Respuesta :
Answer
given,
angular acceleration
[tex]\alpha= (20 e^{ - 0.6 t}\ rad/s^2[/tex]
Assume radius of fan be 1.75 ft
speed of tip P at t = 3
blade at rest at t = 0 s
[tex]d\omega = \alpha\ dt[/tex]
[tex]d\omega = (20 e^{ - 0.6 t})\ dt[/tex]
integrating both side
[tex]\int_0^{\omega}d\omega = \int_0^t(20 e^{ - 0.6 t})\ dt[/tex]
[tex]\omega =\dfrac{20}{0.6} [-e^{ - 0.6 t}]_0^t[/tex]
[tex]\omega =33.33[1 -e^{ - 0.6 t}][/tex]
at t = 3 s
[tex]\omega =33.33[1 -e^{ - 0.6\times 3}][/tex]
ω = 27.82 rad/s
we know,
v = r ω
v = 27.82 x 1.75
v = 48.69 ft/s
again using formula
[tex]d\theta = \omega dt[/tex]
[tex]d\theta = 33.33[1 -e^{ - 0.6\times t}]\ dt[/tex]
integrating both side
[tex]\int_0^{\theta}d\theta = \int_0^t( 33.33[1 -e^{ - 0.6\times t}])\ dt [/tex]
[tex]\theta =33.33\int_0^t[1 -e^{ - 0.6\times t}]\ dt[/tex]
[tex]\theta =33.33[t + \dfrac{1}{0.6}(e^{ - 0.6\times t} - 1])[/tex]
at t= 3
[tex]\theta =33.33[3 + \dfrac{1}{0.6}(e^{ - 0.6\times 3} - 1])[/tex]
[tex]\theta=53.63\ rad[/tex]
[tex]\theta=\dfrac{53.63}{2\pi}[/tex]
θ = 8.54 rev
The speed of the tip P of one of the blades when t = 3 seconds is 48.69 feet/sec and 8.54 revolutions the blade turned in 3 seconds.
What is angular acceleration?
It is defined as the rate of change in angular velocity with respect to time. It is also known as rotational acceleration.
We have given the angular acceleration equation:
[tex]\rm A = (20e^{-0.6t}) \ rad/s^2[/tex]
Let's suppose the radius of the fan r = 1.75 ft
The speed of the tip P at t = 3 sec
Blade at rest t = 0 sec
We know:
[tex]\rm \frac{dw}{dt} = \alpha[/tex] or
[tex]\rm dw = \alpha \ dt[/tex]
[tex]\rm dw = (20e^{-0.6t}) \ dt[/tex]
Integrating both the sides with limits 0 to w
[tex]\rm \int_{0}^{w}dw = \int_{0}^{t}(20e^{-0.6t}) \ dt[/tex]
[tex]\rm w =33.33(1-e^{-0.6t}})[/tex] (after putting the limits)
At t = 3 sec
[tex]\rm w = 33.33(1-e^{-0.6\times3})\\\\\rm w = 27.82 \ rad/sec[/tex]
Converting it to linear velocity:
v = r×w
v = 1.75×27.82
v = 48.69 feet/sec
From the formula:
[tex]\rm d\theta=w \ dt[/tex]
[tex]\rm d\theta = 33.33(1-e^{-0.6t})dt[/tex]
Again integrating both the sides:
[tex]\rm \int_{0}^{\theta}d\theta = \int_{0}^{t}33.33(1-e^{-0.6t})dt[/tex]
[tex]\rm \theta = 33.33(t+\frac{1}{0.6} (e^{-0.6t-1}))[/tex]
At t = 3 sec
[tex]\rm \theta = 33.33(t+\frac{1}{0.6} (e^{-0.6\times3-1}))[/tex]
[tex]\rm \theta = 55.63 \ radian[/tex]
Converting it to revolution:
Ф = [tex]\frac{53.63}{2\pi}[/tex]
Ф = 8.54 rev
Thus, the speed of the tip P of one of the blades when t = 3 seconds is 48.69 feet/sec and 8.54 revolutions the blade turned in 3 seconds.
Learn more about the angular acceleration here:
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