Question

A model of a helicopter rotor has four blades, each of length 3.0 m from the central shaft to the blade tip. The model is rotated in a wind tunnel at a rotational speed of 560 rev/min.part A :What is the linear speed of the bladetip?part B :What is the radial acceleration of the bladetip expressed as a multiple of the acceleration of gravity,g?

Answer 1

Answer with Explanation:

We are given that

r=3 m

Angular frequency=
`\omega`=560rev/min

A.1 revolution=
`2\pi` radian

560 revolutions=
`560* 2\pi` rad

Angular frequency=
`2* 3.14* (560)/(60)`rad/s

1 min=60 s

`\pi=3.14`

Angular frequency=
`\omega=58.6rad/s`

Linear speed=
`\omega r`

Using the formula

Linear speed=
`58.6* 3=175.8m/s`

Hence, the linear speed of the blade tip=175.8m/s

B.Radial acceleration=
`a_(rad)=(v^2)/(r)`

By using the formula

Radial acceleration=
`a_(rad)=((175.8)^2)/(3)= 10.301* 10^3m/s^2`

Radial acceleration=
`(10.301)/(9.8)g* 10^3=1.05* 10^3 g`

Where
`g=9.8m/s^2`

Hence, the radial acceleration
`=1.05* 10^3 g`