Find the radius of the hole in a uniform cylinder such that the moment of inertia is 1/4 of the original value.
Set up the moment of inertia for a solid cylinder
The moment of inertia of a solid uniform cylinder about its central axis is:
\( I_{\text{solid}} = \frac{1}{2}MR^2 \)
Where M is the mass and R is the radius of the cylinder.
Determine the moment of inertia of a cylinder with a hole
For a cylinder with a hole of radius r, we can think of it as the solid cylinder minus a smaller cylinder of radius r. The moment of inertia will be:
\( I_{\text{with hole}} = \frac{1}{2}M_{\text{total}}R^2 - \frac{1}{2}M_{\text{hole}}r^2 \)
Since the cylinder is uniform, the mass is proportional to the area. If we call the original mass M, then:
\( M_{\text{hole}} = M \cdot \frac{\pi r^2}{\pi R^2} = M \cdot \frac{r^2}{R^2} \)
Therefore:
\( I_{\text{with hole}} = \frac{1}{2}M R^2 - \frac{1}{2} \left(M \cdot \frac{r^2}{R^2}\right) r^2 = \frac{1}{2}M R^2 - \frac{1}{2}M \frac{r^4}{R^2} \)
Apply the condition that the moment of inertia is 1/4 of the original
According to the problem, we have:
\( I_{\text{with hole}} = \frac{1}{4} I_{\text{solid}} \)
Substituting:
\( \frac{1}{2}M R^2 - \frac{1}{2}M \frac{r^4}{R^2} = \frac{1}{4} \cdot \frac{1}{2}M R^2 \)
Simplifying:
\( \frac{1}{2}M R^2 - \frac{1}{2}M \frac{r^4}{R^2} = \frac{1}{8}M R^2 \)
Solve for the radius of the hole
Continuing with our equation:
\( \frac{1}{2}M R^2 - \frac{1}{2}M \frac{r^4}{R^2} = \frac{1}{8}M R^2 \)
Dividing both sides by \(\frac{1}{2}M\):
\( R^2 - \frac{r^4}{R^2} = \frac{1}{4}R^2 \)
Rearranging:
\( R^2 - \frac{1}{4}R^2 = \frac{r^4}{R^2} \)
\( \frac{3}{4}R^2 = \frac{r^4}{R^2} \)
\( \frac{3}{4}R^4 = r^4 \)
Taking the fourth root:
\( r = R \cdot \left(\frac{3}{4}\right)^{1/4} = R \cdot \sqrt[4]{\frac{3}{4}} \)
\( r = R \cdot \sqrt[4]{\frac{3}{4}} \approx R \cdot 0.9306 \)
Therefore, \( r = R \cdot \sqrt[4]{\frac{3}{4}} \)
\(\boxed{r = R \cdot \sqrt[4]{\frac{3}{4}}}\)
\(\boxed{r = R \cdot \sqrt[4]{\frac{3}{4}}}\)
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