A uniform solid cylinder with radius R and length L has moment of inertia I 1 , about the axis of cylinder. A concentric solid cylinder of radius R ' = R 2 and length L ' = L 2 is carved out of the original cylinder. If I 2 is the moment of inertia of the carved out portion of the cylinder…

Question

A uniform solid cylinder with radius R and length L has moment of inertia I 1 , about the axis of cylinder. A concentric solid cylinder of radius R ' = R 2 and length L ' = L 2 is carved out of the original cylinder. If I 2 is the moment of inertia of the carved out portion of the cylinder then I 1 I 2 = __________. (Both I 1 and I 2 are about the axis of the cylinder)

Aicharya · Accepted Answer

Moment of inertia of solid cylinder of radius $R$ and length $L$ about the axis is $I_{1} = \frac{M_{1} R^{2}}{2}$ .

Moment of inertia of carved out portion of the cylinder of radius $\frac{R}{2}$ and length $\frac{L}{2}$ is $I_{2} = \frac{M_{2} {(\frac{R}{2})}^{2}}{2} = \frac{M_{2} R^{2}}{8}$

Now, the ratio $\frac{I_{1}}{I_{2}} = \frac{4 M_{1}}{M_{2}}$

As, $mass = density \times volume$

So, $\frac{I_{1}}{I_{2}} = \frac{4 \times ρ π R^{2} L}{\frac{ρ π R^{2} L}{4 \times 2}} = 32$

A uniform solid cylinder with radius R and length L has moment of inertia I 1 , about the axis of cylinder.…

Solution