1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Design problem on a cement mixer

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant 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)
    3. 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


    T = Fr
    F = (Mc + Ma) g
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted