Search any question & find its solution
Question:
Answered & Verified by Expert
Consider a thin metal strip of mass $1 \mathrm{~kg}$ and length $5 \mathrm{~m}$. Calculate its moment of inertia about an axis perpendicular to strip and located at $100 \mathrm{~cm}$ on strip from one its end. (Assume the breadth as the strip is negligible)
Options:
Solution:
2121 Upvotes
Verified Answer
The correct answer is:
$4.33 \mathrm{~kg}-\mathrm{m}^2$
The given situation is as shown below,
Here, $O$ is the point where centre of mass of whole strip exist.
The point $O^{\prime}$ is $100 \mathrm{~cm}$ or $1 \mathrm{~m}$ away from one end. So, its distance from centre of mass or from point $O$ is $1.5 \mathrm{~m}$.
$\therefore$ Moment of inertia about an axis perpendicular to strip at point $O^{\prime}$ can be calculated using parallel axis theorem as
$I^{\prime}=I_{\mathrm{CM}}+M R^2=\frac{M L^2}{12}+M R^2$
$=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5=2.08+2.25=4.33 \mathrm{~kg} \mathrm{~m}$
Here, $O$ is the point where centre of mass of whole strip exist.
The point $O^{\prime}$ is $100 \mathrm{~cm}$ or $1 \mathrm{~m}$ away from one end. So, its distance from centre of mass or from point $O$ is $1.5 \mathrm{~m}$.
$\therefore$ Moment of inertia about an axis perpendicular to strip at point $O^{\prime}$ can be calculated using parallel axis theorem as
$I^{\prime}=I_{\mathrm{CM}}+M R^2=\frac{M L^2}{12}+M R^2$
$=\frac{1 \times 5 \times 5}{12}+1 \times 1.5 \times 1.5=2.08+2.25=4.33 \mathrm{~kg} \mathrm{~m}$
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.