|
|
|
|
|
| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 4 Dec 2007 20:01:15 IST
|
|
|
a uniform thin bar of mass 6m & length 12L is bent to make a regular hexagon.its moment of inertia about an axis passing through the centre of mass & perpendicular to the plane of the hexagon is
|
|
|
|
|
|
|
|
the answer is 20 ml^2/3
find the moment of inertia due to one rod passing through its centre perpendicular to its axis ....
thn find it along the centre of mass of the system by using the perpendicular axis theorem....
and thn multiply it by 6 as thr r six such rods...
nudge me if u hv any doubts
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
4 Dec 2007 22:52:08 IST
|
|
|
Answer is 20ml2.
Now, once it is bent into a hexagon, mass of each rod = m, and length = 2L.
So, MI of one rod about an axis passing through its center = m(2l)2/12
= ml2/3.
MI of the rod about an axis passing through the center of mass of the hexagon =
ml2/3 + m( 3l)2 (As distance of the center of each rod from the COM is 3l, which can be found by using elementary trigonomentry.)
So, MI of one rod about desired axis = ml2/3 + 3ml2 = 10ml2/3.
MI of 6 rods would be 20ml2.
Hence the required MI = 20ml2
|
Will nip in at times to solve problems :)
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
4 Dec 2007 23:44:38 IST
|
|
|
Is your query resolved? Or do you have any more doubts?
|
Will nip in at times to solve problems :)
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
