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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)*

**Given:**

Initial angular displacement of ball at start θ

_{0}=0 rad

Initial angular velocity of ball at start ω

_{0}=0 rad/s

Angular displacement of ball in one revolution θ

_{1}= 2∏ rad = 6.28318548 rad

Angular displacement of ball in initial six revolutions θ

_{6}= 6 * θ

_{1}= 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}-θ

_{6}=76.4171 rad

Time taken by ball to complete initial six revolutions t

_{6}= 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}-t

_{6}=18.604 s

**2. Homework Equations**

ω

_{2}=ω

_{1}+σt ---------- eq:(1)

θ=ω

_{1}t+(1/2σt

^{2}) ---------- eq:(2)

ω

_{2}

^{2}=ω

_{1}

^{2}+2σθ ---------- eq:(3)

where,

ω

_{1}and ω

_{2}are initial and final angular velocities respectively in rad/s

σ is angular acceleration in rad/s

^{2}

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:

ω

_{6}=θ

_{6}/t

_{6}=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 ω

_{6}>ω

_{f}, angular acceleration σ must be negative..

Using eq:(1)

σ=(ω

_{f}-ω

_{6})/t

Here, t would be time taken by ball to reach ω

_{f}from ω

_{6}i.e. t

_{tot}-t

_{6=}t

_{f}

∴σ=(ω

_{f}-ω

_{6})/t

_{f}

∴σ=-0.3025 rad/s

^{2}

However if we use eq:(3)

σ=(ω

_{f}

^{2}-ω

_{6}

^{2})/2θ

Here, θ would be angular displacement of ball to reach ω

_{f}from ω

_{6}= θ

_{tot}-θ

_{6}=θ

_{f}

∴σ=(ω

_{f}

^{2}-ω

_{6}

^{2})/2θ

_{f}

∴σ=-0.5093 rad/s

^{2}

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?