a uniform rod of mass m and length l is hinged at upper end the rod is

1.	a uniform rod of mass m and length l is hinged at upper end the rod is free to rotate in vertical plane a ball of mass m moving horizontally with velocity v colloides at lower end of rod perpendicularly to it and stick to it the minimum velocity of ball such that combined system just complete the vertical circle will be
Answer» The Minimum velocity of the ball such that combined system just COMPLETE the vertical circle is $2\sqrt{\frac{gl}{3}}$ Explanation: We know that for a rod of mass m and length l is $\frac{1}{3}ml^2$ Initially the angular velocity of the rod is zero If after impact the combined angular velocity of the rod and the ball is $\omega$ Then, Initial Kinetic Energy of the ball $=\frac{1}{2}mv^2$ This will comvert into ROTATIONAL kinetic energy of the system (Rod + Ball) after impact Therefore, the kinetic energy of the system (Rod + Ball) after impact $=\frac{1}{2}I_{Sys}\omega^2$ Where, $I_{Sys}$ is the moment of inertia of rod + ball system $I_{Sys}=\frac{1}{3}ml^2+ml^2$ $\implies I_{Sys}=\frac{4}{3}ml^2$ From the conservation of energy The kinteic energy before the impact and after the impact will be same Hence, $\frac{1}{2}mv^2=\frac{1}{2}I_{Sys}\omega^2$ $\implies mv^2=\frac{4}{3}ml^2\times \omega^2$ $\implies \omega^2=\frac{3v^2}{4l^2}$ $\implies \omega=\sqrt{\frac{3}{4}}\frac{v}{l}$ THUS, the linear velocity of the system $V=\omega\times l$ $\implies V=\sqrt{\frac{3}{4}}v$ For the system to just complete a full circle $V=\sqrt{gl}$ $\implies \sqrt{\frac{3}{4}}v=\sqrt{gl}$ $\implies v^2=\frac{4}{3}gl$ $\implies v=2\sqrt{\frac{gl}{3}}$ Hope this answer is helpful. Know More: SIMILAR question Click Here: brainly.in/question/9258976