Simple roulette in terms of circular motion

[Mahowny]

Homework Statement

I am trying to understand game of roulette through kinematic equations of circular motion. Roulette is game where a ball spins in circular motion. Initially it is accelerated such that velocity of ball increases with time however after reaching its peak value it starts to decelerate until it finally leaves its circular track and falls in rotar. (ref: attached image)

Spoiler

Given:
Initial angular displacement of ball at start θ₀=0 rad
Initial angular velocity of ball at start ω₀=0 rad/s

Angular displacement of ball in one revolution θ₁= 2∏ rad = 6.28318548 rad
Angular displacement of ball in initial six revolutions θ₆= 6 * θ₁= 37.6991 rad
Total angular displacement of ball before it leaves the outer track θ_tot=114.1162 rad (i.e. slightly more than 18 revolutions)
Angular displacement of ball after its initial six revolutions before it leaves the outer track θ_f= θ_tot-θ₆=76.4171 rad

Time taken by ball to complete initial six revolutions t₆= 3.875 s
Total time taken by ball before it leaves the outer track t_tot=22.479 s
Time taken by ball after its initial six revolutions before it leaves the outer track t_f=t_tot-t₆=18.604 s

Homework Equations

ω₂=ω₁+σt ---------- eq:(1)
θ=ω₁t+(1/2σt²) ---------- eq:(2)
ω₂²=ω₁²+2σθ ---------- eq:(3)

where,
ω₁ and ω₂ are initial and final angular velocities respectively in rad/s
σ is angular acceleration in rad/s²
θ is angular displacement in radians
t is time in seconds

The Attempt at a Solution

Angular velocity of ball at sixth revolution can be calculated as follows:
ω₆=θ₆/t₆=9.7288 rad/s

Similarly,
Angular velocity of ball before it leaves the outer track can be calculated as follows:
ω_f=θ_f/t_f=4.10 rad/s

Since ω₆>ω_f, angular acceleration σ must be negative..

Using eq:(1)
σ=(ω_f-ω₆)/t
Here, t would be time taken by ball to reach ω_f from ω₆ i.e. t_tot-t₆₌t_f
∴σ=(ω_f-ω₆)/t_f
∴σ=-0.3025 rad/s²

However if we use eq:(3)
σ=(ω_f²-ω₆²)/2θ
Here, θ would be angular displacement of ball to reach ω_f from ω₆ = θ_tot-θ₆=θ_f
∴σ=(ω_f²-ω₆²)/2θ_f
∴σ=-0.5093 rad/s²

My question! How do i get two different values of angular deceleration constant σ?
The values in "Given:" section are real time observations still why there is difference in σ when it is calculated by different equations?

 

[CWatters]

There are several issue with this question...

Initially it is accelerated such that velocity of ball increases with time however after reaching its peak value it starts to decelerate until it finally leaves its circular track and falls in rotar. (ref: attached image)

Not sure I agree with that. The ball will start to decelerate the moment it leaves the hand. So if time starts at that moment there won't be any acceleration phase.

Angular velocity of ball at sixth revolution can be calculated as follows:
ω6=θ6/t6=9.7288 rad/s

That's not correct. That's an average for the velocity over the first six revolutions not the final velocity after six revolutions. The ball is decelerating so the velocity after six revolutions will be slower.

Similarly,
Angular velocity of ball before it leaves the outer track can be calculated as follows:
ωf=θf/tf=4.10 rad/s

That's similarly incorrect for the same reason.

There are two time intervals ...

t₀ to t₆
and
t₆ to t_f

You need to apply some of these equations to each interval to generate a set of simultaneous equations.

ω_f = ω_i + σt
θ = (ω_i + ω_f)t/2
ω_f² = ω_i² + 2σθ
θ = ω_it + 0.5σt²

where
ω_i is the initial angular velocity of that interval
ω_f is the final angular velocity of that interval
θ is the angular displacement in that interval
σ is the angular acceleration/deceleration (assumed constant for both intervals)
t is the duration of that interval

It should be possible to solve those simultaneous equations to give values for the unknowns ω₀ ω₆ ω_f and σ

I've not attempted to do that!

Aside: There is some limited similarity with the linear problem of an object falling past a window - where you are given the time as it goes past the top and bottom of the window, the height of the window etc.

 

[Tony Carson]

You need to take into account the ball deceleration isn't actually constant. This is because the wheel is often placed on a slightly angle table and the ball track, where the ball rolls is not completely flat. So it has a kind of wave effect. I can send you some information about it if you want just message me.