JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A thin rod of mass \(0.9\, kg\) and length \(1 \,m\) is suspended, at rest, from one end so that it can freely oscillate in the vertical plane. A particle of move \(0.1\, kg\) moving in a straight line with velocity \(80\,m / s\) hits the rod at its bottom most point and sticks to it (see figure). The angular speed (in \(rad/s\)) of the rod immediately after the collision will be
- A \(30\)
- B \(28\)
- C \(20\)
- D \(25\)
Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ L }_{ i }=\overrightarrow{ L }_{ f }\) \(mvL = I \omega\) \(mvL =\left(\frac{ ML ^{2}}{3}+ mL ^{2}\right) \omega\) \(0.1 \times 80 \times 1=\left(\frac{0.9 \times 1^{2}}{3}+0.1 \times 1^{2}\right)\) \(8=\left(\frac{3}{10}+\frac{1}{10}\right) \omega\)…
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