Question

If two thin uniform rod of lengths L1 and L2 of same material are joined to form ‘T’ shape as shown in the figure, then the distance of centre of mass of the system from centre of mass of first rod of length L1 is given by

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{y_{C.O.M}=\frac{l_{2}}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Length \: of \: rods = l_{1} \: and \: l_{2} \\ \\ \red{\underline \bold{To \: find :}} \\ \tt: \implies Distance \: between \: C.O.M \: of \: l_{1} \: rod \: to \: system \: C.O.M =?$

• According to given question :

$\tt \circ \: Centre \: of \: mass \:of \: uniform \: rod = \frac{l}{2} \\ \\ \tt \circ \: Let \: C.O.M \: of \: rod \: l_{1} \: is \: at \: point \:A\: and \: is \: connected \: with \: rod \: l_{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies y_{C.O.M} = \frac{ m_{1} y_{C.O.M_{1} } + m_{2} y_{C.O.M_{2} } }{ m_{1} + m_{2} } \\ \\ \tt: \implies y_{C.O.M} = \frac{ m_{1} \times 0 + m_{2} \times \frac{ l_{2 } }{2} }{ m_{1} + m_{2}} \\ \\ \tt: \implies y_{C.O.M} = \frac{ m_{2} l_{2} }{2( m_{1} + m_{2}) } \\ \\ \tt \circ \: If \: m_{1} = m_{2} = m \\ \\ \tt: \implies y_{C.O.M} = \frac{m l_{2} }{2 \times 2m} \\ \\ \green{\tt: \implies y_{C.O.M} = \frac{ l_{2} }{4}} \\ \\ \green{\tt \therefore C.O.M \: of \: system \: is \: \frac{ l_{2}}{4} \: distance \: from \: point \: A}$