Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip. (a) Calculate the magnitude (in m/s2) of the centripetal acceleration at the tip of a 3.10 m long helicopter blade that rotates at 280 rev/min. m/s2 (b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s). vtip vsound =

Respuesta :

Explanation:

It is given that,

Length of the helicopter, l = 3.1 m

The helicopter rotates, the length of helicopter will become the radius of circular path, r = 3.1 m

Angular speed of the helicopter, [tex]\omega=280\ rev/min=29.32\ rad/s[/tex]

(a) The centripetal acceleration in terms of angular velocity is given by :

[tex]a_c=r\times \omega^2[/tex]

[tex]a_c=3.1\times (29.32)^2[/tex]

[tex]a_c=2664.95\ m/s^2[/tex]

(b) Let v is the linear speed of the tip. The relation between the linear and angular speed is given by :

[tex]v=r\times \omega[/tex]

[tex]v=3.1\times 29.32[/tex]

v = 90.89 m/s

[tex]\dfrac{v_{tip}}{v_{sound}}=\dfrac{90.89}{340}=0.267[/tex]

Hence, this is the required solution.

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