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arsen [322]
3 years ago
5

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid ra

tes and experience large centripetal accelerations, especially at the tip. (a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min. (b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).
Advanced Placement (AP)
2 answers:
Radda [10]3 years ago
7 0
NOTE: Check your formula sheet if you have one these are often there.

Ac = \frac{V^{2} }{r} = rω^2

We know r = 4.00m

but what is ω?

ω = \frac{2π}{T} = 2π*f

f is our frequency. we want it in seconds so we can divide by 60.

300/60 = 5rev/s

Ac = rω^2 = (4)(5)^2 = 100m/s

(b) Linear speed or v is equal to v = ω = (5)(4) = 20m/s
      340 / 20 = 17

      The speed of sound is 17 times faster.

NOTE: finished physics so i might be rusty and use an equation wrong. Tell me if something doesnt make sense. Im still new to this myself.


erastova [34]3 years ago
3 0

Answer:

Centripetal acceleration, a=3946.35\ m/s^2

\dfrac{v}{v'}=0.36

Explanation:

(a) The length of the helicopter blade, l = r = 4 m

Angular velocity of the blade, \omega=300\ rev/min = 31.41\ rad/s

The centripetal acceleration at the tip of the helicopter is given by :

a=\dfrac{v^2}{r}

Since, v=r\omega

a=\dfrac{(r\omega)^2}{r}

a=r\omega^2

a=4\times (31.41)^2

a=3946.35\ m/s^2

(b) The linear speed of the tip of helicopter blade is given by :

v=r\omega

v=4\times (31.41)

v = 125.64 m/s

If the speed of sound is taken t o be 340 m/s, v' = 340 m/s

Comparison between linear speed of the tip to the speed of sound is :

\dfrac{v}{v'}=\dfrac{125.64}{340}

\dfrac{v}{v'}=0.36

Hence, this is the required solution.

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