Question

Questions 10-12 A uniform rod of mass M-0.6 kg and length -1 m with a point mass m0.3 kg at its free end is rotating with angular speed s-10.0 rad/s about the s-axis, as shown in the M.L 2m moving in the plane of rotation collides perpendicularly in the direction of the rotation of the rod and sticds to the rod at a distance 2L/3 form point P with a linear speed Jup.. What is the angular monentum vector relative to point P just after the collision in units of kgm2/s 7 (a) irk 0) 15 () k() z1k () 23k

Answer 1

Consider an axis of rotation through point P (Z-axis)

For the system of rod with the point mass and the colliding mass , angular momentum is conserved about this axis.

That is $\vec{L}_i=\vec{L}_f$

$\vec{L}_i=\vec{L}_{rod+point -mass}+\vec{L}_{colliding- mass}$

$\vec{L}_i=\left [ \left ( \frac{ML^2}{3}+mL^2 \right )\omega_0+(2m)(3\omega_0L)\left ( \frac{2L}{3} \right ) \right ]\hat{k}$

Hence $\vec{L}_f=\left [ \left ( \frac{ML^2}{3}+mL^2 \right )\omega_0+(2m)(3\omega_0L)\left ( \frac{2L}{3} \right ) \right ]\hat{k}$

$\vec{L}_f=\left [ \left ( \frac{0.6*1^2}{3}+0.3*1^2 \right )10+(2*0.3)(3*10*1)\left ( \frac{2*1}{3} \right ) \right ]\hat{k}$

$\vec{L}_f=17\hat{k}$

Answer is $a)\,17\hat{k}$