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Finding moment of inertia

  1. May 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A pipe made of aluminum with a density of 2.7 g/cm3 is a right cylinder 43 cm long whose outer diameter is 4.8 cm and whose inner diameter is 2.4 cm. What is the rotational inertia about the central axis of the pipe? Note that the rotational inertia of the thick cylinder can be expressed as the rotational inertia for a solid cylinder of radius R2 minus the rotational inertia of the solid cylinder of radius R1.

    2. Relevant equations
    D= m/v I= mr^2 for a solid cylinder volume of a cylinder = piR^2H

    3. The attempt at a solution

    what i basically did is find the mass of the two cylinders

    D=m/v outer cylinder = 2.7g/cm^2 = m/v and inner cylinder = 2.7g/cm^2 = m/v

    mass of the outer cylinder comes out to be 2100.89g and inner cylinder comes out to be 525.22g

    then i put them into the moment of inertia equation I= mr^2

    ((2100.89g)(2.4cm)^2) - ((525.2g)(1.2cm)^2) = 11344.81 g*cm^2, then i convert it into kg*m^2 and i get .001134 kg*m^2, anything wrong im doing here?
  2. jcsd
  3. May 8, 2008 #2


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    Homework Helper

    Hi Oomair,

    The moment of inertia of a solid cylinder about its central axis is:

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