Question

Find the maximum angular displacement of the rod. if particle sticks after collision?

Answer 1

ANSWER.

$\sf \to \: maximum \: angular \: displacement \: of \: rod \: =63 \degree$

Option [ 1 ] is correct answer.

EXPLANATION.

$\sf \to \: as \: we \: know \: that \: \\ \\ \sf \to \: angular \: momentum \implies \: l_{i} = l_{f} \\ \\ \sf \to \: mvl \: = i \omega \\ \\ \sf \to \: mvl \: = (ml {}^{2} + \frac{(2m {l}^{2} )}{3} \omega) \\ \\ \sf \to \: mvl \: = \frac{5m {l}^{2} }{3} \times \omega \\ \\ \sf \to \: \omega \: = \frac{3v}{5l}$

$\sf \to \: \dfrac{1}{2} i \omega {}^{2} = mgl(1 - \cos( \theta) ) + 2mg \dfrac{l}{2} (1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{1}{2} \times ( \frac{5}{3} m {l}^{2} ) \times ( \frac{3v}{5l} ) {}^{2} = mgl(1 - \cos( \theta) ) + mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{1}{2} \times ( \frac{5}{3}ml {}^{2} ) \times ( \frac{9v {}^{2} }{25 {l}^{2} } ) = 2mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{ 3mv {}^{2} }{2 \times 5} = 2mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{3 \times 36}{10} = 2gl(1 - \cos( \theta) )$

$\sf \to \: \dfrac{3 \times 36}{10 \times 2 \times 10} = 1 - \cos( \theta) \\ \\ \sf \to \: \frac{27}{50} = 1 - \cos( \theta) \\ \\ \sf \to \: 1 - \frac{27}{50} = \cos( \theta) \\ \\ \sf \to \: \frac{50 - 27}{50} = \cos( \theta) \\ \\ \sf \to \: \frac{23}{50} = \cos( \theta) \\ \\ \sf \to \: \cos( \theta) \approx \: 63 \degree$

Answer 2

ANSWER.

$\sf \to \: maximum \: angular \: displacement \: of \: rod \: =63 \degree$

Option [ 1 ] is correct answer.

EXPLANATION.

$\sf \to \: as \: we \: know \: that \: \\ \\ \sf \to \: angular \: momentum \implies \: l_{i} = l_{f} \\ \\ \sf \to \: mvl \: = i \omega \\ \\ \sf \to \: mvl \: = (ml {}^{2} + \frac{(2m {l}^{2} )}{3} \omega) \\ \\ \sf \to \: mvl \: = \frac{5m {l}^{2} }{3} \times \omega \\ \\ \sf \to \: \omega \: = \frac{3v}{5l}$

$\sf \to \: \dfrac{1}{2} i \omega {}^{2} = mgl(1 - \cos( \theta) ) + 2mg \dfrac{l}{2} (1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{1}{2} \times ( \frac{5}{3} m {l}^{2} ) \times ( \frac{3v}{5l} ) {}^{2} = mgl(1 - \cos( \theta) ) + mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{1}{2} \times ( \frac{5}{3}ml {}^{2} ) \times ( \frac{9v {}^{2} }{25 {l}^{2} } ) = 2mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{ 3mv {}^{2} }{2 \times 5} = 2mgl(1 - \cos( \theta) ) \\ \\ \sf \to \: \frac{3 \times 36}{10} = 2gl(1 - \cos( \theta) )$

$\sf \to \: \dfrac{3 \times 36}{10 \times 2 \times 10} = 1 - \cos( \theta) \\ \\ \sf \to \: \frac{27}{50} = 1 - \cos( \theta) \\ \\ \sf \to \: 1 - \frac{27}{50} = \cos( \theta) \\ \\ \sf \to \: \frac{50 - 27}{50} = \cos( \theta) \\ \\ \sf \to \: \frac{23}{50} = \cos( \theta) \\ \\ \sf \to \: \cos( \theta) \approx \: 63 \degree$

#BE BRAINLY.