A hoop of radius 0.130 m and mass 0.420 kg is suspended by a point on its perimeter as shown in the figure. If the hoop is allowed to oscillate side to side as a pendulum, what is the period of small oscillations? T = 2pi (mgl/I)^.5 I assume that l = r, since center of mass is in the middle of the hoop. I = 3mr^2/2 So T = 2pi(2g/3r) Also tried with I = mr^2/2 in case I'm misunderstanding the problem and it is oscillating in the other direction, but this did not work either. Anyone see my error? Next: A mass M is suspended from a spring and oscillates with a period of 0.860 s. Each complete oscillation results in an amplitude reduction of a factor of 0.965 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 0.500 of its initial value. Tried .965^t = .5, then t = log(.5)/log(.965) but this didn't work... any other suggestions?