Question

Two particles of masses m1 and m2 are connected by a rigid massless rod of length land move freely in a plane. Show that moment of inertia of the system about an axis perpendicular to the plane and passing through the c.m Is nl2 where n=m1*m2/m1+m2

Answer 1

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Let the center of mass of the system be at a distance $\sf{\bar x}$ from the particle of mass $\sf{m_1,}$ which is given by,

$\longrightarrow\sf{\bar x=\dfrac{m_1(0)+m_2(l)}{m_1+m_2}}$

$\longrightarrow\sf{\bar x=\dfrac{m_2l}{m_1+m_2}\quad\quad\dots(1)}$

The moment of inertia of the system, about an axis perpendicular to plane of rod and passing through center of mass will be,

$\longrightarrow\sf{I=m_1(\bar x)^2+m_2(l-\bar x)^2}$

From (1),

$\longrightarrow\sf{I=m_1\left(\dfrac{m_2l}{m_1+m_2}\right)^2+m_2\left(l-\dfrac{m_2l}{m_1+m_2}\right)^2}$

$\longrightarrow\sf{I=m_1\left(\dfrac{m_2l}{m_1+m_2}\right)^2+m_2\left(\dfrac{(m_1+m_2)l-m_2l}{m_1+m_2}\right)^2}$

$\longrightarrow\sf{I=m_1\left(\dfrac{m_2l}{m_1+m_2}\right)^2+m_2\left(\dfrac{m_1l+m_2l-m_2l}{m_1+m_2}\right)^2}$

$\longrightarrow\sf{I=m_1\left(\dfrac{m_2l}{m_1+m_2}\right)^2+m_2\left(\dfrac{m_1l}{m_1+m_2}\right)^2}$

$\longrightarrow\sf{I=\dfrac{m_1(m_2l)^2}{(m_1+m_2)^2}+\dfrac{m_2(m_1l)^2}{(m_1+m_2)^2}}$

$\longrightarrow\sf{I=\dfrac{m_1(m_2)^2l^2+m_2(m_1)^2l^2}{(m_1+m_2)^2}}$

$\longrightarrow\sf{I=\dfrac{m_1m_2l^2(m_1+m_2)}{(m_1+m_2)^2}}$

$\longrightarrow\underline{\underline{\sf{I=\dfrac{m_1m_2l^2}{m_1+m_2}}}}$

Taking $\sf{\dfrac{m_1m_2}{m_1+m_2}=n,}$

$\longrightarrow\underline{\underline{\sf{I=nl^2}}}$

Hence Proved!