Question

A particle having mass 2 kg is moving along straight line 3 x + 4 y equal 5 with speed 8 metre per second find angular momentum of the particle about origin x and y are in metre

Answer 1

Answer:

Explanation:

16 kg m2 /sec

Answer 2

Answer:

The angular momentum of the particle about origin is  $16kg.m^{2} /s$  

Explanation:

Given mass of the particle, $m=2kg$

direction of motion of particle is along a straight line, $3x+4y=5$

speed of the particle, $v=8m/s$

Drop a perpendicular to origin from the line, $3x+4y-5=0$

Therefore , the perpendicular distance of a line from a point is given by

                                 $r=\Big | \frac{Ax^{'}+By^{'}+C }{\sqrt{A^{2}+B^{2} } } \Big|$

For the line $3x+4y-5=0$ from origin(0.0), the perpendicular distance is

                    $r=\Big | \frac{3\times 0+4\times 0+(-5) }{\sqrt{3^{2}+4^{2} } } \Big|=\Big | \frac{(-5) }{\sqrt{9+16 } } \Big|$

                       $=\Big | \frac{(-5) }{\sqrt{25 } } \Big|=1$

Angular momentum is given by $L=mvr$

                                                        $=2\times 8\times 1=16kg.m^{2} /s$

Hence the angular momentum of the particle is  $16kg.m^{2} /s$