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Bead Dynamics

Tanisha Singh Dhami

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Index

Conclusion

Experiment

Quantitative model

Qualitative analysis

Parameters

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Problem statement

“A circular hoop rotates about a vertical diameter. A small bead is allowed to roll in a groove on the inside of the hoop.

Investigate the relevant parameters affecting the dynamics of the bead.”

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Experiment

Objectives

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Experimental Set-up

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A circular hoop rotates about a vertical diameter. A small bead is allowed to roll in a groove on the inside of the hoop

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Classification

Mechanics

Fluids

Rigid bodies

Statics

Dynamics

Deformable bodies

Particle dynamics

Rigid body dynamics

Bead dynamics

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Part One:

Qualitative Analysis

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Notations - introduction

mg

mg cos ϕ

mg sin ϕ

ϕ

r

m → mass of bead

r → radius of hoop

g → acceleration due to gravity

ϕ→ acute angle from axis of hoop

We will use ω as the angular velocity of the hoop

Assumptions:

The acceleration due to gravity is a constant=9.81.

The bead is an exact sphere for qualitative theory.

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Angular velocity of hoop
Mass of bead
The radius of the hoop
The initial angle
Friction between hoop and bead
Acceleration due to gravity

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Parameters

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FORCES

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Initially, the bead remains at rest.

When some angular velocity is provided, it starts rising up along the hoop, constrained to one side.

Then, it comes to rest at a point midway through the height of the hoop. This point is called an equilibrium point, that we will be looking into.

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What the terms are

The forces acting on the bead:

Centrifugal force (Fc)
Acceleration due to gravity (mg)
Normal (N)
Friction (f)

Force due to gravity

Centrifugal term

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EQUATIONS AND CALCULATION

OR

Force due to gravity

Friction term

Centrifugal term

Friction constant

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EQUILIBRIUM POSITIONS

Two equilibrium positions occur when ϕ = 0°, 180° :

(without the friction term)

Centrifugal force

At ϕ = 0° or 180°, sin ϕ = 0, thus the net force on the bead equals zero. Although the 180 degree theoretically is an equilibrium solution, the bead cant stay stationary there.

Similarly, When the second term = 0,

As rω^2 approaches ∞ the equilibrium position approaches ± 90°.

Which side is chosen (for the bead to rise to about 90°) depends on the initial disturbance.

Positions where there is no net force acting on the bead.

Force due to gravity

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EQUILIBRIUM POSITIONS

Positions where there is no net force acting on the bead.

r

N

R

mg

θ

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Stable/ unstable equilibrium

The equilibrium position at 0° is a stable equilibrium position whereas the other equilibrium positions are unstable.

Stable equilibrium position: A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite the direction of the displacement. This means that stable equilibrium is one, such that the system comes back to equilibrium even if it is slightly displaced.

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Symmetry-broken solutions

In physics, symmetry breaking is a phenomenon in which (infinitesimally) small fluctuations acting on a system crossing a critical point decide the system's fate, by determining which branch of a bifurcation is taken. To an outside observer unaware of the fluctuations (or "noise"), the choice will appear arbitrary.

Which of these two equilibrium points is actually selected depends on the initial disturbance. Even though the two fixed points are entirely symmetrical, a slight asymmetry in the initial conditions will lead to one of them being chosen

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When does the bead start to rise?

We can see that the bead does not move from the equilibrium position at all.

To understand why this happens, we need to consider the equation:

Only if ω is greater than will the bead

start moving. This is because, at ω =

the angle is zero. As ω increases, the

angle increases.

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slow spin

𝛄 ≤ 1

𝛄 > 1

𝛄 >> 1

Fast spin

spin needed to reach 90 degree

Frictionless surface

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Bifurcation diagram

A graphical depiction of the relationship between the values of one parameter and the behaviour of the system.

When γ < 1, the hoop is rotating slowly and the centrifugal force is too weak to balance the force of gravity. Thus the bead stays at the bottom φ = 0. But if γ > 1, the hoop is spinning fast enough that the bottom becomes unstable. Since the centrifugal force grows as the bead moves farther from the bottom, any slight displacement of the bead will be amplified. The bead is therefore pushed up the hoop until gravity balances the centrifugal force. As mentioned earlier, which of these paths is chosen is a symmetry broken solution.

This is called a supercritical pitchfork bifurcation.

So which side does it go?

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Part Two:

Parameters

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ANGULAR VELOCITY OF THE HOOP

The equilibrium position of the bead approaches ± 90° as the angular velocity of the hoop increases.

At very small angular velocities, the bead remains stationary

As ω increases, the acceleration of the bead up the hoop increases.

As the hoop continues to spin, the bead moves from one side to another (not reaching ± 90°) this change in sides happens because once the bead starts rolling down one side, it has an inertia to climb the other side.

Once ω is high enough for the bead to reach the ± 90° positions, the bead directly climbs up till 90° on a single side of the hoop.

Further increasing the angular velocity does not change the bead’s position.

PARAMETER ONE

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MASS OF BEAD

If we do not consider friction, the mass of the bead has no effect on its dynamics or equilibrium positions. Thus, although we had hypothesized that the mass of the bead will affect its dynamics, we realize now that it is not an essential factor.

PARAMETER TWO

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RADIUS OF HOOP

Critical angular velocity will decrease with increase of radius. It is inversely proportional to the square of the radius. ω_o=g

r

Thus, as the radius of the hoop is increase, we will have to provide lesser angular velocity to the hoop to make the bead rise.

We could not experimentally verify the effect due to lack of apparatus to keep angular velocity constant and change the radius of hoop conveniently.

PARAMETER THREE

√

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INITIAL ANGLE OF BEAD

We have used simulations to observe the effects of changing the initial angle of the bead.

The initial angle of the bead has no impact on the dynamics. The bead slides down to zero degrees (whatever initial angle we start from) and then follows the same pattern (keeping other factors constant). Attaching the pictures of simulations:

PARAMETER FOUR

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Friction

Initially, the bead remains at rest.

When some angular velocity is provided, it starts rising up along the hoop, constrained to one side.

Then, it comes to rest at a point midway through the height of the hoop. This point is called an equilibrium point, that we will be looking into.

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PARAMETER FIVE

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FRICTION

Friction helps in raising the bead and also maintaining it at the equilibrium position.

At the initial stationary position, friction has no significant role.

While the bead is rising, lesser the friction, higher the acceleration (rate of change of angle with time). The bead will rise faster.

As we can see, when the friction constant (b) decreases, the net force increases.

PARAMETER FIVE

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ACCELERATION DUE TO GRAVITY

As acceleration due to gravity increases,

If other parameters are kept constant: The bead can still reach the same equilibrium position (with enough angular velocity) but the time it takes when oscillating will decrease.

The critical angular velocity (that we had calculated earlier) needed to displace the bead from its stable equilibrium position increases. Thus, we will have to give it higher angular velocity to move the bead, the velocity needed is proportional to the root of g.

PARAMETER SIX

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Part Three:

Quantitative model

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SIMULATION

Since the friction could not be wholly neglected, i have used some simulations to work around that problem.

One was using java (ejs) and the other was using python (Trinket - GlowScript)

The software used is ejs (easy java simulations).

It allows me to choose a radius, and then change the angular velocity and the initial angle. I can play the simulation and see how the angle of the bead changes instantaneously.

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Angular velocity

I have varied angular velocity and plotted the maximum angle θ reached by the bead.

As seen in the graph angle increases with the increase in angular velocity till the angle reaches 90 degree.

90 degrees in the equilibrium position and even if we further increase the angular velocity, the position angle does not increase.

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Change in Radius

I have kept angular velocity ω constant and gradually increased the radius.

The maximum angle reached by bead increases with the increase in radius.

As you can see, there is an asymptote at the 90 degree position.

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Friction

Using the simulation, I can completely remove friction. Upon doing so, we can see that the observations and hypothesis previously made were correct.

When the hoop was given ω > ω_o (angular velocity needed for the bead to reach equilibrium position), it moved to one side and moved up and down on the same side. (as can be seen in the graph of angle plotted against time, since all the values are above 0 (which is the bottom most point of the hoop).

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ACCELERATION DUE TO GRAVITY

PARAMETER SIX

g=60 m/s²

g=10 m/s²

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Part Four:

Experiment

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EXPERIMENTAL SET-UP

A motor connected to the hoop

Plastic sheet to keep the bead inside + block the wind

Clay + hot glue to attach the hoop onto the motor

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The critical velocity

The radius of the hoop was 8 cm.

Thus, critical angular velocity (ω_o) can be calculated using:

ω_o=√(g/r)

= √(9.81 m/s² /8 cm )

= √(9.81 m/s² / 0.08 m )

= √122.625

= 11.07 Hz or 69.58 rad/s (correct to four significant figures)

Thus, only if the hoop is rotating at 69.58 rad/s will the bead move from the initial equilibrium position.

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The critical velocity

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Angular velocity

RPM	Rad per second	Maximum angle	Angle calculated
125	13	45	44.32
250	26.18	80	79.70
500	52.36	90	87.44

Critical angular velocity calculated to be 11.07 rad/s or 105.75 RPM.

Experimentally too, the bead would only move if about 108 RPM was provided

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EXTENDED RESEARCH: Multiple beads system

When multiple beads were used in the same experiment, approximately half of them moved to one side whereas other half moved to the other.

The ones that were to the right, rose up to the right side of the hoop, and the one which were slightly towards the left rise up on the left side of the hoop.

This proves our theory that the side to which the bead rises depends on the slight difference in its position from the wanted equilibrium position at an angle of zero degrees,

The rotation was tried multiple times, the beads rose up to a similar height everytime. When the set of beads start sliding down the side that they initially rose to, they cross besides one another and move to the opposite side. This proves the theory that the bead moves from side to side after initially rising to one side because of its inertia.

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Lagrangian equations

The same results can be arrived at when using Lagrangian equations of motion too, thus, I will not be going into it in detail.

x = r sin θ cos φ ; x˙ = r˙ sin θ cos φ + r cos θ cos φ ˙θ − r sin θ sin φ φ˙

y = r sin θ sin φ ; y˙ = r˙ sin θ sin φ + r cos θ sin φ ˙θ + r sin θ cos φ φ˙

z = r cos θ ; z˙ = r˙ cos θ − r sin θ ˙θ

T = ½ m (x˙ ² + y˙ ² + z˙ ² )

T = ½ m (r˙² + r² ˙θ² + (r sin θ)² φ˙² )

T = 1 2m R˙ 2 + R 2 ˙θ 2 + (R sin θ) 2φ˙2

But R˙ = 0 , φ˙ = ω = constant

T = ½ m(R ² ˙θ ² + (R sin θ) ² ω ² ) (NB. R˙ = 0)

U = −mgR cos θ (U = 0 at θ = 90◦ )

L = T − U

L = ½ m(R ² ˙θ ² + (R sin θ) ² ω ² ) + mgR cos θ

One single generalized coordinate : θ

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CONCLUSION

What I did:

Observed the motion of the bead when the hoop was rotated, and tried to find out the different parameters that affect this motion.

How i did it:

I performed the experiment to observe the changes due to different parameters, and used computer simulations to predict what would happen in cases that i could not achieve at home.

What i learnt:

The motion is mainly dependant on radius of hoop, mass of hoop and the friction in the hoop. The bead rises because it tries to go further from the centre.