Rotational Motion Question 66
Question: A thin rod of length L and mass M is bent at its midpoint into two halves so that the angle between them is $ 90^{o} $ . The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is
[AIPMPT (S) 2008]
Options:
A) $ \frac{ML^{2}}{24} $
B) $ \frac{ML^{2}}{12} $
C) $ \frac{ML^{2}}{6} $
D) $ \frac{\sqrt{2}ML^{2}}{24} $
Show Answer
Answer:
Correct Answer: B
Solution:
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Since rod is bent at the middle, so each part of it will have same length $ ( \frac{L}{2} ) $ and mass $ ( \frac{M}{2} ) $ as shown.
Moment of inertia of each part through its one end $ =\frac{1}{3}( \frac{M}{2} ){{( \frac{L}{2} )}^{2}} $
Hence, net moment of inertia through its middle point O is
$ I=\frac{1}{3}( \frac{M}{2} ){{( \frac{L}{2} )}^{2}}+\frac{1}{3}( \frac{M}{2} ){{( \frac{L}{2} )}^{2}} $
$ =\frac{1}{3}[ \frac{ML^{2}}{8}+\frac{ML^{2}}{8} ]=\frac{ML^{2}}{12} $
BETA
