Total Moment of Inertia of Two Rods

Click For Summary
SUMMARY

The total moment of inertia of two rods, each 2 meters long and 20 kg in mass, joined at their ends to form a V shape with a 60-degree angle, can be calculated using the parallel axis theorem. The moment of inertia at the edge of a rod is given by the formula 1/3 ML². To find the total moment of inertia, first determine the center of mass (CoM) of the system, then apply the parallel axis theorem to compute the moment of inertia about the CoM before adjusting it to the point where the rods are joined. This method ensures accurate results, as simply adding the moments of inertia does not account for the angular configuration of the rods.

PREREQUISITES
  • Understanding of moment of inertia concepts
  • Familiarity with the parallel axis theorem
  • Knowledge of calculating the center of mass (CoM)
  • Basic integration techniques for physics applications
NEXT STEPS
  • Study the application of the parallel axis theorem in complex systems
  • Learn how to calculate the center of mass for composite shapes
  • Explore integration methods for calculating moment of inertia
  • Review examples of moment of inertia calculations for rods at angles
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and rotational dynamics, as well as engineers and professionals involved in structural analysis and design.

Raios168
Messages
14
Reaction score
0

Homework Statement


The rods of length 2 meters and and mass 20 kg are joined at their ends to form a V shape. What is the total moment of inertia measured from the reference point perpendicular to the plane of the paper and at the point where the two rods are joined. (So find total moment of inertia at the edge of the V shape, but with respect to the plane perpendicular to the page). Also the rods are 60 degree apart.

Homework Equations


Moment of Inertia at the edge of a rod = 1/3 ML^2

The Attempt at a Solution


Can I not just add the two moments of inertia to get 2/3 ML^2 as the total?

EDIT: I think the correct way to do this is to find the CoM of the system (which will be the mid point of the line connecting the midpoints of the two rods). And then use parallel axis theorem to find the total moment of inertia about the CoM. And then apply parallel axis theorem again to find the MI about the the point connecting the two rods. Can anyone confirm this?
 
Last edited:
Physics news on Phys.org
Either method should bring you to the same result. I know which way I'd choose if I was in a hurry :smile:
 
Raios168 said:

Homework Statement


The rods of length 2 meters and and mass 20 kg are joined at their ends to form a V shape. What is the total moment of inertia measured from the reference point perpendicular to the plane of the paper and at the point where the two rods are joined. (So find total moment of inertia at the edge of the V shape, but with respect to the plane perpendicular to the page). Also the rods are 60 degree apart.

Homework Equations


Moment of Inertia at the edge of a rod = 1/3 ML^2

The Attempt at a Solution


Can I not just add the two moments of inertia to get 2/3 ML^2 as the total?

EDIT: I think the correct way to do this is to find the CoM of the system (which will be the mid point of the line connecting the midpoints of the two rods). And then use parallel axis theorem to find the total moment of inertia about the CoM. And then apply parallel axis theorem again to find the MI about the the point connecting the two rods. Can anyone confirm this?

The only problem with the latter plan is how to calculate the MOI of the rods about the C.o.M. Because the rods are fixed together at one end, I think you have to calculate the MOI using integration or use the formula for calculating the MOI of a rod about a rotated coordinate axis.
 
SteamKing said:
The only problem with the latter plan is how to calculate the MOI of the rods about the C.o.M. Because the rods are fixed together at one end, I think you have to calculate the MOI using integration or use the formula for calculating the MOI of a rod about a rotated coordinate axis.
Hence my comment above...
 
Schaum's- 3000 Solved problem in Physics. page 212.
Rule: Moments of inertia about an axis are added algebraically.
 
azizlwl said:
Schaum's- 3000 Solved problem in Physics. page 212.
Rule: Moments of inertia about an axis are added algebraically.
I don't think this applies to this particular case.

The individual rods are not joined at their centroids, but at one end of each rod and at an angle to one another, in the shape of a V when looking parallel to the presumed axis of rotation.

You can calculate the MOI of each rod about one end, but combining the two is a bit trickier than you are led to believe by this 'Rule'.
 
11.33
Four coplanar, large, irregular masses are held by a rigid frame of negligible mass, as shown in figure 11-6. Taking an axis through P and perpendicular to the page, show that I=I1 +I2+I3+I4 where I1 is the moment of inertia of object 1 alone about the axis and similarly for the others.

p is intersection of lines joining the masses.
 
Last edited:
azizlwl said:
11.33
Four coplanar, large, irregular masses are held by a rigid frame of negligible mass, as shown in figure 11-6. Taking an axis through P and perpendicular to the page, show that I=I1 +I2+I3+I4 where I1 is the moment of inertia of object 1 alone about the axis and similarly for the others.

p is intersection of lines joining the masses.

It's not clear what I1 thru I4 represent. Are they supposed to be the moments of inertia for the individual masses referred to the axis thru P?

How exactly to you 'add algebraically' the MOI of one mass which is placed at an angle to another mass?
 
Last edited:
We can go to the basics for moment of inertia about an axis and write the integral for the given scenario. In the figure the z-axis projects out of the page:
Fig1.png

L is the length of each rod of mass M, the distance from the z-axis of a mass element is r. The domain of r is 0 to L, and a differential mass element is ##dm = (M/L)dr##.
$$ I = 2\int_0^L r^2 \frac{M}{L}~dr$$
It should be clear from the symmetry that the moment of inertia will be twice that of a single rod.
 

Similar threads

Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
3K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
6K