A thin spherical shell of mass M = 2.00 kg is released from rest at the top of an incline of height H=1.31 m and rolls without slipping to the bottom. The ramp is at an angle of θ = 24.1o to the horizontal. Calculate the speed of the sphere's CM at the bottom of the ramp.

Answer 1

What I understand is as follows:
mgH = (mv^2)/2 + (Iw^2)/2
In this case, I = (mR^2)/3. hence, mgH = (mv^2)/2 +m(Rw)^2/3. But Rw = v so mgH =5(mv^2)/6 or v^2 = 6gH/5 = 9.81*1.81*6/5 or v ~= 4.616 met/sec

Answer 2

H=1.31 met and not 1.81met. Sorry for the error. Accordingly, v~= 3.927 met/sec.