Rotational Motion Question 189
Question:If $I _{xy}$ is the moment of inertia of a ring about a tangent in the plane of the ring and ${{I} _{x’y’}}$ is the moment of inertia of a ring about a tangent perpendicular to the plane of the ring then
Options:
A) $I _{xy}={{I} _{x’y’}}$
B) $I _{xy}=\frac{1}{2}{{I} _{x’y’}}$
C) ${{I} _{x’y’}}=\frac{3}{4}I _{xy}$
D) $I _{xy}=\frac{3}{4}{{I} _{x’y’}}$
Show Answer
Answer:
Correct Answer: D
Solution:
[d]
$ I _{x y}$, moment of inertia of a ring about its tangent in the plane of the ring
$I _{x^{\prime} y}=\frac{3}{2} M R^2$ .
Moment of inertia about a tangent perpendicular to the plane of ring
$I _{x y}=2 M R^2$
$ \therefore I _{x y}=\frac{3}{4}\left(2 M R^2\right) or I _{x y}=\frac{3}{4} I _{x y} $
BETA
