Acceleration of beads on a ring

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The discussion focuses on calculating the accelerations of two beads, A and B, on a massless ring under the influence of gravity. The position vector for bead B is defined as $$r_{B/O}(t)=\left[r\cos(\theta+\beta)-r\sin\theta\right]e_1-\left[r\cos\theta+\sin(\theta+\beta)\right]e_2$$, and the angular velocity is given by $$\omega_{1/F}=-\dot{\theta(t)}e_3$$. The initial conditions indicate that bead A is fixed on the ring, while bead B is initially at rest with no normal force acting on it. The analysis reveals that the accelerations of A and B are initially independent, with A moving in a circular arc and B moving vertically downward.

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Bling Fizikst
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There seems to be too many things going on at the same time , including center of mass shifts . I am trying to find the accelerations as functions of time .
I tried to write the position vector of B : $$r_{B/O}(t)=\left[r\cos(\theta+\beta)-r\sin\theta\right]e_1-\left[r\cos\theta+\sin(\theta+\beta)\right]e_2$$

$$\omega_{1/F}=-\dot{\theta(t)}e_3$$

$$a_{B/F}(t)=a_{B/1}+a_{O/F}+\dot{\omega_{1/F}}\times r_{B/O}(t) -\omega_{1/F}^2 r_{B/O}(t)$$
##a_{B/1}=-ge_2, a_{O/F}=0##
I am really confused how to proceed . I tried writing the moment wrt ##O## to find ##\dot{\omega_{1/F}}## . But then it involves both ##\beta ,\theta## . Both unknowns , i am unable to put all of these together .
 
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So, the bead at A is fixed on the ring. The ring is considered massless. The initial torque from the weight of A about O is counterclockwise at ##t=0##. The bead at B is initially at rest, there is no Normal force from the ring yet acting on it at ##t=0## I would think, nor is there a torque on the ring yet acting from ##B## by Newtons 3rd ( no velocity, nor any frictional force). So it feels like to me at ##t=0## the acceleration of A and B are initially independent of each other ( A accelerating on a circular arc about O - counterclockwise, and B accelerating linear path vertically - down- just initially).
 
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erobz said:
So, the bead at A is fixed on the ring. The ring is considered massless. The initial torque from the weight of A about O is counterclockwise at ##t=0##. The bead at B is initially at rest, there is no Normal force from the ring yet acting on it at ##t=0## I would think, nor is there a torque on the ring yet acting from ##B## by Newtons 3rd ( no velocity, nor any frictional force). So it feels like to me at ##t=0## the acceleration of A and B are initially independent of each other ( A accelerating on a circular arc about O - counterclockwise, and B accelerating linear path vertically - down- just initially).
If A accelerates initially as you say then B must have some rightward acceleration.
 
haruspex said:
If A accelerates initially as you say then B must have some rightward acceleration.
So you are saying ##A## does not initially accelerate the way I think, because I see no possible contact force on ##B## that could hold it in place and not accelerate it vertically without the presence of friction between the rod and bead. When you said this my thought was "oh, they are initially pressing the bead radially I guess", but what force is balancing the initial weight of the bead without accelerating it upward.

Scratch that, Ok, I see they could be applying a constraint force of magnitude ## \sqrt {2} mg ## initially to constrain the system, so they are in contact initially…um…I think.

I saw this as the bead and the ring are held initially in this position by two distinct forces requiring no contact between rod and bead and released together.
 
Last edited:
erobz said:
they could be applying a constraint force of magnitude ## \sqrt {2} mg ## initially to constrain the system, so they are in contact initially…um…I think.
Yes.
 
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Bling Fizikst said:
I am trying to find the accelerations as functions of time .
You are only asked for the accelerations immediately upon release, so just think about the forces on the masses and torques on the hoop immediately before release.
 

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