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A particle of mass m is projected with a...

A particle of mass `m` is projected with a speed `u` at an angle `theta` to the horizontal at time `t = 0`. Find its angular momentum about the point of projection `O` at time `t`, vectorially. Assume the horizontal and vertical lines through `O` as `X` and `Y` axes, respectively.

Text Solution

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Initial velocity is `vecu=u(costhetahati=sinthetahatj)`
`:.` Velocity after time `t` will be `v=vecu +vecat`
`=ucosthetahati+(usintheta-"gt")hatj`
If ` vecr` be the radius vector of the particle `P` at time `t` (as shown in figure then from `vecs=vecut+vecat^(2)//2`

`:. vecr=ut(costhetahati=sinthetahatj)-("gt"^(2))/2hatj`
`=utcosthetahati+(utsintheta-("gt"^(2))//2)hatj`
`:.` Angular momentum `vecL` of the particle at time `t` will be
`vecL=m(vecrxxvecv)`
i.e. `vecL=m[utcosthetai+(utsintheta-("gt"^(2))/2)hatj]`
`xx[utcosthetahati+(usintheta-"gt")hatj]`
`=(-mut^(2)gcosthetahatk)/2`
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