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Rotational Mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsRotational Mechanics5 Marks
Question
A metre stick weighing 240g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end (figure). A particle of mass 100g is attached to the upper end of the stick through a light string of length 1m. Initially, the rod is kept vertical and the string horizontal when the system is released from rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.
✓

Answer

$\frac{1}{2}\text{l}\omega^2-0=0.1\times10\times1$
$\Rightarrow\omega=\sqrt{20}$
For collision
$0.1\times1^2\times\sqrt{20}+0=\Big[\Big(\frac{0.24}{3}\Big)\times1^2+(0.1)^21^2\Big]\omega'$
$\Rightarrow\omega'=\frac{\sqrt{20}}{10.(0.18)}$
$\Rightarrow0-\frac{1}{2}\omega'^2=-\text{m}_1\text{gl}(1-\cos\theta)-\text{m}_2\text{g}\frac{\text{l}}2{}(1-\cos\theta)$
$=0.1\times10(1-\cos\theta)=0.24\times10\times0.5(1-\cos\theta)$
$\Rightarrow\frac{1}{2}\times0.18\times\Big(\frac{20}{3.24}\Big)=2.2(1-\cos\theta)$
$\Rightarrow(1-\cos\theta)=\frac{1}{(2.2\times1.8)}$
$\Rightarrow1-\cos\theta=0.252$
$\Rightarrow\cos\theta=1-0.252=0.748$
$\Rightarrow\omega=\cos^{-1}(0.748)=41^\circ$

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