Answer:
Hello From your question  you already have the answers for  ( a ) and ( b ) hence I will provide the solution to option ( c )
answer :
a) 5.88 N
b) 0.452 s
c) 27.68 rad/s
Explanation:
Hoop radius ( R ) = 0.08 m
mass of hoop = 1.2 kg
height hoop descended ( h ) = 0.5 m
Calculate The angular velocity of the rotating hoop after it descends 0.5 m
since it is released from rest we will apply the law of conservation
mgh = 1/2 mv^2 + 1/2 * Ic * ω^2 ----- ( 1 )
where :  Ic (moment of inertia ) = m*R^2  ,  ω = angular velocity , v = Rω
Back to equation 1 above
mgh = m*R^2*ω^2  ------- ( 2 )
therefore angular velocity ( ω ) = [tex]\sqrt{\frac{gh}{R^2} }[/tex]  ,   h = 0.5 m,  g = 9.81,  R = 0.08 m
∴ ω = 27.68 rad/s
