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Archived Simple roulette in terms of circular motion

  • Thread starter Mahowny
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1. 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)

6525086-roulette-casino-on-white-background-vector.jpg

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 θtot=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 θf= θtot6=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 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

2. Homework Equations
ω21+σt ---------- eq:(1)
θ=ω1t+(1/2σt2) ---------- eq:(2)
ω2212+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

3. The Attempt at a Solution
Angular velocity of ball at sixth revolution can be calculated as follows:
ω66/t6=9.7288 rad/s

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

Since ω6f, angular acceleration σ must be negative..

Using eq:(1)
σ=(ωf6)/t
Here, t would be time taken by ball to reach ωf from ω6 i.e. ttot-t6=tf
∴σ=(ωf6)/tf
∴σ=-0.3025 rad/s2

However if we use eq:(3)
σ=(ωf262)/2θ
Here, θ would be angular displacement of ball to reach ωf from ω6 = θtot6f
∴σ=(ωf262)/2θf
∴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

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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 tf

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

ωf = ωi + σt
θ = (ωi + ωf)t/2
ωf2 = ωi2 + 2σθ
θ = ωit + 0.5σt2

where
ωi is the initial angular velocity of that interval
ωf 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 ωf 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.
 
You need to take into account the ball deceleration isnt 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.
 

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