Three solid spheres each of mass $m$ and diameter $d$ are stuck together such that the lines connecting the centres form an equilateral triangle of side of length $d$. The ratio $I_0/I_A$ of moment of inertia $I_0$ of the system about an axis passing the centroid and about center of any of the spheres $I_A$ and perpendicular to the plane of the triangle is :

Question

A thin circular disc of mass $ M $ and radius $ R $ is rotating in a horizontal plane about an axis passing through its center and perpendicular to its plane with angular velocity $ \omega $. If another disc of the same dimensions but of mass $ M/2 $ is placed gently on the first disc co-axially, then the new angular velocity of the system is:

Answer 1

To solve the problem of finding the ratio $I_0/I_A$, we need to calculate the moment of inertia of the system about two different axes: one passing through the centroid of the equilateral triangle formed by the centers of the spheres ($I_0$), and another passing through the center of any one of the spheres and perpendicular to the plane of the triangle ($I_A$).

Calculate the moment of inertia $I_0$ of the system about the centroid of the triangle:
- The centroid of an equilateral triangle formed by spheres is at equal distance from all vertices. The distance of the centroid from any of the vertices (sphere centers) is $\frac{d}{\sqrt{3}}$. This is due to geometric properties of an equilateral triangle.
- The moment of inertia $I_0$ about the centroid can be calculated using the parallel axis theorem:
- The formula for the moment of inertia of a single sphere about its center is $I_{\text{sphere}} = \frac{2}{5}mr^2 = \frac{2}{5}m\left(\frac{d}{2}\right)^2 = \frac{md^2}{10}$.
- Using the parallel axis theorem for one sphere with the distance $\frac{d}{\sqrt{3}}$ from the centroid, the moment of inertia is $I_{\text{parallel}} = I_{\text{sphere}} + m\left(\frac{d}{\sqrt{3}}\right)^2$.
- Therefore, $I_{\text{parallel}} = \frac{md^2}{10} + m\frac{d^2}{3} = \frac{md^2}{10} + \frac{md^2}{3} = \frac{13md^2}{30}$.
- Since there are three spheres arranged symmetrically: $I_0 = 3 \cdot \frac{13md^2}{30} = \frac{13md^2}{10}$.
Calculate the moment of inertia $I_A$ of the system about the center of one sphere:
- The moment of inertia $I_A$ for one sphere about its own center is $I_{\text{sphere}} = \frac{md^2}{10}$.
- For the other two spheres, each is at a distance $d$ from this center. Using the parallel axis theorem for each of these two spheres, the moment of inertia is $I_{\text{parallel, other}} = \frac{md^2}{10} + md^2 = \frac{11md^2}{10}$ for each of these spheres.
- Summing all contributions, $I_A = \frac{md^2}{10} + 2 \cdot \frac{11md^2}{10} = \frac{23md^2}{10}$.
Calculate the ratio $I_0/I_A$: $$\frac{I_0}{I_A} = \frac{\frac{13md^2}{10}}{\frac{23md^2}{10}} = \frac{13}{23}.$$

Therefore, the ratio of the moments of inertia is $\frac{13}{23}$, which is the correct answer.

Answer 2

To solve the problem of finding the ratio $I_0/I_A$, we need to calculate the moment of inertia of the system about two different axes: one passing through the centroid of the equilateral triangle formed by the centers of the spheres ($I_0$), and another passing through the center of any one of the spheres and perpendicular to the plane of the triangle ($I_A$).

Calculate the moment of inertia $I_0$ of the system about the centroid of the triangle:
- The centroid of an equilateral triangle formed by spheres is at equal distance from all vertices. The distance of the centroid from any of the vertices (sphere centers) is $\frac{d}{\sqrt{3}}$. This is due to geometric properties of an equilateral triangle.
- The moment of inertia $I_0$ about the centroid can be calculated using the parallel axis theorem:
- The formula for the moment of inertia of a single sphere about its center is $I_{\text{sphere}} = \frac{2}{5}mr^2 = \frac{2}{5}m\left(\frac{d}{2}\right)^2 = \frac{md^2}{10}$.
- Using the parallel axis theorem for one sphere with the distance $\frac{d}{\sqrt{3}}$ from the centroid, the moment of inertia is $I_{\text{parallel}} = I_{\text{sphere}} + m\left(\frac{d}{\sqrt{3}}\right)^2$.
- Therefore, $I_{\text{parallel}} = \frac{md^2}{10} + m\frac{d^2}{3} = \frac{md^2}{10} + \frac{md^2}{3} = \frac{13md^2}{30}$.
- Since there are three spheres arranged symmetrically: $I_0 = 3 \cdot \frac{13md^2}{30} = \frac{13md^2}{10}$.
Calculate the moment of inertia $I_A$ of the system about the center of one sphere:
- The moment of inertia $I_A$ for one sphere about its own center is $I_{\text{sphere}} = \frac{md^2}{10}$.
- For the other two spheres, each is at a distance $d$ from this center. Using the parallel axis theorem for each of these two spheres, the moment of inertia is $I_{\text{parallel, other}} = \frac{md^2}{10} + md^2 = \frac{11md^2}{10}$ for each of these spheres.
- Summing all contributions, $I_A = \frac{md^2}{10} + 2 \cdot \frac{11md^2}{10} = \frac{23md^2}{10}$.
Calculate the ratio $I_0/I_A$: $$\frac{I_0}{I_A} = \frac{\frac{13md^2}{10}}{\frac{23md^2}{10}} = \frac{13}{23}.$$

Therefore, the ratio of the moments of inertia is $\frac{13}{23}$, which is the correct answer.

The Correct Option is C

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