Question

If the moment of inertia of solid sphere of mass 2 kg and radius 10 cm about its tangent is I, then the moment of inertia of a uniform disc of mass 3.5 kg and radius 20 cm about its diameter is:

• A. 3.75 I
• B. 2.25 I
• C. 1.25 I
• D. 1.75 I

Hint:

Always simplify equations in terms of base variables before doing heavy numeric calculations. This helps prevent decimal rounding errors when working with small values like centimeters or grams.

Answer 1

The Correct Option is C

Concept: We can determine the moments of inertia for standard symmetric shapes by using the Parallel Axis Theorem and standard rotational formulas.

Step 1: Formulating the expression for the solid sphere (I).
The moment of inertia of a solid sphere about a central axis passing through its center of mass is \( \frac{2}{5}MR^2 \). Applying the parallel axis theorem to find the moment of inertia about a tangent line: \[ I = \frac{2}{5}M_s R_s^2 + M_s R_s^2 = \frac{7}{5}M_s R_s^2 \] Substitute the given values \( M_s = 2 \, \text{kg} \) and \( R_s = 10 \, \text{cm} = 0.1 \, \text{m} \): \[ I = \frac{7}{5} (2) (0.1)^2 = \frac{14}{5} (0.01) = 0.028 \, \text{kg} \cdot \text{m}^2 \]

Step 2: Formulating the expression for the uniform disc (\( I_d \)).
The moment of inertia of a flat circular disc about its central perpendicular axis is \( \frac{1}{2}MR^2 \). Using the perpendicular axis theorem, the moment of inertia about any diameter axis is exactly half of that value: \[ I_d = \frac{1}{4}M_d R_d^2 \] Substitute the given values \( M_d = 3.5 \, \text{kg} \) and \( R_d = 20 \, \text{cm} = 0.2 \, \text{m} \): \[ I_d = \frac{1}{4} (3.5) (0.2)^2 = \frac{1}{4} (3.5) (0.04) = 3.5 \times 0.01 = 0.035 \, \text{kg} \cdot \text{m}^2 \]

Step 3: Finding the final scaling ratio.
Let us express \( I_d \) in terms of \( I \): \[ \frac{I_d}{I} = \frac{0.035}{0.028} = \frac{35}{28} = \frac{5}{4} = 1.25 \]