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
The angular acceleration required to bring the wheel to rest is -16.91 rad/s².
The given parameters;
The initial angular velocity of the wheel when it started moving is calculated as;
[tex]\omega_i = \omega _0 + \alpha t\\\\\omega_i = 0 + 7.36 \times 5.73\\\\\omega_i = 42.17 \ rad/s[/tex]
The angular acceleration required to bring the wheel to rest is calculated by using third kinematic equation;
[tex]\omega _f^2 = \omega _i^2 + 2\alpha \theta[/tex]
when the wheel stops, the final angular velocity = 0
[tex]0 = (42.17)^2 + 2(8.37 \times 2\pi)\times \alpha\\\\0 = 1778.31 + 105.19 \alpha\\\\105.19 \alpha\ = - 1778.31\\\\\alpha = \frac{- 1778.31 }{105.19} \\\\\alpha = -16.91 \ rad/s^2[/tex]
Thus, the angular acceleration required to bring the wheel to rest is -16.91 rad/s².
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