A small steel ball rolls counterclockwise around the inside of a 30.0 cm diameter roulette wheel. The ball completes exactly 2 revolutions in 1.20 seconds. What is the ball's angular displacement after 1.24 seconds, assuming the initial angular position is 0 (in radians, round to 1 decimal).

The ball's angular displacement after 1.24 seconds is 13.0 rad

Explanation:

The ball's angular velocity is

ω = Δθ/Δt

Since the ball completes 2 revolutions in 1.20 s and each revolution corresponds to an angular displacement Δθ = 2π. Thus,

ω = (2 (2π rad) / 1.20 s

= 10.5 rad/s

The ball moves with a constant angular velocity, so its angular displacement at 1.24 s is

θ_f = θ_i + ωΔt

θ_f = 0 rad + (10.47 rad/s) (1.24 s)

θ_f = 13.0 rad