A rod $A B$ of length $1 \mathrm{~m}$ is placed at the edge of a smooth table as shown. It is hit horizontally at point $B$. If the displacement of centre of mass in 1s is $5 \sqrt{2} \mathrm{~m}$, then the angular velocity of the rod is (Take, $g=10 \mathrm{~ms}^{-2}$ )

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Aicharya · Accepted Answer

Given, length of $\operatorname{rod} A B, l=1 \mathrm{~m}$ Displacement of centre of mass, $s=5 \sqrt{2} \mathrm{~m}$ Time taken, $t=1 \mathrm{~s}$ Acceleration due to gravity, $g=10 \mathrm{~ms}^{-2}$

Let vertical displacement in $1 \mathrm{~s}=y$ Angular velocity $=\omega$ Moment of inertia $=I$ $I($ about centre of rod $)=m l^2 / 12$ $\because \quad y=u t+\frac{1}{2} g t^2$ $\therefore \quad y=\frac{1}{2} \times 10 \times 1^2=5 \mathrm{~m}$ and $\quad s=\sqrt{x^2+y^2}$ $\therefore \quad(5 \sqrt{2})^2=x^2+y^2$ $\Rightarrow \quad(5 \sqrt{2})^2=x^2+5^2$ Displacement along $X$-axis in 1s $ \begin{aligned} & & x^2=50-25 \\ \Rightarrow & & x=\sqrt{25}=5 \mathrm{~m} \\ \therefore & & v_x=\frac{x}{t}=5 \mathrm{~m} / \mathrm{s} \end{aligned} $ Since, angular impulse $=$ change in angular momentum $ \begin{aligned} & & J \cdot l / 2 & =I \omega \\ \Rightarrow & & m v l / 2 & =\frac{m l^2}{12} \omega \\ \Rightarrow & & \omega & =6 \mathrm{v}=6 \times 5=30 \mathrm{rads}^{-1} \end{aligned} $

A rod $A B$ of length $1 \mathrm{~m}$ is placed at the edge of a smooth table as shown. It is hit…

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