Conceptual Moment of Inertia Question

Click For Summary
SUMMARY

The moment of inertia for a composite object consisting of a sphere and a thin rod can be calculated using the formula I = (1/3)mL² + (2/5)MR² + M(L+R)². The rod's moment of inertia is given as 1/3 mL², while the sphere's moment of inertia is 2/5 MR². The additional term M(L+R)² accounts for the parallel axis theorem, which is essential when calculating the moment of inertia about a pivot point. This formula provides a definitive method for determining the moment of inertia of the combined system.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with the parallel axis theorem
  • Knowledge of basic physics principles regarding rotational dynamics
  • Ability to perform algebraic calculations involving mass and distance
NEXT STEPS
  • Study the derivation of the parallel axis theorem
  • Explore advanced applications of moment of inertia in rotational dynamics
  • Learn about the moment of inertia for other geometric shapes
  • Investigate the effects of mass distribution on rotational motion
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding rotational dynamics and calculating moment of inertia for composite objects.

Menisto
Messages
18
Reaction score
0
If I were to attach a sphere (mass M radius R) to the end of a thin rod (mass m length L), the end of the rod being attached to a pivot, how would I calculate the moment of inertia for that object?

The rod: 1/3 mL^2
The sphere: 2/5 MR^2

The object: 1/3 mL^2 +2/5 MR^2 + M(L+R)^2 ?


Thanks for the help in advance...
 
Physics news on Phys.org
yes, I believe that is correct.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
Replies
2
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 10 ·
Replies
10
Views
4K
Replies
11
Views
4K
  • · Replies 28 ·
Replies
28
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K