Question

The M.I of solid cylinder of mass ${}^{\prime }{M}^{\prime }$, length ${}^{\prime }{l}^{\prime }$ and radius ${}^{\prime }{R}^{\prime }$ about a transverse axis passing through its centre is :

• A. $\frac{M{R}^2}{12}+\frac{M{l}^2}{4}$
• B. $\frac{M{R}^2}{4}-\frac{M{l}^2}{12}$
• C. $\frac{M{R}^2}{12}-\frac{M{l}^2}{4}$
• D. $\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$

Answer 1

The correct option is D $\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$

Taking the elements as the thin disks of mass density $\frac{M}{\pi R^{2}l}$
For each and every disk shifting the axis of rotation to the transverse axis and using parallel axis theorem, we get the follwing integration:

$I={\int }_{-l/2}^{l/2}(\frac{M}{4\pi R^{2}l}dx{\times R}^2{\times \pi R}^2+\frac{M}{4\pi R^{2}l}dx\times x^2{\times \pi R}^2)=\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$