A particle of mass 20 g is released with an initial velocity 5 m/s… | JEE Main 2019 (12 Jan, Shift 2)

Q70JEE Main 2019 · 12 JAN · SHIFT 2Medium

Rolling Motion

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/s2)

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Answer: (a)

Frictionless slide: $v_B^2 = v_A^2 + 2gh = 25 + 2(10)(10) = 225 \Rightarrow v_B = 15$ m/s.
At B the velocity is horizontal and B is $a + h = 20$ m below O, so this is the perpendicular distance.
$L = m v_B (a+h) = 0.02 \times 15 \times 20 = 6\ \text{kg m}^2/\text{s}$ .
So the answer is (a) 6 kg-m $^2$ /s.

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Rolling Motion question from JEE Main 2019 (12 Jan, Shift 2)

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