A grinding wheel, initially at rest, is rotated with constant angular acceleration of 7.36 rad/s 2 for 5.73 s. The wheel is then brought to rest with uniform deceleration in 8.37 rev. Find the angular acceleration required to bring the wheel to rest. Note that an increase in angular velocity is consistent with a positive angular acceleration. Answer in units of rad/s 2

First calculate the wheel's angular velocity after having been accelerated:

ω = ω₀ + αt

ω = final angular velocity, ω₀ = initial angular velocity, α = angular acceleration, t = elapsed time

Given values:

ω₀ = 0rad/s (starts at rest), α = 7.36rad/s², t = 5.73s

Plug in and solve for ω:

ω = 0 + 7.36 (5.73)

ω = 42.2rad/s

Then calculate the angular acceleration needed to bring the wheel to a rest in 8.37 revolutions. Use this equation for the wheel's rotational motion:

ω² = ω₀² + 2αθ

ω = final angular velocity, ω₀ = initial angular velocity, α = angular acceleration, θ = angle traveled

Given values:

ω = 0rad/s, ω₀ = 42.2rad/s, θ = 8.37rev = 8.37 (2π) rad = 52.6rad

Plug in and solve for α:

0² = 42.2² + 2α (52.6)

α = - 16.9rad/s²

The wheel must be decelerated at a rate of 16.9rad/s²