Find Speed of Cylinder Rolling Down Loop | m=0.710 kg, h=333 cm, R=49.95 cm

[Punchlinegirl]

A uniform solid cylinder (m=0.710 kg, of small radius) is at the top of a ramp, with height height of 333 cm and which has a loop of radius, R = 49.95 cm at the bottom. Which has friction. The cylinder starts from rest and rolls down the ramp without sliding and goes around the loop. Find the speed of the cylinder at the top of the loop.

I used conservation of energy
$$mgh= (1/2)mv^2 + (1/2)I \omega ^2$$
$$mgh = (1/2)mv^2 + (1/2)(1/2)(MR^2)(v/r)^2$$
$$mgh= (1/2)mv^2 +(1/4)mv^2$$
Solving for v gave me $$\sqrt (4/3)gh$$
So plugging in 9.8 for g and 3.33 for h gave me a speed of 6.60 m/s which wasn't right.
Can someone help me?

 

[Fermat]

h should be the difference in height between the top of the ramp and the top of the loop.

 

[Punchlinegirl]

I got it. Thanks