# A partial of mass m is projected from point O with a horizontal velocity u at time t=0 what is its angular momentum relative to point Is a function of time

Arun
25757 Points
4 years ago
The angular momentum vector v_L = ( v_r ) × ( v_p ) = m ( v_r ) × ( v_u ) …
Set the origin of the coordinate system at point O … the x-component of the
position vector is given by …
… ( v_r )₁ = u₀ t … since the body moves with uniform velocity along the x …
and where … u₀ = initial speed of projection at O …
… the y-component of the position vector is given by …
… ( v_r )₁ = - ½ g t ² … since the body moves in free fall along the y and the
initial y velocity is zero … that is …
… ( v_r ) = ( u₀ t ) e₁ + ( - ½ g t ² ) e₂ … where … e₁ = unit vector along (+x) …
… e₂ = unit vector along (+y) …
For the velocity vector ( v_u ) … the x-component is … ( v_u )₁ = u₀ = constant
… while the y-component is … ( v_u )₂ = - g t … therefore …
… ( v_u ) = u₀ e₁ + ( - g t ) e₂ … it follows that …
… ( v_r ) × ( v_u ) = [ ( u₀ t ) e₁ + ( - ½ g t ² ) e₂ ] × [ u₀ e₁ + ( - g t ) e₂ ]
………………….. = ( u₀ t ) ( - g t ) e₁ × e₂ + ( - ½ g t ² ) u₀ e₂ × e₁
………………….. = - ( u₀ g t ² ) e₃ + ( - ½ u₀ g t ² ) ( - e₃ )
………………….. = - ( u₀ g t ² ) e₃ + ( ½ u₀ g t ² ) e₃
………………….. = - ( ½ u₀ g t ² ) e₃ … where … e₃ = unit vector along (+z) …
Finally … v_L = - ( ½ m u₀ g t ² ) e₃