Question

The rotor blades of a helicopter on the ground with its engine idling rotate at 48rpm. If it takes 12s for the pilot to increase rotor speed to 460rpm when taking off, what is the angular acceleration on the rotor blades?

Answer 1

α = 3.59 [rad/s^2]]

Step-by-step explanation:

First we have to convert the values of the angular speeds of revolutions per minute to radians on second.

wf = final angular velocity = 460 [rpm]

wo= initial velocity = 48 [rpm]

t = time = 12 [s]

`460 [(rev)/(min)]* [(2\pi rad)/(1 rev) ]*[(1min)/(60seg) ] = 48.17[(rad)/(s) ]\\48 [(rev)/(min)]* [(2\pi rad)/(1 rev) ]*[(1min)/(60seg) ] = 5.02[(rad)/(s) ]`

now we can replace, in the following equation:

`w_(o)=w_(i) +\alpha  *t\\where:\\\alpha = angular acceleration [rad/s^(2)]\\replacing:\\\alpha =(w_(o) -w_(i) )/(t) \\\alpha alpha =(48.17-5.02 )/(12) \\\alpha =3.59[(rad)/(s^(2) ) ]`