Calculate Force Applied to Rope for Frictional Torque

[Gillian]

[/PLAIN]
A drum (see the figure) has a radius of 0.40 m and a moment of inertia of 6.9kg⋅m2. The frictional torque of the drum axle is 3.0 N⋅m. A 43m length of rope is wound around the rim. The drum is initially at rest. A constant force is applied to the free end of the rope until the rope is completely unwound and slips off. At that instant, the angular velocity of the drum is 14rad/s. The drum then decelerates and comes to a halt.
In this situation, the constant force applied to the rope is closest to:
a) 11 N
b) 19 N
c) 7.5 N
d) 23 N
e) 15 N

My Attempt

Στ = ΔL/Δt
F⋅r - τ_f = I⋅Δω/Δt
0.4F - 3.0 = 6.9⋅14/Δt

43m/(2π⋅0.4) = 17.109 revolutions (2π) = 107.5 rad
14 = 107.5/t
t = 7.679 s

0.4F - 3.0 = 6.9⋅14/7.679
F = 38.95 N ----> this is not an answer choice and I am not sure where to go next

 

[Hesch]

Look at the energies:

( Rotational energy ) + ( torque friction energy ) = ( rope energy )

( your calculation )

→

F = 23.23 N

 

[Chandra Prayaga]

This line in your calculation was wrong:

14 = 107.5/t

The motion of the wheel is not with constant angular velocity. You cannot relate the final angular velocity to the time by that equation. Since this looks like a homework problem, I would not like to say more.