Question

The moment of inertia of a solid hemisphere (mass 𝑀 and radius 𝑅) about the axis passing through the hemisphere and parallel to its flat surface is \(\frac{2}{5}\) 𝑀𝑅² . The distance of the axis from the center of mass of the hemisphere (in units of 𝑅) is______. (Rounded off to two decimal places)

Answer 1

Correct Answer: 0.36

Given Data

• Moment of inertia about the given axis: \( I = \frac{2}{5} M R^2 \)
• Radius of the hemisphere: \( R \)
• Mass of the hemisphere: \( M \)

Step 1: Applying the Parallel Axis Theorem

Let the distance of the axis from the center of mass of the hemisphere be \( d \). Using the parallel axis theorem, the moment of inertia about the given axis can be expressed as:

\[ I = I_{\text{CM}} + M d^2 \]

where:

\[ I_{\text{CM}} = \text{moment of inertia of the hemisphere about the axis through its center of mass}. \]

Step 2: Moment of Inertia for a Solid Hemisphere

For a solid hemisphere, the moment of inertia about an axis through its center of mass and parallel to the flat surface is:

\[ I_{\text{CM}} = \frac{83}{320} M R^2 \]

Step 3: Substituting into the Parallel Axis Theorem

Substituting \( I = \frac{2}{5} M R^2 \) into the equation:

\[ \frac{2}{5} M R^2 = \frac{83}{320} M R^2 + M d^2 \]

Dividing throughout by \( M R^2 \):

\[ \frac{2}{5} = \frac{83}{320} + d^2 \]

Step 4: Solving for \( d^2 \)

Rearranging:

\[ d^2 = \frac{2}{5} - \frac{83}{320} \]

Converting fractions to a common denominator:

\[ \frac{2}{5} = \frac{128}{320} \]

\[ d^2 = \frac{128}{320} - \frac{83}{320} = \frac{45}{320} \]

Step 5: Solving for \( d \)

Taking the square root:

\[ d = \sqrt{\frac{45}{320}} \]

Breaking it down:

\[ d = \frac{\sqrt{45}}{\sqrt{320}} = \frac{3\sqrt{5}}{8\sqrt{2}} = \frac{3\sqrt{10}}{16} \]

Step 6: Numerical Approximation

Approximating the value:

\[ d \approx 0.375 R \]

Final Answer

Thus, the distance of the axis from the center of mass is \( d \approx 0.375 R \).