Question

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 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).

Answer 1

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.

Answer 2

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.