Question

Find the Moment of Inertia.

Answer 1

Answer:

please see the attached picture

Answer 2

Answer:-

Option (2) is correct.

$\bf{\boxed{moment \: of \: inertia \: (I) = \frac{3}{4} m {r}^{2} }}$

Formula used :-

$\small \: \star \: moment \: of \: inertia \: (I) = m_1 { ( \: r_1\: )}^{2} + m_2 { \:( r_2\:) }^{2}$

Explanation:-

Position of centre of mass from A

$\implies \: \frac{0 \times m + r \times 3m}{m + 3m} \\ \\ \implies \: \frac{3mr}{4m} \\ \\ \bf{\rightarrow \: position \: of \: centre \: of \: mass \: from \: }\\ \: \: \: \: \: \: \bf{ A\: =} \frac{3}{4} r \: \\ \\ \implies m_1= m , m_2 = 3m$

Now moment of inertia of the system of masses ,

$\implies \: I \: = m \: \times { \bigg(\frac{3}{4}r \bigg)}^{2} + 3m \: \times { \bigg(\frac{1}{4} r \bigg)}^{2} \\ \\ \implies \: I= m {r}^{2} \times \frac{9}{16} + m {r}^{2} \times \frac{3}{4} \\ \\ \implies \: I \: = m {r}^{2} \bigg( \frac{9 + 3}{16} \bigg) \\ \\ \implies \: \boxed{I = \frac{3}{4} m {r}^{2} }$

Option (2) is correct.