Moment of inertia about an axis \( AB \) for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass 10 kg and radius \( R \) about an axis parallel to \( AB \) axis with separation of 3 m as shown in the figure below. The value of \( R \) is given as \( \sqrt{\frac{\alpha}{2}} \). The value of \( \alpha \) is _______.
Step 1: Understanding the Concept:
The total moment of inertia of the solid sphere about the axis \(AB\) relies on the parallel axis theorem, as it rotates about an axis separated from its center of mass.
We equate this to the moment of inertia of the rod rotating about one of its ends. Step 2: Key Formula or Approach:
Moment of inertia of a rod about its end: \(I_{rod} = \frac{M L^2}{3}\).
Moment of inertia of a solid sphere about its center of mass: \(I_{cm} = \frac{2}{5} m R^2\).
Parallel axis theorem: \(I_{sphere} = I_{cm} + m d^2\).
Equate the two moments of inertia: \(I_{rod} = I_{sphere}\). Step 3: Detailed Explanation:
First, calculate the moment of inertia of the rod.
Mass of rod \(M = 40 \text{ kg}\), Length \(L = 3 \text{ m}\).
\[ I_{rod} = \frac{40 \times (3)^2}{3} = 40 \times 3 = 120 \text{ kg m}^2 \]
Next, set up the expression for the moment of inertia of the solid sphere.
Mass of sphere \(m = 10 \text{ kg}\), distance from axis \(d = 3 \text{ m}\).
\[ I_{sphere} = \frac{2}{5} m R^2 + m d^2 \]
\[ I_{sphere} = \frac{2}{5}(10) R^2 + 10(3)^2 \]
\[ I_{sphere} = 4 R^2 + 90 \]
Now, equate the two moments of inertia:
\[ 4 R^2 + 90 = 120 \]
\[ 4 R^2 = 120 - 90 = 30 \]
\[ R^2 = \frac{30}{4} = \frac{15}{2} \]
Take the square root to find \(R\):
\[ R = \sqrt{\frac{15}{2}} \]
The problem states \(R = \sqrt{\frac{\alpha}{2}}\). By comparing, we get:
\[ \alpha = 15 \]
Step 4: Final Answer:
The value of \(\alpha\) is \(15\).
