Rotational Motion Part 2

Given, the length of cylinder, L = 80 cm = 0.8 m The radius of cylinder, r = 20 cm = 0.2 m The moment of inertia about $C D, I_{CD}=2.7 \text{kg}-\text{m}^{2}$ Now, the moment of inertia of cylinder about its axis AB is $I_{AB}=\frac{1}{2} M r^{2}$ By parallel axis theorem, the moment of inertia of cylinder about CD will be calculated as $I_{CD}=I_{AB}+M\left(\frac{L}{2}\right)^{2}$ $\Rightarrow 2.7=\frac{M r^{2}}{2}+\frac{M L^{2}}{4}$ $\Rightarrow 2.7=M\left[\frac{(0.2)^{2}}{2}+\frac{(0.8)^{2}}{4}\right]$ $\Rightarrow M=15 \text{kg}$ Density of cylinder is given as $\rho=\frac{M}{V}=\frac{M}{\pi r^{2} L}$ $=\frac{15}{\pi (0.2)^{2}(0.8)}=149.2 \text{kgm}^{-3}$ $=1.49 \times 10^{2} \text{kg} \text{m}^{-3}$ Thus, the density of cylinder is $1.49 \times 10^{2} \text{kgm}^{-3}$.