Question

The moment of inertia of a cylinder of mass$M$, length$L$, and radius$R$ about an axis passing through its center and perpendicular to the axis of the cylinder is$I=M\left(\frac{R^2}{4}+\frac{L^2}{12}\right)$. If such a cylinder is to be made for a given mass of a material, the ratio $\frac{L}{R}$ for it to have the minimum possible I is:

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\sqrt{\frac{2}{3}}$
• D. $\sqrt{\frac{3}{2}}$

Answer 1

The correct option is D

$\sqrt{\frac{3}{2}}$

Step 1. Given data:

1. The mass of the cylinder$=M$
2. The length of the cylinder$=L$
3. The radius of the cylinder$=R$
4. The moment of inertia$I=M\left(\frac{R^2}{4}+\frac{L^2}{12}\right)$

Step 2. Calculating the value$R^2$ in terms of$L$:

As we know,

$$\begin{gathered} mass\left(M\right)=desity\left(d\right)\times volume\left(V\right) \\ \Rightarrow M=dV \\ \Rightarrow M=d\pi R^{2}L(as,V=\pi R^{2}L) \\ \Rightarrow R^2=\frac{M}{d\pi L} \end{gathered}$$

Now putting the value $R^2$in$I$ what we are getting,

$I=M\left(\frac{M}{4\pi dL}+\frac{L^2}{12}\right)$

Step 3. Calculating the ratio $\frac{L}{R}$ for it to have the minimum possible$I$:

For the minimum possible value of$I$,

$$\begin{gathered} \frac{dI}{dL}=0, \\ \Rightarrow \frac{M^2}{4\pi d}\left(-\frac{1}{L^2}\right)+\frac{M}{12}·2L=0 \\ \Rightarrow \frac{M}{12}·2L=\frac{M^2}{4\pi d}\left(\frac{1}{L^2}\right) \\ \Rightarrow \frac{L}{6}=\frac{M}{4\pi d{L}^2} \\ \Rightarrow \frac{L}{6}=\frac{M}{\pi dL}·\frac{1}{4L} \\ \Rightarrow \frac{L}{6}=R^2·\frac{1}{4L} \\ \Rightarrow \frac{R^2}{L^2}=\frac{4}{6} \\ \Rightarrow \frac{R}{L}=\sqrt{\frac{2}{3}} \\ \Rightarrow \frac{L}{R}=\sqrt{\frac{3}{2}} \end{gathered}$$

Thus the required ratio is$\sqrt{\frac{3}{2}}$.

Hence, option D is the correct answer.