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.
Similar Questions
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Important Questions from Mechanics
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Suppose there is a uniform circular disc of mass $M \text{ kg}$ and radius $r \text{ m}$ shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis $A$ of the disc is given by $\frac{x}{256} M r^2$. The value of $x$ is ________.
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