[Mechanics] Friction decelerating a helicopter rotor assembly

[DanRow93]

Homework Statement

A Chinook helicopter’s rotor blades and hub have a top speed of 300 revs/min and a combined mass of 300 kg.
On a maintenance test the blade assembly is allowed to stop without applying the brake, in this condition the blades take 48 seconds to come to a standstill. The effective radius off the rotor is 6.8 m.

I need to calculate the following:

a. The deceleration of the rotor blades, assuming a constant rate of deceleration.

b. The number of turns the rotor assembly will make.

c. The frictional resistance at this deceleration.

Homework Equations

The Attempt at a Solution

I have completed the first two parts:

300 rev/min * (2pi)/60 = 31.416 rad/s

Acceleration = (ω2 - ω1)/(t2 - t1) = (0-31.416)/(48) = -0.6545 rad/s^(2)

ϴ = ω1 * t + 1/2 * a * t^(2) = (31.416*48)+(0.5x(-0.6545)*48^(2)) = 753.984 rads

753.984/(2pi) = 120 revolutions

For the final part, all I have came up with so far is that the friction will be acting downwards as a result of gravity, F = mg = 300*9.81 = 2943N

I'm not sure if this is the right thing to do though, as the question asks the friction 'at this deceleration', so I guess I need an equation that contains acceleration in it.

Thank you for any help!

EDIT: Do I need to use the equation:

Frictional torque = angular acceleration * Inertia?

 

[berkeman]

EDIT: Do I need to use the equation:

Frictional torque = angular acceleration * Inertia?

Yes, or something similar. The force downward due to gravity would not enter into the calculation, IMO.