JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Rotational Mechanics

A particle of mass 20 g is released with an initial velocity

5 m / s

along the curve from the point A, as shown in the figure. The point A is at 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 / s^{2}

]

(A) $2 kg - m^{2} / s$
(B) $8 kg - m^{2} / s$
(C) $6 kg - m^{2} / s$ ✓ Correct
(D) $3 kg - m^{2} / s$

Answer: (C)

Solution

Take the drop

h = 10

m and the perpendicular distance of

B

from

O

r = 20

m from the figure. Energy conservation on the frictionless curve gives

v_{B}^{2} = v_{A}^{2} + 2 g h = 5^{2} + 2 (10) (10) = 225

, so

v_{B} = 15

m/s. The angular momentum about

O

L = m v_{B} r = 0.020 \times 15 \times 20 = 6 kg m^{2} /s

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option

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