Ask Question
30 October, 16:57

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

+4
Answers (1)
  1. 30 October, 17:20
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “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 ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers