A particle of mass 20 g is released with an initial velocity 5m/s along the curve from point A, as shown in the figure. The point A is at a height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : 

(take g = 10 m/s2 )

Answers (1)

Angular momentum -

\vec{L}=\vec{r}\times \vec{p}

- wherein

\vec{L}  represent angular momentum of a moving particle about a point.

it can be calculated  as L=r_1\, P=r\, P_1

r_1 = Length of perpendicular on line of motion

P_1 = component of momentum along perpendicualar to r

Appyl work energy theorem from A to B

W_{f} = \Delta KE

mgh = \frac{1}{2}mv^{2}_{B}-\frac{1}{2}mv_{A}^{2}

v^{2}_{B}-v_{A}^{2}= 2gh

v^{2}_{B} = 2\times 10\times 10+25=225

v_{B} = 15m/s

Angular momentum about 0=L

L = mvr_{ob}=\frac{20}{10^3}\times 15\times 20 = 6kg .m^{2}/s

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