# 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

That was really a helpful reply. Thanks buddy.

Mine advice - try to find relationship between ball speed in some ball rounds and left time till dropp and left distance which make ball. That relationship also is not constant value and not linear, but still possible to use because you can find some relationship which will cover bigest part of all possible cases.