., without the information that the shaft is rotating, a solution would be that P is at the top of the shaft and the rod is vertical.

Think about the fact that the line of action of the weight is vertically through the mass. Where must it lie in relation to P?

 

[Alif Yasa]

So, if i make the angle between vertical axis and PC is y, and the equilibrium angle x

The net torques are

mg sin(x)(R cos(x+y)+R+L)-mg cos(x)(R sin(x+y)) =0
sin(x)(R+L) = R sin(y)

and from the forces

tan(y)=u
sin(y)=u/sqrt(1+u^2)

then i get
sin(x) = (Ru)/((R+L)sqrt(1+u^2))

Is that correct ?

 

[haruspex]

So, if i make the angle between vertical axis and PC is y, and the equilibrium angle x

The net torques are

mg sin(x)(R cos(x+y)+R+L)-mg cos(x)(R sin(x+y)) =0
sin(x)(R+L) = R sin(y)

and from the forces

tan(y)=u
sin(y)=u/sqrt(1+u^2)

then i get
sin(x) = (Ru)/((R+L)sqrt(1+u^2))

Is that correct ?

Looks right!
(What I was trying to get you to spot is that the mass must be vertically under P.)

 

[Alif Yasa]

Looks right!
(What I was trying to get you to spot is that the mass must be vertically under P.)

Thanks !