JEE Advanced 2022 Paper 2 · Q01 · Angular Momentum
A particle of mass $1\,\text{kg}$ is subjected to a force which depends on the position as $\vec{F} = -k(x\hat{i} + y\hat{j})\,\text{kg m s}^{-2}$ with $k = 1\,\text{kg s}^{-2}$. At time $t = 0$, the particle's position $\vec{r} = \left(\dfrac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}\right)\,\text{m}$ and its velocity $\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \dfrac{2}{\pi}\hat{k}\right)\,\text{m s}^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and $y$ components of the particle's velocity, respectively. Ignore gravity. When $z = 0.5\,\text{m}$, the value of $(x v_y - y v_x)$ is _____ $\text{m}^2\,\text{s}^{-1}$.
Reveal answer + step-by-step solution
Correct answer:3
Solution
The force $\vec{F} = -k(x\hat{i}+y\hat{j})$ is central in the $xy$-plane (passes through the origin in $xy$), so torque about the $z$-axis vanishes and the $z$-component of angular momentum $L_z = m(xv_y - yv_x)$ is conserved. The motion in $z$ is force-free (no $\hat{k}$ component in $\vec{F}$, gravity ignored), so $z$ does not affect $L_z$. Evaluate $L_z$ at $t = 0$: $L_z/m = xv_y - yv_x = (1/\sqrt{2})(\sqrt{2}) - (\sqrt{2})(-\sqrt{2}) = 1 + 2 = 3\,\text{m}^2\,\text{s}^{-1}$. Since $L_z$ is conserved, the value at $z = 0.5\,\text{m}$ is also $3$.
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