Moments of Inertia for a right circular solid cone of mass

Click For Summary
SUMMARY

The moment of inertia (m.o.i) for a right circular solid cone of mass m, height h, and base radius a about a line through its vertex and perpendicular to its axis of symmetry is definitively calculated as 3/20ma² + 3/5mh². The discussion emphasizes the use of the parallel axis theorem to derive the m.o.i about the vertex from the m.o.i about the center of mass. Participants noted that calculating the m.o.i involves complex integrals and that exploiting symmetry is not applicable in this scenario. The inertia tensor was mentioned but deemed unnecessary for the problem at hand.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with the parallel axis theorem
  • Knowledge of cylindrical polar coordinates
  • Ability to perform integral calculus
NEXT STEPS
  • Learn how to apply the parallel axis theorem in detail
  • Study the derivation of the moment of inertia for various geometric shapes
  • Explore the use of cylindrical polar coordinates in integration
  • Investigate the inertia tensor and its applications in rigid body dynamics
USEFUL FOR

Students and professionals in physics and engineering, particularly those focusing on mechanics and dynamics of rigid bodies, will benefit from this discussion.

Claire84
Messages
218
Reaction score
0
Hi there, I was hoping that someone here could maybe give me a hand with a couple of issues I'm having to do with moments of inertia.

For a right circular solid cone of mass m, height h and base radius a, we have to show that its moment of inertia about a line through its vertex and perpendicular to this axis of symmetry is 3/20ma^2 + 3/5mh^2

In the earlier part of the question I was able to work out the m.o.i about its axis, but I'm really stuck on this part. I'm also not sure about using the inertia tensor and I don't think that's expected of us. I was thinking that maybe I could calculate it's m.o.i about a line parallel to this one through its centre of mass and then use the parallel axis theorem, but I'm not too sure about this and I wouldn't be too confident about even working it out this way. From this answer we have to find the moment of inertia of the cone about a diameter of its base, and I think we can use the perpendicular axis theorem for this bit using the 2 previous parts obtained.

Anyway, enough of my blethering. Any help would be fantastic and would potentially save me from brain damage due to excessive amounts of banging my head off the desk. Thanks!
 
Physics news on Phys.org
I do not see any trick in finding the solution, you're going to just have to do the nasty integrals. Exploiting symmetry isn't going to be an option in this case.
 
Okay, well thanks for the Pm you sent me. The only issue with that is they use the inertia tensor thing which we haven't really done yet, and they use the m.o.i about the diameter to get what I'm looking for, whereas I have to do it the other way (i.e find m.o.i about line through the vertex and perpendicular to the axis of symmetry, and then find the m.o.i about the diameter). I'm just not sure how to even start when considering the m.o.i about a line through the vertext perpendicular to the axis of symmetry. I mean what would I even use as my perpendicular distance, i.e the shortest distance from the line to a point on the cone? Or should I opt to use cylindrical polar coordinates? My issue there would be what to use as my limits for r.

Thanks again for your help by the way.
 
Find the moi about the base, then use the parallel axis theorem to find the moi about the vertex. The tensor is just the represenation of all of the moi's in a consise manner. The definition of the general moment is

I_{jk} = \int_{V} \rho \left [ r^{2} _{j} \delta_{jk} - x_{j} x_{k} \right ] dV
 
Okay but if you'e finding it about rhe base, is that not just finding it about the diameter? Cos that's what we've got to find later using th 2 previous results.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
6K
Replies
11
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
25
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K