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kirkfras
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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