# Understanding roulette in terms of circular motion

I am trying to understand game of roulette through kinematic equations of circular motion.

Given:
Initial angular displacement of ball at start θ[SUB]0[/SUB]=0 rad
Initial angular velocity of ball at start ω[SUB]0[/SUB]=0 rad/s

Angular displacement of ball in one revolution θ[SUB]1[/SUB]= 2∏ rad = 6.28318548 rad
Angular displacement of ball in initial six revolutions θ[SUB]6[/SUB]= 6 * θ[SUB]1[/SUB]= 37.6991 rad
Total angular displacement of ball before it leaves the outer track θ[SUB]tot[/SUB]=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[SUB]f[/SUB]= θ[SUB]tot[/SUB]-θ[SUB]6[/SUB]=76.4171 rad

Time taken by ball to complete initial six revolutions t[SUB]6[/SUB]= 3.875 s
Total time taken by ball before it leaves the outer track t[SUB]tot[/SUB]=22.479 s
Time taken by ball after its initial six revolutions before it leaves the outer track t[SUB]f[/SUB]=t[SUB]tot[/SUB]-t[SUB]6[/SUB]=18.604 s

Relevant equations
ω[SUB]2[/SUB]=ω[SUB]1[/SUB]+σt ---------- eq:(1)
θ=ω[SUB]1[/SUB]t+(1/2σt[SUP]2[/SUP]) ---------- eq:(2)
ω[SUB]2[/SUB][SUP]2[/SUP]=ω[SUB]1[/SUB][SUP]2[/SUP]+2σθ ---------- eq:(3)

where,
ω[SUB]1[/SUB] and ω[SUB]2[/SUB] are initial and final angular velocities respectively in rad/s
σ is angular acceleration in rad/s[SUP]2[/SUP]
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:

Similarly,
Angular velocity of ball before it leaves the outer track can be calculated as follows:

Since ω[SUB]6[/SUB]>ω[SUB]f[/SUB], angular acceleration σ must be negative…

Using eq:(1)
σ=(ω[SUB]f[/SUB]-ω[SUB]6[/SUB])/t
Here, t would be time taken by ball to reach ω[SUB]f[/SUB] from ω[SUB]6[/SUB] i.e. t[SUB]tot[/SUB]-t[SUB]6=[/SUB]t[SUB]f[/SUB]
∴σ=(ω[SUB]f[/SUB]-ω[SUB]6[/SUB])/t[SUB]f[/SUB]

However if we use eq:(3)
σ=(ω[SUB]f[/SUB][SUP]2[/SUP]-ω[SUB]6[/SUB][SUP]2[/SUP])/2θ
Here, θ would be angular displacement of ball to reach ω[SUB]f[/SUB] from ω[SUB]6[/SUB] = θ[SUB]tot[/SUB]-θ[SUB]6[/SUB]=θ[SUB]f[/SUB]
∴σ=(ω[SUB]f[/SUB][SUP]2[/SUP]-ω[SUB]6[/SUB][SUP]2[/SUP])/2θ[SUB]f[/SUB]

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?

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

In every spin theese values can be diferent till +/- more than 1 sec. And really all calculattions from this moment are very low in value.

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

This way you find something like averidge velocity duering six rounds , but not at sixth revoluttion.

My question! How do i get two different values of angular deceleration constant σ?

Because angular decelerattion is not constant

Thanks bebediktus