What is the Moment of Inertia of a Cone about its Longitudinal Axis?

Click For Summary
SUMMARY

The moment of inertia of a solid cone about its longitudinal axis (z-axis) is calculated using the triple integral I_z = ∫∫∫_T (x² + y²) dxdydz. The cone is defined by the equations x² + y² ≤ z² and 0 ≤ z ≤ h. The correct evaluation of the integral in cylindrical coordinates leads to the result I_z = πh⁵/10, contrary to the incorrect evaluation yielding πh⁵/5. The error arises from using the moment arm r² instead of z² in the integral setup.

PREREQUISITES
  • Understanding of cylindrical coordinates
  • Knowledge of triple integrals in calculus
  • Familiarity with the concept of moment of inertia
  • Basic geometry of cones
NEXT STEPS
  • Review the derivation of moment of inertia for different geometric shapes
  • Learn about the application of triple integrals in physics
  • Study the differences between Cartesian and cylindrical coordinate systems
  • Explore advanced topics in solid mechanics related to rotational dynamics
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and calculus, as well as educators teaching the concepts of moment of inertia and integration techniques.

c0der
Messages
52
Reaction score
0

Homework Statement


Find the moment of inertia of a solid cone about its longitudinal axis (z-axis)

The cone: x^2+y^2<=z^2, 0<=z<=h

I_z = \int\int\int_T(x^2+y^2)dxdydz

Homework Equations


Representing the cone in cylindrical coords:

x=zcos\theta
y=zsin\theta
z=z

The Attempt at a Solution


The integral in cylindrical coords is:

I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz

Evaluating the triple integral gives:
\pi h^5/5

But the answer in the book is:
\pi h^5/10

I don't see what I did wrong
 
Physics news on Phys.org
c0der said:

Homework Statement


Find the moment of inertia of a solid cone about its longitudinal axis (z-axis)

The cone: x^2+y^2<=z^2, 0<=z<=h

I_z = \int\int\int_T(x^2+y^2)dxdydz

Homework Equations


Representing the cone in cylindrical coords:

x=zcos\theta
y=zsin\theta
z=z

The Attempt at a Solution


The integral in cylindrical coords is:

I_z = \int_0^h \int_0^{2\pi} \int_0^z (z^2)rdrd\theta dz

The moment arm for the second moment from the z axis is ##r^2##, not ##z^2##.

Evaluating the triple integral gives:
\pi h^5/5

But the answer in the book is:
\pi h^5/10

I don't see what I did wrong
 
I see:

x=rcos\theta, y=rsin\theta where 0<=r<=z

Thanks a lot
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
1K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
52
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K