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 =
`(V^(2) )/(r)` = rω^2

We know r = 4.00m

but what is ω?

ω =
`(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

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

`(v)/(v')=0.36`

Step-by-step 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=(v^2)/(r)`

Since,
`v=r\omega`

`a=((r\omega)^2)/(r)`

`a=r\omega^2`

`a=4* (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* (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 :

`(v)/(v')=(125.64)/(340)`

`(v)/(v')=0.36`

Hence, this is the required solution.