Question

Moment of inertia ($M.I$) of four bodies, having the same mass and radius, are reported as:

$I_1$ = $M.I$of a thin circular ring about its diameter,

$I_2$ = $M.I$of a circular disc about an axis perpendicular to the disc and going through the center,

$I_3$ =$M.I$of a solid cylinder about its axis and

$I_4$ = $M.I$of a solid sphere about its diameter.

• A. $I_1=I_2=I_3<I_4$
• B. $I_1+I_2=I_3+\left(\frac{5}{2}\right)I_4$
• C. $I_1+I_3<I_2+I_4$
• D. $I_1=I_2=I_3>I_4$

Answer 1

The correct option is D

$I_1=I_2=I_3>I_4$

Step 1. Given data:

1. $I_1$ = $M.I$of a thin circular ring about its diameter,
2. $I_2$ = $M.I$of a circular disc about an axis perpendicular to the disc and passes through the center,
3. $I_3$ =$M.I$of a solid cylinder about its axis and
4. $I_4$ = $M.I$of a solid sphere about its diameter.

Step 2. Comparing the moment of inertia of all given bodies:

According to the question all bodies are having same mass$\left(M\right)$ and radius$\left(R\right)$.

The moment of inertia is defined as a quantity that expresses the tendency of a body to withstand angular acceleration, which is equal to the sum of the products of the mass of each particle within the body and the square of its distance from the axis of rotation.

Then,

$$\begin{gathered} I_1=\frac{M{R}^2}{2} \\ {I}_2=\frac{M{R}^2}{2} \\ {I}_3=\frac{M{R}^2}{2} \\ {I}_4=\left(\frac{2}{5}\right)M{R}^2 \end{gathered}$$

Then, $I_1=I_2=I_3>I_4$

Hence, option D is the correct answer.