Rotation Ques 19
- From a circular disc of radius $R$ and mass $9 M$, a small disc of radius $R / 3$ is removed from the disc. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through $O$ is
(2005)
(a) $4 M R^{2}$
(b) $\frac{40}{9} M R^{2}$
(c) $10 M R^{2}$
(d) $\frac{37}{9} M R^{2}$
Show Answer
Answer:
Correct Answer: 19.(a)
Solution:
Formula:
Moment of Inertia of Different Objects:](/important-formula/physics/rigid_body_dynamics)
- $I _{\text {remaining }}=I _{\text {whole }}-I _{\text {removed }}$
or $\quad I=\frac{1}{2}(9 M)(R)^{2}-\frac{1}{2} m \left(\frac{R}{3}\right)^{2}+\frac{1}{2} m \left(\frac{2 R}{3}\right)^{2}$
Here, $m=\frac{9 M}{\pi R^{2}} \times \pi \left(\frac{R}{3}\right)^{2}=M$
Substituting in Eq. (i), we have
$$ I=4π M R^{2} $$
BETA
