JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2020
- Hard
- Numerical
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
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Answer & Solution
Correct Answer
(C) \(20\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ L }_{ i }=\overrightarrow{ L }_{ f }\)
\(mvL = I \omega\)
\(mvL =\left(\frac{ ML ^{2}}{3}+ mL ^{2}\right) \omega\)
\(0.1 \times 80 \times 1=\left(\frac{0.9 \times 1^{2}}{3}+0.1 \times 1^{2}\right)\)
\(8=\left(\frac{3}{10}+\frac{1}{10}\right) \omega\)
\(8=\frac{4}{10} \omega\)
\(\omega=20\, rad \frac{ rad }{ sec }\)
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