Rotation 1 Question 10
13. The moment of inertia of a uniform cylinder of length $l$ and radius $R$ about its perpendicular bisector is $I$. What is the ratio $l / R$ such that the moment of inertia is minimum?
(a) $\frac{\sqrt{3}}{2}$
(b) 1
(c) $\frac{3}{\sqrt{2}}$
(d) $\sqrt{\frac{3}{2}}$
(2017 Main)
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Answer:
Correct Answer: 13. (d)
Solution:
- MI of a solid cylinder about its perpendicular bisector of length is
$$ \begin{aligned} I & =M \frac{l^{2}}{12}+\frac{R^{2}}{4} \\ \Rightarrow I & =\frac{m R^{2}}{4}+\frac{m l^{2}}{12}=\frac{m^{2}}{4 \pi \rho l}+\frac{m l^{2}}{12} \quad\left[\because \rho \pi r^{2} l=m\right] \end{aligned} $$
For $I$ to be maximum,
$$ \begin{aligned} & \frac{d I}{d l}=-\frac{m^{2}}{4 \pi \rho} \quad \frac{1}{l^{2}}+\frac{m l}{6}=0 \\ & \Rightarrow \quad \frac{m^{2}}{4 \mu \pi \rho}=\frac{m l^{3}}{6} \Rightarrow l^{3}=\frac{3 m}{2 \pi \rho} \\ & \Rightarrow \quad l=\frac{3}{2}^{1 / 3} \frac{m}{\pi \rho}^{1 / 3} \\ & \rho=\frac{m}{\pi R^{2} l} \Rightarrow R^{2}=\frac{m}{\pi \rho l} \\ & \Rightarrow \quad R^{2}=\frac{m}{\pi \rho} \frac{2}{3}^{1 / 3} \frac{\pi \rho}{m}{ }^{1 / 3}=\frac{m}{\pi \rho}^{2 / 3} \frac{2}{3}^{1 / 3} \\ & \Rightarrow \quad R=\frac{m}{\pi \rho}^{1 / 3} \frac{2}{3}^{1 / 6} \end{aligned} $$
$$ \begin{aligned} & \therefore \quad \frac{l}{R}=\sqrt{\frac{3}{2}} \end{aligned} $$
