A point mass m_A is connected to a point mass | vedclass

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Question

$\mathrm{I}_{\mathrm{AA}}=\mathrm{m}_{\mathrm{B}} l^{2}$

$\mathrm{I}_{\mathrm{BB}}=\mathrm{m}_{\mathrm{A}} l^{2}$

$\frac{\mathrm{I}_{\mathrm{BB}

Vedclass · Accepted Answer

$(1/4)\, l$

Vedclass · Answer

$\mathrm{I}_{\mathrm{AA}}=\mathrm{m}_{\mathrm{B}} l^{2}$

$\mathrm{I}_{\mathrm{BB}}=\mathrm{m}_{\mathrm{A}} l^{2}$

$\frac{\mathrm{I}_{\mathrm{BB}}}{\mathrm{I}_{\mathrm{AA}}}=3$

$\frac{\mathrm{m}_{\mathrm{A}}}{\mathrm{m}_{\mathrm{B}}}=3$

$\Rightarrow \mathrm{m}_{\mathrm{A}}=3 \mathrm{m}_{\mathrm{B}}$

But $\mathrm{m}_{\mathrm{A}} \mathrm{x}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}\left(l-\mathrm{x}_{\mathrm{A}}\right)$

$3 \mathrm{m}_{\mathrm{B}} \mathrm{x}_{\mathrm{A}}=\mathrm{m}_{\mathrm{B}}\left(l-\mathrm{x}_{\mathrm{A}}\right)$

$3 x_{A}=l-x_{A}$

$4 x_{A}=l$

$\mathrm{x}_{\mathrm{A}}=\frac{l}{4}$

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