JEE Advanced 2021 Paper 2 · Q09 · Angular Momentum

Correct answer:0.18

Solution

When the string just becomes taut, the bob is on the floor moving horizontally with momentum $P = 0.2$ kg m/s (no friction, no vertical motion yet). The string makes an angle with the vertical such that the horizontal distance from the suspension point equals $\sqrt{L^2 - H^2} = \sqrt{1 - 0.81} = \sqrt{0.19}$ m. Angular momentum about suspension point $J = |\vec{r} \times \vec{p}| = P \times H = 0.2 \times 0.9 = 0.18$ kg m$^2$/s (only the perpendicular distance from the suspension point to the line of momentum matters, which is $H = 0.9$ m). So $J = 0.18$.

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