24calculate the inertia abt axis perpendicular to the plane....let it be I
now the inertia abt axis inclined at angle θ to the lane wil be Isin2θ
A rod of mass 2m and length 2l is bent at its midpoint making 90° angle
find the moment of inertia of the system about an axis passing through the point of bent symmetrically and making minimum angle of 45° with the plane of rod........
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16 Answers
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1i have got my answer but looking for a much simpler method.......
62The angle made is 45 degree.. so from the top view the lenght of the rod is 2L/√2
also the mass distribution is same as a rod
so the moment of inertia is 2m(2l/√2)2/12 = ml2/3
Arshad can you give your source..
1sorry sir but the correct answer is ml^2/2
this is the answer
first find the angle the axis is making with each side of the rod
using direction cosines
as this is a symmetrical case therefore the axis will make equal angles with both sides of the rod
assuming that angle to be @
we get
(cos45°)^2+(cos@)^2+(cos@)^2=1
(cos@)^2=1/4
@=60°
so moment of inertia about this axis
(2m)*(2l)^2*(sin60°)^2/24
=ml^2/2
1@ eureka
i told u it was an iit type of question.....they all involve deep thinking....dont go by first looks....
24hmm yeah...............
thanx for posting such a ques....keep up the good work..[1]
1i hav to ask some ques.
1) when the axis is not perpendicular to the plane of a structure and at an angle θ to it then the moment of inertia is given by I*sin2θ where I is the moment of inertia for the perpendicular axis
is this a general formula and how did u arrive at this
2) nishant bhaiya when u saw from the top view y did the lenghth of the rod become 2L√2
3) arshad in ur soln. can u explain the formula u hav used for moment of inertia and
y did u divide by 24 and not by 12