Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 2 (Q.No. 18)
18.
M.I. of a thin ring (external diameter D, internal diameter d) about an axis perpendicular to the plane of the ring, is
(D4 + d4)
(D4 - d4)(D4 + d4)(D4 x d4).Discussion:
6 comments Page 1 of 1.
Jero M said:
2 years ago
The given answer is correct since the MI is about an axis perpendicular to the plane.
Izz=Ixx + Iyy according to the perpendicular axis theorem.
So, we get, Izz= πd^4/64 + πd^4/64 = πd^4/32.
Izz=Ixx + Iyy according to the perpendicular axis theorem.
So, we get, Izz= πd^4/64 + πd^4/64 = πd^4/32.
Chintan Patel said:
4 years ago
It should be π/64.
VIMAL PRAKASH said:
6 years ago
I think the guys saying π/64 (D^4-d^4) must be in any confusion, because the same would be the answer if the question would have been said that the axis is along the diameter (xx or yy). But if it is perpendicular then, π/32 (D^4-d^4) ....means option B will be the right answer. Correct me if I'm wrong but with a proper explanation.
Alok said:
6 years ago
There is perpendicular of plane mean a polar moment of inertia asking so Ip= Ixx + Iyy.
So, B is correct.
So, B is correct.
Vishu tyagi said:
7 years ago
Yes, it is π/64(D^4-d^4).
Baloch said:
8 years ago
It must be &pie; D^4-d^4 over 64.
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