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A particle of mass 20 gm) slides along the frictionless surface. When the particle reaches point B, its angular momentum (in kg m2/s) about O will be 


[Take g=10 m/s2]


Solution

Let the speed of particle be v when it reaches point B.


As there are no non-conservative forces acting on the particle, hence on applying law of conservation of mechanical energy,
M.Ei=M.Ef
KEi+PEi=KEf+PEf
Taking PE=0 at the reference level,
(12×0.02×52)+(0.02×10×10)=12×0.02×v2+0
v=15 m/s     ....(1)
Angular momentum about O
L=mvr
Since at B, r=20 m about O.
L=0.02×15×20
L=6 kg m2/s
The angular momentum of the particle about point O when it reaches B is 6 kg m2/s.

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