Moment of inertia of a cylinder

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SUMMARY

The moment of inertia of a solid cylinder is derived using the formula I=∫ r² dm, where dm is expressed as dv multiplied by density (ρ). To arrive at the correct moment of inertia, I=1/2mr², the differential volume element must be defined as dv=2πrh dr, rather than the incorrect dv=πr²h. This distinction is crucial for accurately calculating the moment of inertia by considering the cylinder's geometry and density.

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  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concepts of moment of inertia and its physical significance.
  • Knowledge of geometric properties of cylinders, including radius and height.
  • Basic principles of density and its application in physics.
NEXT STEPS
  • Study the derivation of moment of inertia for different shapes, such as spheres and rods.
  • Explore the application of integration in physics, particularly in calculating volumes and areas.
  • Learn about the physical significance of density in material science and engineering.
  • Investigate the use of visual aids, such as diagrams and models, in understanding complex physical concepts.
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Students of physics, mechanical engineers, and anyone involved in the study of rotational dynamics and solid mechanics.

gboff21
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I have been deriving the moment of inertia of a solid cylinder and have got this far:
I=∫ r2 dm
and dm=dv *density

h=height
r=radius
\rho=density

To get to the correct I=1/2mr2. you need to make dv=2\pirh dr
why isn't it dv= dv=\pir2h
as in volume=cross-sectional area*height
 
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Consider a circle on the end of the cylinder at radius r. If it goes all the way through the circle, its area is 2πrh. Let dV be this area times an incremental radius, dr:
dV = 2πrh*dr
Imagine it "unrolled" into a rectangular solid to see it clearly.

Don't forget the density.
 
Thanks I've been struggling to get my head round that for a while!
 
Most welcome. Always draw a picture!
 

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