Moment of Inertia for Three Point Masses

AI Thread Summary
The discussion centers on calculating the moment of inertia for three particles located at specific coordinates. The user initially computes the moment of inertia as I = 5ma² based on their calculations of the distances from the z-axis. However, they express confusion as the expected answer is 4.12ma². Other participants suggest that the discrepancy may stem from the reference material being used, indicating that the issue may not be with the user's calculations. The conversation highlights the importance of verifying the source of expected answers in physics problems.
atlantic
Messages
21
Reaction score
0

Homework Statement



Three particles of mass m are placed at A=(-a, -a), B=(a, -a) and C=(0, a)

Find the moment of inertia for an axis along the z-axis through the origin



Homework Equations



I = m((rA)2 + (rB)2 + (rC)2)



The Attempt at a Solution


I calculate that:
(rA)2 = (sqrt[(a)2 + (a)2])2 = 2a2
(rB)2 = (sqrt[(-a)2 + (a)2])2 = 2a2
(rC)2 = (a)2 = a2

So that:
I = m(a2 + 2a2 + 2a2) = 5ma2

But the answer is supposed to be 4.12ma2. What am I doing wrong?
 
Physics news on Phys.org
atlantic said:
So that:
I = m(a2 + 2a2 + 2a2) = 5ma2
Looks good to me.

But the answer is supposed to be 4.12ma2. What am I doing wrong?
I don't think you're doing anything wrong. What book are you using?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Correct Use of the Parallel Axis Theorem for Moment of Inertia
Replies
2
Views
2K
Moment of inertia of a disk about an axis not passing through its CoM
Replies
11
Views
3K
Parallel Axis Theorem Not Working
Replies
28
Views
1K
Moment of inertia of a rectangular prism parallel to one face
Replies
11
Views
950
Solving for 'a' with Torque, Force, and Mass Moment of Inertia
Replies
3
Views
2K
Moment of inertia of T bar about 3 axes
Replies
21
Views
2K
Moment of inertia of a disc using rods as differential elements
Replies
8
Views
2K
Moment of inertia of a rod bent into a square
Replies
12
Views
2K
Moment of inertia of a cuboid parallel to one of its faces (repost with diagram)
Replies
25
Views
2K
What is the "r" in the moment of inertia?
Replies
13
Views
2K

Hot Threads

Recent Insights

Back
Top