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# a particle of mass m is projected from origin with speed u at an angle theta with positive x asis find angular momentum of particle at any time t about origin

Vikas TU
14149 Points
4 years ago
Angular momentum is mvr where r is the perpendicular distance to momenta.
In x – direcn. at any time t,
Vx = ucos(thetha).....................................(1)
x = ucos(thetha)*t.............................(2)
In y – direc. at any time t,
Vy  =  usin(thetha) – gt...............................(3)
y = usin(thetha) – 0.5gt^2......................(4)
from eqn. (2) and (4) we have got the coordinates.
Now distance b/w origin to the those coordinates =>
that is r = > root( (usin(thetha) – 0.5gt^2)^2  +   (ucos(thetha)*t)^2)
and v = root(Vx^2 + Vy^2)

Hence the final angular momentum at time t => mvr
that is=>    m * root(Vx^2 + Vy^2) *  root( (usin(thetha) – 0.5gt^2)^2  +   (ucos(thetha)*t)^2)
or
=> m * root((ucos(thetha))^2 +  (usin(thetha) – gt)^2) *  root( (usin(thetha) – 0.5gt^2)^2  +   (ucos(thetha)*t)^2)