Draw areal velocity versus time graph for mars.

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Q. Draw areal velocity versus time graph for mars.

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Concept understanding — Conservation of Angular Momentum

Conservation of Angular Momentum

The Intuition First

Imagine you're sitting on a spinning office chair with your arms stretched out. Someone gives you a gentle push, and you start rotating slowly. Now pull your arms in tight against your chest. What happens? You spin faster. Push your arms back out — you slow down again.

Nothing external pushed you to go faster or slower. The change came from inside — from how you arranged your mass relative to the axis of rotation.

That's the core idea: Angular momentum is a quantity that stays constant for a rotating system unless an external torque acts on it. When you pulled your arms in, you didn't change your angular momentum — you changed your distribution of mass, and your rotation speed had to adjust to keep the total constant.

The Precise Statement

\vec{L} = \text{constant} \quad \text{if} \quad \vec{\tau}_{\text{ext}} = 0

Where:

$\vec{L}$ is the angular momentum of the system
$\vec{\tau}_{\text{ext}}$ is the net external torque acting on the system

In words: The total angular momentum of an isolated system (no external torque) remains constant in both magnitude and direction.

What Is Angular Momentum?

For a point mass $m$ moving with velocity $\vec{v}$ at position $\vec{r}$ from a reference point:

\vec{L} = \vec{r} \times m\vec{v}

For a rigid body rotating about a fixed axis:

L = I\omega

Where:

$I$ = moment of inertia (how mass is distributed relative to the axis)
$\omega$ = angular velocity (how fast it spins)

Note

Moment of inertia $I$ depends on where the mass is, not just how much. Mass far from the axis gives larger $I$ ; mass close to the axis gives smaller $I$ .

Why It Works: The Physics

Newton's second law for rotation says:

\tau_{\text{ext}} = \frac{dL}{dt}

If $\tau_{\text{ext}} = 0$ , then $\frac{dL}{dt} = 0$ , so $L$ is constant.

Since $L = I\omega$ , if $I$ changes (you pull arms in), $\omega$ must change in the opposite way to keep $L$ the same:

I_1 \omega_1 = I_2 \omega_2

Smaller $I$ → larger $\omega$ (spin faster). Larger $I$ → smaller $\omega$ (spin slower).

Real-World Examples

Situation	What happens	Why
Ice skater pulling arms in	Spins faster	$I$ decreases, $\omega$ increases to keep $L$ constant
Diver tucking into a ball	Rotates faster in midair	Same principle — no external torque during flight
Cat falling upside-down	Twists body to land on feet	Changes $I$ of different body parts to rotate without external torque
Planet orbiting the Sun	Speeds up when closer, slows down when farther	Gravitational force is central (torque = 0), so $L$ is constant

Watch out

Angular momentum is a vector. Its direction matters too. If no external torque acts, the axis of rotation stays fixed in space. This is why a spinning gyroscope or a bicycle wheel resists being tilted.

Common Mistake to Avoid …

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For a planet in elliptical orbit, areal velocity (area swept per unit time) is constant due to conservation of angular momentum. The graph of areal velocity vs. time for Mars is a horizontal straight line.

The Concept: Why Areal Velocity is Constant

The key idea here is Kepler's Second Law — often called the Law of Equal Areas. It states that a line joining a planet and the Sun sweeps out equal areas in equal intervals of time. This isn't a guess; it's a direct consequence of conservation of angular momentum.

Think about it: as Mars moves in its elliptical orbit, it speeds up when closer to the Sun (perihelion) and slows down when farther away (aphelion). The areal velocity — the area swept per unit time — stays exactly the same throughout the orbit. Why? Because the Sun's gravitational force is always directed toward the Sun (a central force), so it exerts zero torque on the planet. No torque means angular momentum is conserved, and constant angular momentum implies constant areal velocity.

\frac{dA}{dt} = \frac{L}{2m} = \text{constant}

Drawing the Graph

Identify the quantity on each axis. The question asks for areal velocity (on the y-axis) versus time (on the x-axis). Areal velocity is the rate at which area is swept, so its SI unit would be $\text{m}^2/\text{s}$ .
Recognize the constancy. Since areal velocity doesn't change during the orbit, its value is the same at every instant — whether Mars is at perihelion, aphelion, or anywhere in between.
Plot the graph. On a graph with time $t$ on the horizontal axis and areal velocity $\frac{dA}{dt}$ on the vertical axis, you draw a horizontal straight line parallel to the time axis. The height of this line depends on Mars's specific angular momentum and mass, but the shape is always a flat line. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
$4\times10^{16}\ m^2s^{-1}$
›Reveal solutionSolution
Tests Kepler's second law (constant areal velocity) and the fact that speed is minimum at the farthest point (aphelion). Answer: $20\ km/s$ .

Concept and Intuition

For a planet orbiting under a central (gravitational) force, angular momentum about the sun is conserved, which is equivalent to saying the line joining planet and sun sweeps out equal areas in equal times — Kepler's Second Law. Mathematically, $\dfrac{dA}{dt} = \dfrac{L}{2m}$ , a constant for the whole orbit.

At the two special points of the elliptical orbit — perihelion (closest) and aphelion (farthest) — the planet's velocity vector is exactly perpendicular to the radius vector (the radial component of velocity is momentarily zero there, since $r$ is at a turning point). At these two points only, the instantaneous rate of area sweep simplifies exactly to $\dfrac{dA}{dt} = \dfrac{1}{2} r v$ , with no need for a general dot-product treatment. Since $r$ and $v$ are inversely related ( $rv = \text{const}$ at the apsides, from angular momentum conservation), the minimum speed occurs at the maximum distance (aphelion) — a very intuitive coupling of "far means slow, near means fast."

Step-by-Step Solution

Areal velocity is constant: $\dfrac{dA}{dt} = 4\times10^{16}\ m^2s^{-1}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$(\hat{i} + 2\hat{j})$
›Reveal solutionSolution
Computing $\vec L=\vec r\times\vec p$ and reading off the z-component gives $L_z=-9\ \mathrm{kgm^2s^{-1}}$ .

Concept and Intuition

Angular momentum about the origin is $\vec L = \vec r\times\vec p$ , where $\vec p = m\vec v$ . We only need the z-component of the resulting cross product, which (using the standard determinant expansion) comes only from the x- and y- components of $\vec r$ and $\vec p$ (the z-components of $\vec r$ and $\vec p$ do not affect $L_z$ ).

Step-by-Step Solution

Momentum: $\vec p = m\vec v = 3\,(2\hat i+\hat j+2\hat k) = 6\hat i+3\hat j+6\hat k$ , i.e. $(p_x,p_y,p_z)=(6,3,6)$ .

Position vector: $\vec r = \hat i+2\hat j$ , i.e. $(r_x,r_y,r_z)=(1,2,0)$ .

Compute $\vec L=\vec r\times\vec p$ via the determinant:

$\vec L = \begin{vmatrix}\hat i&\hat j&\hat k\\ 1&2&0\\ 6&3&6\end{vmatrix} = \hat i(2\cdot6-0\cdot3)-\hat j(1\cdot6-0\cdot6)+\hat k(1\cdot3-2\cdot6)$
$=\hat i(12)-\hat j(6)+\hat k(3-12) = 12\hat i-6\hat j-9\hat k$ …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$Q. The figure shows the revolution of the planet (P) around the sun (S) in elliptical orbit. Which of the following statements is true? [FIGURE] (an ellipse with the sun S at a point left of centre; the ellipse is divided by two perpendicular diameters into quadrant points B at top, D at bottom, A at left end, C at right end; the planet P is shown between D and C moving anticlockwise toward C) (A) Time taken by the planet to travel BAD is less than that for DCB (B) Time taken by the planet to travel BAD is greater than that for DCB (C) Time taken by the planet to travel ADC is less than that for CBA (D) Time taken by the planet to travel ADC is greater than that for CBA$
›Reveal solutionSolution
Because the sun sits at a focus offset toward $A$ , the area swept by the sun–planet line over arc $BAD$ is less than over arc $DCB$ ; by Kepler's equal-area law this means BAD takes less time than DCB.

Concept and Intuition

Kepler's second law says the line joining the sun to the planet sweeps out equal areas in equal times — this is a direct consequence of angular momentum conservation about the focus. The chord $BD$ (the minor axis) passes through the geometric centre $O$ of the ellipse, so it splits the ellipse into two geometrically equal halves. But the area swept from the focus $S$ is not the same as the geometric half-ellipse area, because $S$ is not at $O$ — it's offset toward $A$ .

Step-by-Step Solution

Set up coordinates: ellipse centred at $O$ , major axis along $A$ – $C$ , minor axis along $B$ – $D$ (through $O$ ). Focus $S$ lies on the major axis, between $O$ and $A$ (closer to $A$ , consistent with $A$ being the near/perihelion point).

The chord $BD$ divides the ellipse into two geometrically equal halves (each area $= \tfrac12\pi ab$ ), since any line through the centre bisects an ellipse's area.

However, the area swept by the radius vector from $S$ as the planet moves along arc $B\to A\to D$ is bounded not by the chord $BD$ but by the two segments $SB$ and $SD$ . Since $S$ lies inside the $A$ -side half (between $O$ and $A$ , i.e. on the same side as $A$ relative to chord $BD$ ), the swept region is the half-ellipse area minus the triangle $DSB$ : $\text{Area}_{swept}(BAD) = \tfrac12\pi ab - T$ , where $T=\text{area}(\triangle DSB) > 0$ . …
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AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
$(2\hat{i}+3\hat{j}+4\hat{k})\ \mathrm{N}$
›Reveal solutionSolution
Constant angular momentum $\Rightarrow$ zero torque $\Rightarrow$ $\vec r\times\vec F=0$ ; solving component-wise gives $b=9$ .

Concept and Intuition

Angular momentum $\vec L=\vec r\times\vec p$ changes according to $\dfrac{d\vec L}{dt}=\vec\tau=\vec r\times\vec F$ . If $\vec L$ is constant, the torque about the origin due to the applied force must be zero. This happens precisely when $\vec r$ and $\vec F$ are parallel (or anti-parallel), i.e. the force line passes through the origin (a central force situation).

Step-by-Step Solution

Given $\vec F=2\hat i+3\hat j+4\hat k$ and $\vec r=6\hat i+b\hat j+12\hat k$ .

Torque $\vec\tau=\vec r\times\vec F=\begin{vmatrix}\hat i&\hat j&\hat k\\6&b&12\\2&3&4\end{vmatrix}$ .

$\hat i$ component: $b(4)-12(3)=4b-36$ .

$\hat j$ component: $-\big(6(4)-12(2)\big)=-(24-24)=0$ — automatically zero.

$\hat k$ component: $6(3)-b(2)=18-2b$ . …
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AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
$\left(\dfrac{m}{M+m}\right)v$
›Reveal solutionSolution
This is a combined linear + angular momentum conservation problem for a free rod struck at one end by a sticking particle; solving for the centre-of-mass velocity and the rotation about the new centre of mass, and adding their contributions at point $B$ , gives option (C).

Concept and Intuition

When an object is struck by another object and they stick together (perfectly inelastic collision) with no external force acting during the (very short) collision, two conservation laws apply for the combined system:

Linear momentum is conserved — this fixes the velocity of the new centre of mass.

Angular momentum about any fixed point in space is conserved — this fixes the angular velocity of the combined body about its own (new) centre of mass.

The velocity of any point on the rigid body afterward is then the vector sum of the centre-of-mass translational velocity and the rotational velocity ( $\vec\omega \times \vec r$ ) of that point relative to the new centre of mass.

Step-by-Step Solution

Centre of mass velocity. Before collision only the particle has momentum $mv$ (rod at rest). So

$V_{cm} = \frac{mv}{M+m}$

Location of the new centre of mass. Taking the rod's own centre $O$ ( $x=0$ ) and end $B$ at $x=L$ : the particle (mass $m$ ) sticks at $B$ , so

$d = \frac{mL}{M+m}\quad\text{(distance of new CM from }O\text{, towards }B\text{)}$
The distance from the new CM to $B$ is then $L-d = \dfrac{ML}{M+m}$ .

Moment of inertia about the new CM. The rod's own moment of inertia about $O$ is $\tfrac13 ML^2$ (rod of length $2L$ about its centre). Using the parallel-axis theorem for the rod's shift, plus the particle's contribution at its own distance from the new CM:

$I_{cm} = \frac13 ML^2 + Md^2 + m(L-d)^2 = ML^2\left(\frac13 + \frac{m}{M+m}\right)$
(the last two terms combine to the standard two-body reduced-mass form $\frac{Mm}{M+m}L^2$ ).

Angular momentum conservation. Taking angular momentum about the fixed point $O$ (the rod's original centre, an inertial point unaffected by the collision):

$L_{initial} = mvL \quad(\text{particle's angular momentum about }O)$
Equating this to the final angular momentum (rotation about new CM + orbital contribution of the CM's motion about $O$ ) and solving gives the angular speed:

$\omega = \frac{mv}{L(M+m)\left(\frac13+\frac{m}{M+m}\right)}$ …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$\frac{1}{64}$
›Reveal solutionSolution
Conservation of angular momentum with $R\to R/4$ (so $I\to I/16$ ) speeds up rotation 16×, cutting the day from 24 h to 1.5 h.

Concept and Intuition

This is exactly like a figure skater pulling in their arms: shrinking the radius reduces the moment of inertia, and with no external torque acting, angular momentum $L=I\omega$ stays constant, so $\omega$ must increase to compensate.

Step-by-Step Solution

Volume $V=\frac43\pi R^3$ . If $V_{new}=V_{old}/64$ , then $R_{new}^3=R_{old}^3/64\Rightarrow R_{new}=R_{old}/4$ .

Moment of inertia of a uniform sphere: $I=\frac25MR^2$ . Mass is unchanged, so $I_{new}/I_{old}=(R_{new}/R_{old})^2=(1/4)^2=1/16$ . …
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AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
$Q.A wheel is rotating freely at some angular speed. A second wheel initially at rest and with thrice the rotational inertia of the first, is suddenly coupled to the first wheel. The fraction of the original rotational kinetic energy lost is (A) 0.50 (B) 0.25 (C) 0.66 (D) 0.75$
›Reveal solutionSolution
Coupling conserves angular momentum but not KE; working through it, $3/4$ of the original rotational KE is lost.

Concept and Intuition

When two rotating bodies are suddenly coupled (like an inelastic collision but rotational), angular momentum is conserved because the coupling forces are internal, but kinetic energy is not conserved — some is dissipated as heat/sound in the coupling process, just like inelastic linear collisions.

Step-by-Step Solution

Let first wheel: inertia $I_1=I$ , initial angular speed $\omega$ . Second wheel: inertia $I_2=3I$ , at rest.

Conservation of angular momentum: $I\omega+3I(0)=(I+3I)\omega'\Rightarrow \omega'=\dfrac{I\omega}{4I}=\dfrac{\omega}{4}$ .

Initial KE: $KE_i=\dfrac12 I\omega^2$ .

Final KE: $KE_f=\dfrac12(4I)\left(\dfrac{\omega}{4}\right)^2=\dfrac12(4I)\dfrac{\omega^2}{16}=\dfrac{I\omega^2}{8}$ . …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$Q. Assertion (A): Angular speed, linear speed as Kinetic energy change with time but angular momentum remains constant for a planet orbiting the sun. Reason (R): Angular momentum is constant as no torque acts on the planet. (A) Both A and R are true and R is a correct explanation for A (B) Both A and R are true but R is not a correct explanation for A (C) A is true, R is false (D) A is false, R is false$
›Reveal solutionSolution
A central force exerts no torque about the force centre, so angular momentum is conserved even as speed and KE vary — both statements are true and R explains A.

Concept and Intuition

Gravitational force on a planet is always directed toward the Sun (a central force). Torque $\vec\tau = \vec r \times \vec F$ vanishes whenever $\vec F$ is parallel (or anti-parallel) to $\vec r$ , which is exactly the case here. Since $\vec\tau = d\vec L/dt$ , zero torque means angular momentum $\vec L$ is strictly constant — this is precisely why planets sweep equal areas in equal times (Kepler's 2nd law), speeding up near perihelion and slowing down near aphelion, so their linear/angular speed and KE do change even though $L$ doesn't.

Step-by-Step Solution

Assertion states: angular speed, linear speed, KE change with time, but angular momentum is constant. This is a correct description of planetary motion (true, due to varying orbital radius in an ellipse).

Reason states: angular momentum is constant because no torque acts on the planet. This is the correct physical cause (true). …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
$\frac{M}{A}$
›Reveal solutionSolution
Areal velocity of a planet is directly related to its angular momentum by $A = L/(2M)$ — this relation is Kepler's second law written in terms of angular momentum.

Concept and Intuition

Kepler's second law says a planet sweeps out equal areas in equal times. This is really a statement about conservation of angular momentum: for a particle at position $\vec r$ moving with momentum $\vec p = m\vec v$ , the infinitesimal area swept in time $dt$ is $dA = \tfrac12 |\vec r \times \vec v|\,dt$ , and $|\vec r \times m\vec v| = L$ is exactly the angular momentum. So the constant areal velocity is nothing but angular momentum divided by twice the mass.

Step-by-Step Solution

Areal velocity: $A = \dfrac{dA_{swept}}{dt} = \dfrac12 |\vec r \times \vec v|$ .

Angular momentum: $L = |\vec r \times \vec p| = M|\vec r \times \vec v|$ (mass of the planet is $M$ ).

Comparing the two: $|\vec r \times \vec v| = \dfrac{L}{M}$ , so …
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←Previous questionQ18 — Give one example each of central force and non-central force.→Next questionQ20 — What is the direction of areal velocity of the earth around the sun?

Q.Draw areal velocity versus time graph for mars.

The Concept: Why Areal Velocity is Constant

Drawing the Graph

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