Calculating Moment of Inertia Ratio

The problem asks for the ratio of the moment of inertia ($I_1$) of a solid cylinder to that ($I_2$) of a smaller, co-centric cylinder carved from it.

Moment of Inertia Formula

The moment of inertia ($I$) for a uniform solid cylinder of mass $M$ and radius $R$ about its central axis is given by the formula:

$I = \frac{1}{2}MR^2$

We assume the original cylinder has uniform density ($\rho$). The mass ($M$) can be expressed as $M = \rho V$, where $V$ is the volume. For a cylinder, $V = \pi R^2 L$. Thus, $M = \rho \pi R^2 L$. Substituting this into the inertia formula gives:

$I = \frac{1}{2}(\rho \pi R^2 L)R^2 = \frac{1}{2}\rho \pi R^4 L$

Calculating $I_1$

For the original solid cylinder:

• Length = $L$
• Radius = $R$
• Mass = $M_1 = \rho \pi R^2 L$
• Moment of Inertia $I_1 = \frac{1}{2}M_1 R^2 = \frac{1}{2}(\rho \pi R^2 L) R^2 = \frac{1}{2}\rho \pi R^4 L$

Calculating $I_2$

For the smaller carved cylinder:

• Length = $L_2 = L/2$
• Radius = $R_2 = R/3$
• Mass = $M_2 = \rho \pi R_2^2 L_2 = \rho \pi \left(\frac{R}{3}\right)^2 \left(\frac{L}{2}\right)$
• $M_2 = \rho \pi \left(\frac{R^2}{9}\right) \left(\frac{L}{2}\right) = \frac{1}{18} \rho \pi R^2 L$
• Moment of Inertia $I_2 = \frac{1}{2}M_2 R_2^2 = \frac{1}{2} \left(\frac{1}{18} \rho \pi R^2 L\right) \left(\frac{R}{3}\right)^2$
• $I_2 = \frac{1}{2} \left(\frac{1}{18} \rho \pi R^2 L\right) \left(\frac{R^2}{9}\right) = \frac{1}{2} \left(\frac{1}{162} \rho \pi R^4 L\right)$

Finding the Ratio $I_1/I_2$

Now, we find the ratio of the moments of inertia:

$ \frac{I_1}{I_2} = \frac{\frac{1}{2}\rho \pi R^4 L}{\frac{1}{2} \left(\frac{1}{162} \rho \pi R^4 L\right)} $

Cancel out the common terms ($\frac{1}{2}\rho \pi R^4 L$):

$ \frac{I_1}{I_2} = \frac{1}{\frac{1}{162}} = 162 $

The ratio $I_1/I_2$ is 162.