the angular velocity of a rotating wheel as a function of time
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There are a few ways to do this.
i) observe that the acceleration is 9rad/s / 3s = 3 rad/s², so
Θ = ω₀*t + ½αt² = -9rad/s*10s + ½ * 3rad/s² * (10s)² = 60 rads
ii) It gained 15 rad/s over 5 s, so by 10 s it will gain another 15 rad/s with a final velocity of 21 rad/s. Then
Θ = ωavg * t = ½(21 - 9)rad/s * 10s = 60 rads
iii) the area under the t-axis is ½ * -9rad/s * 3s = -13.5 rads
and the area over is ½ * 21rad/s * 7s = 73.5 rads
so the net area = angular distance = 60 rads
i) observe that the acceleration is 9rad/s / 3s = 3 rad/s², so
Θ = ω₀*t + ½αt² = -9rad/s*10s + ½ * 3rad/s² * (10s)² = 60 rads
ii) It gained 15 rad/s over 5 s, so by 10 s it will gain another 15 rad/s with a final velocity of 21 rad/s. Then
Θ = ωavg * t = ½(21 - 9)rad/s * 10s = 60 rads
iii) the area under the t-axis is ½ * -9rad/s * 3s = -13.5 rads
and the area over is ½ * 21rad/s * 7s = 73.5 rads
so the net area = angular distance = 60 rads
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