Finding moment of inertia

  • Thread starter Oomair
  • Start date
  • #1
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Homework Statement


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.



Homework Equations


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


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?
 

Answers and Replies

  • #2
alphysicist
Homework Helper
2,238
2
Hi Oomair,

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

[tex]
I=\frac{1}{2}MR^2
[/tex]
 

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