Q. A particle of mass $m = 5\ units$ is moving with a uniform speed $v = 3\sqrt{2}\ units$ in the $XOY$ plane along the straight line $Y = X + 4$. The magnitude of the angular momentum about origin is
A) Zero
B) $60\ units$
C) $75\ units$
D) $40\sqrt{2}\ units$
Explanation
(b) Form the triangle $OAC$
$d = OC\sin{45{^\circ} = 4 \times \frac{1}{\sqrt{2}} = 2\sqrt{2}}$
Angular momentum = Linear momentum $x$ Perpendicular distance of line of action of linear momentum from the point of rotation
$L = p \times d = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 60\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.