The answer is: Torque = Force x Distance = (Mg) x ((4Pi/3)R sin(Theta)).
M = Mass. For simplicity, assume the barrel weight is 0 and the CONTENTS are 50kgs.
g = acceleration due to gravity = 9.8 m/(s^2)
Force = Mg = 490N.
R = max radius of the INSIDE of the barrel, which I assume to be 350mm = 0.35m.
'Theta' is angle of rotation. Theta = Angular displacement from load level in the barrel. For the puposes of this problem, we constrain Theta to the interval 0 to 90 degrees.
Distance = horizontal displacement of the Center of Gravity (CG) of the load from the center of the barrel.
The TOTAL distance of the CG of the load from the axis of barrel rotation is (4/3Pi)(R).
The HORIZONTAL distance of the CG of the load from the axis of barrel rotation is the TOTAL distance (which is (4/3Pi)(R)) times COS(90-Theta). On our 0 to 90 degree interval, COS(90-Theta) is the same as SIN(Theta). So the horizontal torque arm length is ((4/3Pi)(R))(SIN(Theta)).
Mark is correct: the theoretical torque to start the barrel moving (disregarding friction) is zero. The max torque to keep it moving is given by using the formula above (first line of this post) and choosing the appropriate Theta. I would pick Theta = 90 degrees, which makes SIN(Theta) = 1, and collapses the formula to:
Max torque = (Mg)(4/3Pi)R = 490N x (4/3Pi)(0.35m) = 72.8 (Nm) of torque. This is roughly 53.6 foot-lbs.