Concept Card: Angular Momentum & Fundamental Relation with Torque
1. Physical Framework & Definitions:
Angular momentum ($\vec{L}$), also termed the moment of linear momentum, quantifies the rotational inertia and momentum of a particle or rigid body about a specified reference point or rotation axis.
- Vector Cross-Product Definition: For a particle of mass $m$ with position vector $\vec{r}$ relative to an origin $O$ moving with linear momentum $\vec{p} = m\vec{v}$:
$$\mathbf{\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})}$$ - Vector Classification: An axial vector (pseudo-vector) perpendicular to the instantaneous plane containing $\vec{r}$ and $\vec{v}$, governed by the Right-Hand Rule.
- SI Unit & Dimensions: $\text{J}\cdot\text{s} = \text{kg}\cdot\text{m}^2/\text{s}$. Dimensional formula: $[M L^2 T^{-1}]$ (identically matching the dimensions of Planck’s constant $h$).
- Scalar Magnitude Methods:
- Angle Formulation: $L = r p \sin\theta = m v r \sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{v}$.
- Lever Arm Method ($r_{\perp}$): $\mathbf{L = p \cdot r_{\perp} = m v \cdot r_{\perp}}$, where $r_{\perp} = r\sin\theta$ is the perpendicular distance from the origin to the line of velocity.
- Transverse Momentum Method ($p_{\perp}$): $\mathbf{L = r \cdot p_{\perp} = m r v_{\perp} = m r^2 \omega}$.
2. Fundamental Dynamical Relation with Torque ($\vec{\tau} = \frac{d\vec{L}}{dt}$)
Differentiating the angular momentum vector with respect to time:
$$\frac{d\vec{L}}{dt} = \frac{d}{dt}(\vec{r} \times \vec{p}) = \left(\frac{d\vec{r}}{dt} \times \vec{p}\right) + \left(\vec{r} \times \frac{d\vec{p}}{dt}\right)$$
- Since $\frac{d\vec{r}}{dt} = \vec{v}$ and $\vec{p} = m\vec{v}$, the first term vanishes: $\vec{v} \times (m\vec{v}) = \vec{0}$.
- Since $\frac{d\vec{p}}{dt} = \Sigma \vec{F}_{\text{net}}$ (Newton’s Second Law):
$$\mathbf{\vec{\tau}_{\text{net}} = \frac{d\vec{L}}{dt}}$$ - For a System of Particles: Internal mutual central forces cancel pairwise in torque ($\Sigma \vec{\tau}_{\text{int}} = \vec{0}$):
$$\mathbf{\vec{\tau}_{\text{ext}} = \frac{d\vec{L}}{dt}}$$ - For Fixed-Axis Rotation with Constant $I$:
$$\vec{\tau}_{\text{ext}} = \frac{d}{dt}(I\vec{\omega}) = I \frac{d\vec{\omega}}{dt} = \mathbf{I \vec{\alpha}}$$
3. König’s Theorem for Angular Momentum (Rolling & Combined Motion)
For an extended rigid body executing general planar motion:
$$\mathbf{\vec{L}_O = \vec{L}_{\text{cm}} + \vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}}}$$
- $\vec{L}_{\text{cm}} = I_{\text{cm}}\vec{\omega}$ is the spin angular momentum (intrinsic rotation about the centre of mass).
- $\vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}}$ is the orbital angular momentum (motion of the centre of mass treated as a point mass about origin $O$).
4. Angular Momentum of a Particle Moving in a Straight Line
If a particle moves with constant velocity $\vec{v}$ along a straight line at a perpendicular distance $d$ from point $O$:
$$L = m v d = \text{constant!}$$
Key Takeaway: A particle does not need to move in a circle or curved trajectory to possess angular momentum. Straight-line motion possesses constant non-zero angular momentum about any reference point not lying on the trajectory line.
5. Common JEE Pitfalls & Traps
- Trap 1 (Reference Point Dependency): Angular momentum $\vec{L}$ is strictly relative to a specified reference point. Writing $\vec{L}$ without specifying the reference point is meaningless.
- Trap 2 (Valid Points for $\vec{\tau} = \frac{d\vec{L}}{dt}$): The rotational second law holds strictly true ONLY if computed about:
- A fixed point in an inertial reference frame, OR
- The centre of mass of the system (even if accelerating!), OR
- A point whose acceleration vector is directed towards/away from the centre of mass.
- Trap 3 (Direction in Vector Addition for Rolling Bodies): In $\vec{L}_O = \vec{L}_{\text{cm}} + \vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}}$, pay close attention to signs. About a ground point, spin and orbital terms add in the same direction ($-\hat{k}$); about an elevated point above the body, they subtract!
Solved Examples
Example 1 (Direct Conceptual Application – 3D Vector Calculus & Torque Verification):
A particle of mass $m = 2.0\text{ kg}$ moves in the $xy$-plane with position vector $\vec{r}(t) = (3.0 t^2\hat{i} + 4.0 t\hat{j})\text{ m}$.
(a) Find the velocity vector $\vec{v}(t)$ and linear momentum $\vec{p}(t)$.
(b) Determine the angular momentum vector $\vec{L}(t)$ of the particle about the origin $O$.
(c) Find the net force $\vec{F}(t)$ and torque $\vec{\tau}(t)$ about the origin, and verify explicitly that $\vec{\tau} = \frac{d\vec{L}}{dt}$.
Solution:
(a) Differentiating position:
$$\vec{v}(t) = \frac{d\vec{r}}{dt} = (6.0 t\hat{i} + 4.0\hat{j})\text{ m/s}$$
Linear momentum:
$$\vec{p}(t) = m\vec{v}(t) = 2.0(6.0 t\hat{i} + 4.0\hat{j}) = (12.0 t\hat{i} + 8.0\hat{j})\text{ kg}\cdot\text{m/s}$$
(b) Angular momentum about origin $O$:
$$\vec{L}(t) = \vec{r}(t) \times \vec{p}(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3.0 t^2 & 4.0 t & 0 \\ 12.0 t & 8.0 & 0 \end{vmatrix} = [ (3.0 t^2)(8.0) – (4.0 t)(12.0 t) ]\hat{k} = \mathbf{-24.0 t^2\hat{k}\text{ J}\cdot\text{s}}$$
(c) Acceleration and net force:
$$\vec{a}(t) = \frac{d\vec{v}}{dt} = 6.0\hat{i}\text{ m/s}^2 \implies \vec{F}(t) = m\vec{a}(t) = 2.0(6.0\hat{i}) = 12.0\hat{i}\text{ N}$$
Torque about the origin:
$$\vec{\tau}(t) = \vec{r}(t) \times \vec{F}(t) = (3.0 t^2\hat{i} + 4.0 t\hat{j}) \times (12.0\hat{i}) = 48.0 t(\hat{j} \times \hat{i}) = \mathbf{-48.0 t\hat{k}\text{ N}\cdot\text{m}}$$
Differentiating $\vec{L}(t)$:
$$\frac{d\vec{L}}{dt} = \frac{d}{dt}(-24.0 t^2\hat{k}) = -48.0 t\hat{k}\text{ N}\cdot\text{m} = \vec{\tau}(t)$$
Verified!
Example 2 (Mathematical Derivation – Angular Momentum of a Projectile):
A projectile of mass $m$ is launched from origin $O$ with initial speed $u$ at an angle $\theta$ to the horizontal.
(a) Derive the angular momentum $\vec{L}(t)$ of the projectile about the launch point $O$ as a function of time.
(b) Evaluate $\vec{L}$ at the highest point (apex) of the trajectory.
(c) Evaluate $\vec{L}$ just before landing, and verify that $\vec{\tau} = \frac{d\vec{L}}{dt}$.
Solution:
(a) Position and momentum vectors:
$$\vec{r}(t) = (u\cos\theta t)\hat{i} + \left(u\sin\theta t – \frac{1}{2}gt^2\right)\hat{j}$$
$$\vec{p}(t) = m(u\cos\theta)\hat{i} + m(u\sin\theta – gt)\hat{j}$$
Cross product $\vec{r} \times \vec{p} = (x p_y – y p_x)\hat{k}$:
$$x p_y = m u^2 \sin\theta\cos\theta t – m g u\cos\theta t^2$$
$$y p_x = m u^2 \sin\theta\cos\theta t – \frac{1}{2}m g u\cos\theta t^2$$
Subtracting yields:
$$\mathbf{\vec{L}(t) = -\frac{1}{2}m g u\cos\theta\,t^2\hat{k}}$$
(b) At the apex, $t_{\text{apex}} = \frac{u\sin\theta}{g}$:
$$\vec{L}_{\text{apex}} = -\frac{1}{2}m g u\cos\theta \left(\frac{u\sin\theta}{g}\right)^2\hat{k} = \mathbf{-\frac{m u^3 \sin^2\theta\cos\theta}{2g}\hat{k}}$$
(c) At the landing point, $T = \frac{2u\sin\theta}{g}$:
$$\vec{L}_{\text{landing}} = -\frac{1}{2}m g u\cos\theta \left(\frac{2u\sin\theta}{g}\right)^2\hat{k} = \mathbf{-\frac{2 m u^3 \sin^2\theta\cos\theta}{g}\hat{k}} = 4\vec{L}_{\text{apex}}$$
*Torque verification:* $\vec{\tau}(t) = \vec{r} \times (-m g\hat{j}) = -m g(u\cos\theta t)\hat{k} = \frac{d\vec{L}}{dt}$.
Example 3 (Standard JEE Advanced Scenario – Combined Translation & Rotation of a Rolling Sphere):
A uniform solid sphere of mass $M = 4.0\text{ kg}$ and radius $R = 0.50\text{ m}$ rolls without slipping on a horizontal floor with linear speed $v_{\text{cm}} = 6.0\text{ m/s}$ along $+x$ ($I_{\text{cm}} = \frac{2}{5}M R^2$).
(a) Find the angular velocity $\vec{\omega}$ of the sphere.
(b) Calculate the spin angular momentum $\vec{L}_{\text{cm}}$ of the sphere about its centre of mass.
(c) Calculate the total angular momentum $\vec{L}_O$ about a fixed point $O$ on the floor.
(d) Calculate the total angular momentum $\vec{L}_P$ about a point $P$ fixed at height $h = 2R = 1.0\text{ m}$ directly above the floor.
Solution:
(a) Pure rolling without slipping: $\omega = \frac{v_{\text{cm}}}{R} = \frac{6.0}{0.50} = 12.0\text{ rad/s}$. Rolling clockwise: $\vec{\omega} = -12.0\hat{k}\text{ rad/s}$.
(b) Spin angular momentum: $I_{\text{cm}} = \frac{2}{5}M R^2 = \frac{2}{5}(4.0)(0.25) = 0.40\text{ kg}\cdot\text{m}^2$.
$$\vec{L}_{\text{cm}} = I_{\text{cm}}\vec{\omega} = (0.40)(-12.0\hat{k}) = \mathbf{-4.80\hat{k}\text{ J}\cdot\text{s}}$$
(c) Total angular momentum about floor point $O$:
$$\vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}} = (x\hat{i} + R\hat{j}) \times (M v_{\text{cm}}\hat{i}) = -M v_{\text{cm}} R\hat{k} = -(4.0)(6.0)(0.50)\hat{k} = -12.0\hat{k}\text{ J}\cdot\text{s}$$
$$\vec{L}_O = \vec{L}_{\text{cm}} + \vec{r}_{\text{cm}} \times M\vec{v}_{\text{cm}} = -4.80\hat{k} – 12.0\hat{k} = \mathbf{-16.80\hat{k}\text{ J}\cdot\text{s}}$$
(d) Total angular momentum about top point $P$ ($y_P = 2R$):
The position of the CM relative to $P$ is $\vec{r}’_{\text{cm}} = x’\hat{i} – R\hat{j}$.
$$\vec{r}’_{\text{cm}} \times M\vec{v}_{\text{cm}} = (x’\hat{i} – R\hat{j}) \times (M v_{\text{cm}}\hat{i}) = +M v_{\text{cm}} R\hat{k} = +12.0\hat{k}\text{ J}\cdot\text{s}$$
$$\vec{L}_P = \vec{L}_{\text{cm}} + \vec{r}’_{\text{cm}} \times M\vec{v}_{\text{cm}} = -4.80\hat{k} + 12.0\hat{k} = \mathbf{+7.20\hat{k}\text{ J}\cdot\text{s}}$$
Example 4 (Edge Case – Conical Pendulum: Constant vs. Precessing Angular Momentum):
A bob of mass $m = 0.20\text{ kg}$ is attached to a light cord of length $L = 1.0\text{ m}$ revolving in a horizontal circle as a conical pendulum with semi-vertical cone angle $\theta = 30^\circ$. Take $g = 10.0\text{ m/s}^2$.
(a) Find the orbital angular velocity $\omega$ and linear speed $v$ of the bob.
(b) Determine the angular momentum $\vec{L}_C$ about the center $C$ of the circular orbit.
(c) Determine the angular momentum $\vec{L}_O$ about the suspension apex $O$, and explain why $\vec{L}_O$ precesses.
Solution:
(a) Radius $r = L\sin 30^\circ = 0.50\text{ m}$. Forces: $T\cos 30^\circ = mg$ and $T\sin 30^\circ = m\omega^2 r$.
$$\omega = \sqrt{\frac{g\tan 30^\circ}{r}} = \sqrt{\frac{10.0 \times \frac{1}{\sqrt{3}}}{0.50}} = \sqrt{\frac{20.0}{\sqrt{3}}} \approx 3.398\text{ rad/s}$$
Linear speed: $v = \omega r = (3.398)(0.50) \approx 1.699\text{ m/s}$.
(b) About orbit center $C$: $\vec{r}_C \perp \vec{v}$ in the horizontal plane.
$$L_C = m v r = m \omega r^2 = (0.20)(3.398)(0.25) \approx \mathbf{0.170\text{ J}\cdot\text{s}}$$
Directed vertically upward ($+\hat{k}$) at all times; strictly constant in magnitude and direction.
(c) About apex $O$: String vector $\vec{r}_O$ has length $L = 1.0\text{ m}$ and is perpendicular to tangent velocity $\vec{v}$.
$$|\vec{L}_O| = m v L = (0.20)(1.699)(1.0) \approx \mathbf{0.340\text{ J}\cdot\text{s}}$$
$\vec{L}_O$ is perpendicular to the string, inclined at $60^\circ$ to the vertical. As the bob circles, $\vec{L}_O$ precesses about the vertical axis. Gravity exerts a horizontal torque about $O$ of magnitude $\tau = mg(L\sin 30^\circ)$, which continually rotates the direction of $\vec{L}_O$ ($\vec{\tau} = \frac{d\vec{L}_O}{dt}$).
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle of mass $m$ moves with constant speed $v$ along a straight line $y = b$ parallel to the $x$-axis. The magnitude of its angular momentum about the origin is:
(A) Zero
(B) $m v b$
(C) $m v x$
(D) $m v \sqrt{x^2 + b^2}$
Problem 2 (JEE Main – Single Correct):
The physical quantity that has the same dimensions as angular momentum is:
(A) Work
(B) Linear momentum
(C) Planck’s constant
(D) Power
Problem 3 (JEE Main – Single Correct):
A particle of mass $m = 1.0\text{ kg}$ is located at position $\vec{r} = (2.0\hat{i} – 5.0\hat{j})\text{ m}$ with velocity $\vec{v} = (3.0\hat{i} + 4.0\hat{j})\text{ m/s}$. The angular momentum of the particle about the origin is:
(A) $+23.0\hat{k}\text{ J}\cdot\text{s}$
(B) $-23.0\hat{k}\text{ J}\cdot\text{s}$
(C) $+7.0\hat{k}\text{ J}\cdot\text{s}$
(D) $-7.0\hat{k}\text{ J}\cdot\text{s}$
Problem 4 (JEE Main – Single Correct):
A constant torque of magnitude $\tau = 20.0\text{ N}\cdot\text{m}$ acts on a body rotating about a fixed axis for a time interval of $\Delta t = 3.0\text{ s}$. The change in the angular momentum of the body is:
(A) $20.0\text{ J}\cdot\text{s}$
(B) $60.0\text{ J}\cdot\text{s}$
(C) $180.0\text{ J}\cdot\text{s}$
(D) $6.67\text{ J}\cdot\text{s}$
Problem 5 (JEE Main – Single Correct):
For a particle moving under a central force field (where $\vec{F} = f(r)\hat{r}$ is directed always toward or away from origin $O$):
(A) Linear momentum is conserved
(B) Angular momentum about $O$ is strictly conserved
(C) Kinetic energy is strictly zero
(D) Linear acceleration is zero
Problem 6 (JEE Advanced – One or More Correct):
For a particle moving in the $xy$-plane, which of the following statements is/are correct?
(A) The relation $\vec{\tau} = \frac{d\vec{L}}{dt}$ holds strictly if both torque and angular momentum are evaluated about the same fixed point in an inertial frame.
(B) If a particle moves in a straight line with constant speed, its angular momentum about any point lying on that straight line is zero.
(C) If the net external torque about a fixed axis is zero, the component of angular momentum along that axis is conserved.
(D) For a rigid body rotating about an axis of symmetry, the angular momentum vector $\vec{L}$ is strictly parallel to the angular velocity vector $\vec{\omega}$.
Problem 7 (JEE Advanced – One or More Correct):
A projectile of mass $m$ is launched from the origin with initial velocity $u$ at an angle $\theta$ to the horizontal:
(A) The angular momentum of the projectile about the launch origin is zero at $t = 0$.
(B) The magnitude of the angular momentum about the launch origin increases proportionally to $t^2$.
(C) The magnitude of the torque of gravity about the launch origin increases proportionally to $t$.
(D) The angular momentum about the apex of the trajectory is zero at the instant the projectile is at the apex.
Problem 8 (JEE Advanced – One or More Correct):
A uniform circular disc of mass $M$ and radius $R$ rolls without slipping with constant forward speed $v$ along a horizontal ground:
(A) The spin angular momentum about its centre of mass has magnitude $\frac{1}{2}M v R$.
(B) Its total angular momentum about the instantaneous point of contact on the ground has magnitude $\frac{3}{2}M v R$.
(C) Its total angular momentum about any fixed point lying on the ground line is constant in time with magnitude $\frac{3}{2}M v R$.
(D) The net external torque acting on the disc about the instantaneous point of contact is zero.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A particle of mass $m = 0.50\text{ kg}$ is dropped from rest from point $(x = 4.0\text{ m}, y = 20.0\text{ m})$ at $t = 0$. Taking $g = 10.0\text{ m/s}^2$, find the magnitude of the angular momentum of the particle about the origin at $t = 2.0\text{ s}$ in $\text{J}\cdot\text{s}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A heavy flywheel of moment of inertia $I = 5.0\text{ kg}\cdot\text{m}^2$ is rotating about its fixed central axis with an initial angular speed of $\omega_0 = 40.0\text{ rad/s}$. A constant frictional braking torque of $\tau = 10.0\text{ N}\cdot\text{m}$ is applied. Calculate the time (in seconds) required for the flywheel to come to rest.
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (A) | 4. (B) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C, D) | 9. 40 | 10. 20
Solution 1:
$\vec{L} = \vec{r} \times \vec{p} = (x\hat{i} + b\hat{j}) \times (m v\hat{i}) = -m v b\hat{k}$. Magnitude is $m v b$, which is constant since lever arm $r_{\perp} = b$ is constant.
Correct Option: (B)
Solution 2:
$[L] = [M L^2 T^{-1}]$. From $E = h\nu$, $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$. Identical to Planck’s constant.
Correct Option: (C)
Solution 3:
$\vec{L} = (2\hat{i} – 5\hat{j}) \times 1.0(3\hat{i} + 4\hat{j}) = [2(4) – (-5)(3)]\hat{k} = (8 + 15)\hat{k} = +23.0\hat{k}\text{ J}\cdot\text{s}$.
Correct Option: (A)
Solution 4:
$\Delta L = \tau \Delta t = (20.0)(3.0) = 60.0\text{ J}\cdot\text{s}$.
Correct Option: (B)
Solution 5:
Central force has $\vec{r} \parallel \vec{F} \implies \vec{\tau} = \vec{r} \times \vec{F} = \vec{0} \implies \vec{L} = \text{constant}$.
Correct Option: (B)
Solution 6:
All statements (A, B, C, D) are verified theorems of rotational dynamics.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: At $t = 0$, $\vec{r} = \vec{0} \implies \vec{L} = \vec{0}$.
– (B) True: $\vec{L}(t) = -\frac{1}{2}m g u\cos\theta t^2\hat{k} \propto t^2$.
– (C) True: $\vec{\tau}(t) = -m g (u\cos\theta t)\hat{k} \propto t$.
– (D) True: At the apex, relative position vector from the apex is $\vec{0}$, so $\vec{L}_{\text{about apex}} = \vec{0}$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: $L_{\text{cm}} = I_{\text{cm}}\omega = \frac{1}{2}M R^2(v/R) = \frac{1}{2}M v R$.
– (B) True: $L_{\text{contact}} = I_{\text{contact}}\omega = \frac{3}{2}M R^2(v/R) = \frac{3}{2}M v R$.
– (C) True: $L_O = L_{\text{cm}} + R M v = \frac{3}{2}M v R$ (constant for any fixed ground point).
– (D) True: All forces pass through contact point or through CM vertically aligned with contact.
Correct Options: (A, B, C, D)
Solution 9:
$v = gt = (10.0)(2.0) = 20.0\text{ m/s}$ along vertical line $x = 4.0\text{ m}$. Lever arm from origin is $r_{\perp} = 4.0\text{ m}$.
$L = m v r_{\perp} = (0.50)(20.0)(4.0) = 40.0\text{ J}\cdot\text{s}$.
Correct Answer: 40
Solution 10:
$\alpha = \frac{\tau}{I} = \frac{10.0}{5.0} = 2.0\text{ rad/s}^2$.
$t = \frac{\omega_0}{\alpha} = \frac{40.0}{2.0} = 20.0\text{ seconds}$.
Correct Answer: 20