How Do You Calculate the Moment of Inertia for a Cone?

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SUMMARY

The moment of inertia for a right circular cone can be calculated using the integral I = ∫r² dm, where dm is derived from the cone's density and volume. The volume of the cone is given by V = (1/3)πr²h. The discussion highlights a common mistake of neglecting the r² factor in the integral, leading to an incorrect simplification of I = m. To accurately compute the moment of inertia, the problem should be set up as a double integral with respect to the cone's longitudinal axis.

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  • Understanding of integral calculus
  • Familiarity with the concepts of density and volume
  • Knowledge of the geometric properties of cones
  • Experience with setting up double integrals
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Students in physics or engineering courses, particularly those studying mechanics, as well as educators looking for examples of calculating moments of inertia for geometric shapes.

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Homework Statement


Find the moment of inertia of a right circular cone of radius r and height h and mass m


Homework Equations



I = ∫r2 dm
V = 1/3*π*r2*h

The Attempt at a Solution


Assume density is p

dm = p dv
divide both sides by dr
dm/dr = p dv/dr

dm/dr = p (d/dr * 1/3*π * r2*h)
so
dm/dr = p(2/3)*πrh
so:
dm = (2/3)pπrh dr

Sub that into the moment of inertia equation

∫(2/3)pπrh dr = I

I = (1/3) pπr2h
p = m/v
I = (1/3)(m/v)πr2h
I = v*(m/v)
I = m

What am I doing wrong?
 
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Hint: Set the problem up as a double integral problem. Lay the problem out with the cone's longitudinal axis being the x-axis. Use a ring for your dV.

If you need more hints, let me know.
 
You forgot the factor of r2 multiplying dm. What you found was [itex]\int dm[/itex], which unsurprisingly turns out to equal the mass of the cone.

About what axis are you supposed to be calculating the moment of inertia?
 

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