Answer
a) V(cm) = 1.56 m/2
b) K Tot = 5.359 J
c) From Ground Ref Frame = (i) = 3.12 m/2, (ii) v = 0 m/2, (iii) = 2.206 m/s
d) From CM ref Frame = (i)= v = 1.56 m/s (right), (ii)= -v= -1.56 m/s(left), (iii) v= 1.56 (down).
Work Step by Step
a) V(cm) = r\omega = (1.2/2)(2.60) = 1.56 m/2
b) K Tot = K rot + k trans = 1/2 MV(cm)^2 + 1/2I\omega^2. I of a hoop = MR^2. So,
(0.5)(2.20)(1.56)^2 + (0.5)(2.20)(0.6)^2(2.60)^2 = 5.359 J
c) From Ground Ref Frame = (i) = 2V = (2)(1.56) = 3.12 m/2, (ii) v = 0 m/s as hoop is instantaneously at rest at this point on the ground, (iii) = v cos(45) + v sin (45) = (1.56)(cos(45)) + (1.56)(sin(45)) = 2.206 m/s
d) From CM ref Frame = (i)= v = 1.56 m/s (right), (ii)= -v= -1.56 m/s(left), (iii) v= v cos (90) + v sin(90)= (1.56)(cos(90)) + (1.56)(sin(90)) = 1.56 (down).
