JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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
- A \(\vec r\,cm = \frac{{13}}{8}L\hat x + \frac{5}{8}L\hat y\)
- B \(\vec r\,cm = \frac{{5}}{8}L\hat x + \frac{13}{8}L\hat y\)
- C \(\vec r\,cm = \frac{{3}}{8}L\hat x + \frac{11}{8}L\hat y\)
- D \(\vec r\,cm = \frac{{11}}{8}L\hat x + \frac{3}{8}L\hat y\)
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. \({\overrightarrow r _{cm}} = \frac{{2m\,{{\overrightarrow r }_1} + m{{\overrightarrow r }_2} + m{{\overrightarrow r }_3}}}{{4\,m}}\)…
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