At a given instant, an airplane is flying horizontally with speed v0 = 290 mph and acceleration a0 = 12 ft/s². At the same time, the airplane’s propellers rotate at an angular speed ω = 1500 rpm while accelerating at a rate α = 0.3 rad/s² . Knowing that propeller diameter is d = 14 ft, determine

Question

At a given instant, an airplane is flying horizontally with speed [latex]v_{ 0 }[/latex] = 290 mph and acceleration [latex]{ a }_{ 0 }[/latex] = 12 ft/s². At the same time, the airplane’s propellers rotate at an angular speed ω = 1500 rpm while accelerating at a rate α = 0.3 rad/s². Knowing that propeller diameter is d = 14 ft, determine the magnitude of the acceleration of a point on the periphery of the propellers at the given instant.

Accepted Answer

Using a cylindrical coordinate system with origin at the propeller’s axis of rotation and z axis in the direction of motion, the general expression for acceleration is

[latex]\vec{a}=\left(\ddot{R}-R \dot{ heta}^2 ight) \ \hat{u}_R+(R \ddot{ heta}+2 \dot{R} \dot{ heta}) \ \hat{u}_ heta+\ddot{z} \ \hat{u}_z.[/latex]

For the propeller tip we have [latex]R=d / 2, \dot{R}=\ddot{R}=0, \ddot{z}=a_0, \dot{ heta}=\omega, ext { and } \ddot{ heta}=\alpha[/latex]. Thus the acceleration is

[latex]\vec{a}=\frac{-d \omega^2}{2} \hat{u}_R+\frac{d \alpha}{2} \hat{u}_ heta+a_0 \hat{u}_z,[/latex]

and the magnitude of the acceleration is

[latex]|\vec{a}|=\sqrt{\left(\frac{-d \omega^2}{2} ight)^2+\left(\frac{d \alpha}{2} ight)^2+a_0^2}=173,000 \ ft / s ^2.[/latex]

Question 2.258: At a given instant, an airplane is flying horizontally with ...

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