Rotational Motion Question 25
Question: A circular hole of diameter R is cut from a disc of mass M and radius R, -the circumference of the cut passes through the centre of the disc. The moment of inertia of the remaining portion of the disc about an axis perpendicular to the disc and passing through its centre is
[JEE ONLINE 07-05-2012]
Options:
A) $ ( \frac{15}{32} )MR^{2} $
B) $ ( \frac{1}{8} )MR^{2} $
C) $ ( \frac{3}{8} )MR^{2} $
D) $ ( \frac{13}{32} )MR^{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $I_{Total}$ disc= $\frac{MR^2}{2}$
As mass is proportional to area, $M_{Removed}=\frac{M}{4}$
Now, about the same perpendicular axis:
$ I_{Removed} =\frac{M}{4}\frac{(R/2)^2}{2} + \frac{M}{4}(\frac{{R}}{{2}}^2)=\frac{3MR^2}{32}$
⇒$ {I_Remaining} Disc = I_{Total}−I_{Removed}=\frac{MR^2}{2}− \frac{3MR^2}{32}=\frac{13MR^2}{32}$
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