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

Step 1: Plan the two moments of inertia.
We first find I for the sphere about a tangent line, then find the disc's moment about its diameter, and finally take the ratio.
Step 2: Moment of inertia of the sphere about a tangent.
A solid sphere about its centre has $\tfrac{2}{5}MR^{2}$. Using the parallel axis theorem to shift to a tangent line a distance $R$ away, we add $MR^{2}$: \[ I = \frac{2}{5}M_sR_s^{2} + M_sR_s^{2} = \frac{7}{5}M_sR_s^{2} \]
Step 3: Put the sphere numbers in.
With $M_s = 2$ kg and $R_s = 0.1$ m: \[ I = \frac{7}{5}(2)(0.1)^{2} = 0.028\ \text{kg}\,\text{m}^{2} \]
Step 4: Moment of inertia of the disc about a diameter.
A disc about its central axis is $\tfrac{1}{2}MR^{2}$. By the perpendicular axis theorem the moment about a diameter is half of that: \[ I_d = \frac{1}{4}M_dR_d^{2} \]
Step 5: Put the disc numbers in.
With $M_d = 3.5$ kg and $R_d = 0.2$ m: \[ I_d = \frac{1}{4}(3.5)(0.2)^{2} = 0.035\ \text{kg}\,\text{m}^{2} \]
Step 6: Take the ratio.
\[ \frac{I_d}{I} = \frac{0.035}{0.028} = \frac{5}{4} = 1.25 \] \[ \boxed{I_d = 1.25\,I} \]