• Support PF! Buy your school textbooks, materials and every day products Here!

Moments of inertia

  • #1
2
0
Hi,

I need some assistance understanding moments of inertia. I am doing some review for an upcoming exam, however I am slightly stumped by this question. I have already tried googling "moments of inertia" to try and understand the concept better, but I am having issues knowing when to apply which formulas.

I have attached my specific question and the "answer" to the question, hopefully someone can help me understand how to get the answer.

----

Two identical slender rods of length l and mass m are linked together at 90deg, as shown in figure Q4a (image attached), to form a link in a mechanism. Point B is midway between A and C, and points B, D, E, F and G are equally spaced along the lower link.

i) Determine the position of the center of gravity of the link

ii) Find the moment of inertia of the link about the point B

iii) Find the moment of inertia of the link about the point D

iv) Find the radius of gyration about the point G

fxfGv.png




---

ANSWERS:

i) Center of gravity is at D

ii) IB=(5/12)ml2 kg m2

iii) ID=(7/24)ml2 kg m2

iv) k= sqrt(17/24) l m


----

*My attempts*


i)


I know inherently that the centre of gravity is at D. How do i prove it mathematically?

----

ii)


I get the moment of inertial about B like so:

I[subBb[/sub] = (ml2)/3 + ((ml2/3) - (ml2)/4)

But i don't understand why I am supposed to subtract. Is it because the centre of gravity is l/2 below B?

----
iii)


For ID I get the answer by doing:

ID = ((ml2)/3 - (ml2)/4) + ((ml2)/(3*16)) + (9ml2)/(16*3)

Why do I subtract (ml2)/4) if that rod is above D?


----

iv)


I'm trying to use IG to find the radius of gyration:

So for IG

IG = (7/24)ml2 - ml2


Here I am using the parallel axis theorem, but why do I subtract the (md2) of the AC rod from the Moment of inertia about D?
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,356
4,884
the centre of gravity is at D. How do i prove it mathematically?
It is obviously on the line BG, but where? You seem to have forgotten how to find a mass centre. It is the point about which the net first moment is zero. E.g. in the y direction, if D is the mass centre then measuring the yi offset of each element mi from D you want Σyimi=0.
I get the moment of inertial about B like so:

I[subBb[/sub] = (ml2)/3 + ((ml2/3) - (ml2)/4)
By what logic? Normally you would just add up the MoIs of the simple parts. What is the MoI of AC about B? What is the MoI of BG about B?
For ID I get the answer by doing:
Again, how do you explain the logic behind that?
(Can you see how to use the results of (i) and (ii) and the parallel axis theorem to answer (iii) quickly?)
Here I am using the parallel axis theorem, but why do I subtract the (md2) of the AC rod from the Moment of inertia about D?
I have no idea why you would do such a subtraction, especially since it gives the wrong answer.
 

Related Threads for: Moments of inertia

Replies
1
Views
5K
Replies
4
Views
701
  • Last Post
Replies
3
Views
633
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
0
Views
871
Top