A rolling wheel of diameter of 68 cm slows down uniformly from 8.4 m/s to rest over a distance of 115 m. what is the magnitude of its angular acceleration if there was no slipping?

Refer to the diagram shown below.

r = 68/2 = 34 cm = 0.34 m, the radius of the wheel

u = 8.4 m/s, the intial tangential velocity

s = 115 m, the stopping distance.

The circumference of the wheel is

C = 2πr = 2π (0.34) = 0.17π m

A circumference is equivalent to 2π radians, therefore the stopping distance is equivalent to

θ = [115 / (0.17π) ]*2π = 215.327 rad

The intial angular velocity is

ω₀ = u/r = (8.4 m/s) / (0.34 m) = 24.706 rad/s

Let α = the angular acceleration.

Because the wheel comes to rest, therefore

ω₀² + 2*α*θ = 0

(24.706 rad/s) ² + 2 * (α rad/s) * (215.327 rad) = 0

610.3815 + 430.654α = 0

α = - 1.417 rad/s²

Answer: - 1.417 rad/s²