Search any question & find its solution
Question:
Answered & Verified by Expert
The moment of inertia of a circular disc of radius $2 \mathrm{~m}$ and mass $1 \mathrm{~kg}$ about an axis $\mathrm{XY}$ passing through its center of mass and perpendicular to the plane of the disc is $2 \mathrm{~kg} \mathrm{~m}^2$. The moment of inertia about an axis parallel to the axis $\mathrm{XY}$ and passing through the edge of the disc is
Options:
Solution:
2847 Upvotes
Verified Answer
The correct answer is:
$6 \mathrm{~kg} \mathrm{~m}^2$
About XY axis, $\mathrm{I}=\frac{1}{2} \mathrm{MR}^2=\frac{1}{2} \times 1 \times(2)^2=2 \mathrm{~kg} \cdot \mathrm{m}^2$
Moment of inertia about an axis parallel to the axis XY and passing through the edge of the disc,
$$
\mathrm{I}^{\prime}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2=6 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Moment of inertia about an axis parallel to the axis XY and passing through the edge of the disc,
$$
\mathrm{I}^{\prime}=\frac{1}{2} \mathrm{MR}^2+\mathrm{MR}^2=\frac{3}{2} \mathrm{MR}^2=6 \mathrm{~kg} \cdot \mathrm{m}^2
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.