JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2019
- Medium
- MCQ
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.
\({\overrightarrow r _{cm}} = \frac{{2m\,{{\overrightarrow r }_1} + m{{\overrightarrow r }_2} + m{{\overrightarrow r }_3}}}{{4\,m}}\)
\({\overrightarrow r _{cm}} = \frac{{2m\left( {L\hat i + L\hat j} \right) + m\left( {2L\hat i + \frac{L}{2}\hat j} \right) + m\left( {\frac{{5L}}{2}\hat i} \right)}}{{4\,m}}\)
\( = \frac{{\frac{{13}}{2}L\hat i + \frac{5}{2}L\hat j}}{4}\)
\( = \frac{{13}}{8}L\hat i + \frac{5}{8}L\hat j\)
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