Design problem on a cement mixer

[kirkfras]

Homework Statement

Calculating torque on a portable cement mixer which is to be designed as a hollow cylinder with a closed base that rotates on a shaft (bearing). Dimentions being 0.5m diameter (D) and a height of 0.9m (h); the thickness of the cylinder is 0.006m (t). the drum is tilted at an angle of 15 degrees (@) to horizontal. it has aggregate of mass 363 kg (Ma) inside it. the drum operational speed is 10 rpm (sp). the drum is made of mild steel.

Homework Equations

Torque = moment of inertia * angular acceleration
moment of inertia = mass * radius of gyration^2 (for hollow cylinder)
F = mg
density of mild steel (ds) = 7860 kg/m3
final angular velocity = initial angular velocity + (angular acceleration * time)

The Attempt at a Solution

r = D/2
volume of steel used on cylinder (Vc) = [(pi * r^2 * t)] + pi * [ {(r + t)^2} - (r^2)] * h
mass of steel used on cylinder (Mc) = ds * Vc

I = (Mc + Ma) * r^2

assume it takes 2 second to reach speed of 10 rpm on start up

angular acceleration (aa) = {[(sp * 2 * pi) /60] / 2}

T = I * aa

but i know the material in the cylinder will occupy a certain volume and have a centroid where the weigth will act, and where does the effect of the drum being tilted come into affect

OR

T = Fr
F = (Mc + Ma) g

 

[LightHero]

r = r + [tan(@) * h / 2]but i know that the force is due to the weight of the material and steel used, but i am confused as to whether this is the appropiate equation