The position vector of the centre of mass \(\vec r\, cm\) of an asymmetric uniform bar of negligible area of cross-section as shown in figure is

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Answer & Solution

Correct Answer

(A) \(\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y\)

Step-by-step Solution

Detailed explanation

Three parts of rod can be considered as point masses.

\({\overrightarrow r _{cm}} = \frac{{2m\,{{\overrightarrow r }_1} + m{{\overrightarrow r }_2} + m{{\overrightarrow r }_3}}}{{4\,m}}\)

\({\overrightarrow r _{cm}} = \frac{{2m\left( {L\hat i + L\hat j} \right) + m\left( {2L\hat i + \frac{L}{2}\hat j} \right) + m\left( {\frac{{5L}}{2}\hat i} \right)}}{{4\,m}}\)

\( = \frac{{\frac{{13}}{2}L\hat i + \frac{5}{2}L\hat j}}{4}\)

\( = \frac{{13}}{8}L\hat i + \frac{5}{8}L\hat j\)

Same subject

Match List\(-I\) with List\(-II\)

List \(-I\)	List \(-II\)
\((A)\) Moment of inertia of solid sphere of radius \((R)\) about any tangent	\((I)\) \(\frac{5}{3} MR ^{2}\)
\((B)\) Moment of inertia of hollow sphere of radius \((R)\) about any tangent	\((II)\) \(\frac{7}{5} MR ^{2}\)
\((C)\) Moment of inertia of circular ring of radius \((R)\) about its diameter.	\((III)\) \(\frac{1}{4} MR ^{2}\)
\((D)\) Moment of inertia of circular disc of radius \((R)\) about any diameter.	\((IV)\) \(\frac{1}{2} MR ^{2}\)

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