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PhyMedAdv 2022 • Paper 2

Particle Angular Momentum Z Component Calculation

A particle of mass 1 kg1\text{ kg} is subjected to a force which depends on the position as F⃗=−k(xi^+yj^) kg m s−2\vec{F} = -k(x\hat{i} + y\hat{j})\text{ kg m s}^{-2} with k=1 kg s−2k = 1\text{ kg s}^{-2}. At time t=0t = 0, the particle's position r⃗=(12i^+2j^) m\vec{r} = \left(\frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j}\right)\text{ m} and its velocity v⃗=(−2i^+2j^+2πk^) m s−1\vec{v} = \left(-\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}\right)\text{ m s}^{-1}. Let vxv_x and vyv_y denote the xx and yy components of the particle's velocity, respectively. Ignore gravity. When z=0.5 mz = 0.5\text{ m}, the value of (xvy−yvx)(x v_y - y v_x) is _____ m2s−1\text{m}^2\text{s}^{-1}.

Official Numerical Answer3

Topics & Concepts

System of Particles and Rotational MotionAngular MomentumKinematics

Step-by-Step Solution

To find the value of (xvy−yvx)(x v_y - y v_x) when z=0.5 mz = 0.5\text{ m}, we analyze the force acting on the particle and its effect on the components of motion.

Step 1: Equations of Motion and Conservation Law

The given force acting on the particle of mass m=1 kgm = 1\text{ kg} is: F⃗=−k(xi^+yj^)\vec{F} = -k(x\hat{i} + y\hat{j}) with k=1 kg s−2k = 1\text{ kg s}^{-2}.

Using Newton's second law, F⃗=ma⃗\vec{F} = m \vec{a}, we get the acceleration components: ax=Fxm=−kx=−xa_x = \frac{F_x}{m} = -k x = -x ay=Fym=−ky=−ya_y = \frac{F_y}{m} = -k y = -y az=Fzm=0a_z = \frac{F_z}{m} = 0

Now, consider the time derivative of the quantity (xvy−yvx)(x v_y - y v_x): ddt(xvy−yvx)=dxdtvy+xdvydt−dydtvx−ydvxdt\frac{d}{dt}(x v_y - y v_x) = \frac{dx}{dt} v_y + x \frac{dv_y}{dt} - \frac{dy}{dt} v_x - y \frac{dv_x}{dt} ddt(xvy−yvx)=vxvy+xay−vyvx−yax=xay−yax\frac{d}{dt}(x v_y - y v_x) = v_x v_y + x a_y - v_y v_x - y a_x = x a_y - y a_x

Substituting ax=−xa_x = -x and ay=−ya_y = -y: ddt(xvy−yvx)=x(−y)−y(−x)=−xy+xy=0\frac{d}{dt}(x v_y - y v_x) = x(-y) - y(-x) = -xy + xy = 0

Since the time derivative is zero, the quantity (xvy−yvx)(x v_y - y v_x), which corresponds to the zz-component of angular momentum per unit mass (Lz/mL_z / m), is a constant of motion (i.e., conserved at all times).

Step 2: Calculation of the Conserved Value at t=0t = 0

At time t=0t = 0, the initial position and velocity vectors are: r⃗(0)=12i^+2j^\vec{r}(0) = \frac{1}{\sqrt{2}}\hat{i} + \sqrt{2}\hat{j} v⃗(0)=−2i^+2j^+2πk^\vec{v}(0) = -\sqrt{2}\hat{i} + \sqrt{2}\hat{j} + \frac{2}{\pi}\hat{k}

From these, we identify:

  • x(0)=12 mx(0) = \frac{1}{\sqrt{2}}\text{ m}
  • y(0)=2 my(0) = \sqrt{2}\text{ m}
  • vx(0)=−2 m s−1v_x(0) = -\sqrt{2}\text{ m s}^{-1}
  • vy(0)=2 m s−1v_y(0) = \sqrt{2}\text{ m s}^{-1}

Substituting these initial values into (xvy−yvx)(x v_y - y v_x): xvy−yvx=(12)(2)−(2)(−2)x v_y - y v_x = \left(\frac{1}{\sqrt{2}}\right)(\sqrt{2}) - (\sqrt{2})(-\sqrt{2}) xvy−yvx=1−(−2)=3 m2s−1x v_y - y v_x = 1 - (-2) = 3\text{ m}^2\text{s}^{-1}

Conclusion

Since (xvy−yvx)(x v_y - y v_x) remains constant throughout the motion, its value when z=0.5 mz = 0.5\text{ m} is also 33.

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