Statement
\\$2.7.42.\\$ A rod of mass \\$m_1\\$ and length \\$l\\$ is suspended on a hinge. A small piece of plasticine of mass \\$m_2\\$ sticks to the middle of the rod, moving until it collides with it horizontally at a speed of \\$V\\$. Find the maximum angle of deviation of the rod from the vertical. Ignore the friction in the hinge.
Solution
Angular momentum
\\$\\$
\\\\left\\\\\\\\{\\\\begin{matrix} L=m_2v\\\\frac{l}{2} \\\\\\\\\\\\ L=I\\\\omega \\\\end{matrix}\\\\right.
\\$\\$
Parallel axis theorem
\\$\\$
I=I_0+m_2d^2=\\\\frac{1}{3}m_1l^2+m_2\\\\left(\\\\frac{l}{2}\\\\right)^2
\\$\\$
\\$\\$
m_2v\\\\frac{l}{2}=\\\\omega\\\\left(\\\\frac{1}{3}m_1l^2+\\\\frac{1}{4}m_2l^2\\\\right)
\\$\\$
From where
\\$\\$
\\\\omega = \\\\frac{6m_2}{4m_1+3m_2}\\\\frac{v}{l}
\\$\\$
From the figure
\\$\\$
h=\\\\frac{l}{2}(1-\\\\cos\\\\alpha )
\\$\\$
Law of conservation of energy in rotational motion
\\$\\$
\\\\frac{I\\\\omega^2}{2}=(m_1+m_2)\\\\frac{l}{2}(1-\\\\cos\\\\alpha )
\\$\\$
Expressing \\$\\\\cos\\\\alpha\\$
\\$\\$
1-\\\\cos\\\\alpha =\\\\frac{I\\\\omega^2}{(m_1+m_2)l}
\\$\\$
\\$\\$
\\\\boxed{\\\\cos\\\\alpha =1-\\\\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}}
\\$\\$
Answer
\\$\\$
\\\\cos\\\\alpha =1-\\\\frac{3m_2^2v^2}{gl(4m_1+3m_2)(m_1+m_2)}
\\$\\$