amusement park booth gambling wheel with 1-35, finding angular speed

Question

A certain amusement park booth has a gambling wheel that can rotate on a fixed axis. The wheel has a radius of r = 30 cm and contains the numbers 1 through 35 evenly spaced around its edge in sequential order. On one occasion, the wheel is given a spin. As the wheel rotates and the number 1 passes a fixed pointer, its angular speed is 𝜔 = 3.7 rad/s. The wheel comes to rest after an additional 19.2 s.

A. When the when wheel is rotating at half its initial angular speed, what will be the linear speed of a point on its rim?

v =  m/s

B. What is the angular acceleration of the wheel while is is slowing down to a stop?

𝛼 =  rad/s²

C. What angle does the wheel rotate through before coming to a stop?

𝜃 =  rad

D. Just for fun, calculate the number on the wheel that it will be stopped at and where the pointer will be pointing.

The Winning Number is... 

Answer 1

A.

The relationship between v and ω is

v = ωr = 3.7×0.3 = 1.11 m/s

B.

In general, if constant acceleration causes angular speed ω₁ to change to ω₂,

and that change took time Δt, then the acceleration is

a = (ω₂ - ω₁)/Δt

In our case,

ω₁ = 3.7 rad/s

ω₂ = 0

Δt = 19.2 s

So

the acceleration

a = (0 - 3.7)/19.2 = -0.19 rad/s²

C.

There are two equivalent formulas that can be used:

The angle

α = ωΔt/2 = aΔt²/2

Using the first:

α = 3.7×19.2/2 = 35.52 rad

D.

Since there are 35 gaps between the numbers, the angle between two number is

2π/35 rad

So the number of gaps that will pass will be

35.52/(2π/35) = 197.9

Assuming that the wheel rotates in the direction of increasing numbers,

it will point at position

(1 + 197.9) mod 35 = 23.9

So when the wheel stops,

the pointer will point into the gap between numbers 23 and 24, closer to 24.