A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the string is held in place and the hoop is released from rest (?Fig. E10.20?). After the hoop has descended 75.0 cm, calculate (a) the angular speed of the rotating hoop and (b) the speed of its center.

Solution 20E Step 1 of 4: y2= 0, so U =20 because U = Mgy2 2 y = 0.750 m, 1 1 2 The hoop is released from the rest , v 1 0 so K = 01because K = mv 1 2 1 2 vcm = R and I cm = MR Step 2 of 4: a) Conservation of energy is K 1 U =1K + U2 2 0 + U = K + 0 1 2 Mgy = Mv 1 2 + I 2 1 2 cm 2 cm 1 2 2 1 2 2 Mgy =1MR 2 + MR 2 Mgy = MR 2 2 1 gy (9.8 m/s )(0.750 m) = 1 = 0.0800 m R = 33.9 rad/s