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Ask a class whether a ball rolling in a straight line has angular momentum, and most will say no, because nothing is spinning. The answer is yes, about almost every point. This idea sits underneath collisions with rods, satellites, and particles striking discs in JEE problems. This post shows how to calculate it, why it stays constant, and the misunderstandings that research has documented.
What does the syllabus expect?
Angular momentum and its conservation appear in the rotational motion unit of the JEE (Main) 2026 syllabus, the NEET (UG) 2026 syllabus and the JEE (Advanced) 2026 brochure, which also lists "collision of point masses with rigid bodies". The NCERT Class 11 chapter on Systems of Particles and Rotational Motion defines it for a single particle as l = r × p.
Why do students say it is zero?
Close and Heron (2011) asked students in introductory calculus-based physics about the angular momentum of a particle moving in a straight line. Saying it was identically zero was widespread, and few students applied l = r × p correctly. Their errors came from confusing angular momentum with angular velocity or with linear momentum. Many also assumed that linear and angular momentum are always conserved together. The authors found that helping students see why a straight-moving particle has angular momentum worked better than drilling the formula alone.
How do you calculate it?
Choose a reference point O. The magnitude of r × p is mvr sin θ, where θ is the angle between the position vector and the velocity. But r sin θ is just the perpendicular distance d from O to the line along which the particle moves. So:
L = m v d
The direction is perpendicular to the plane containing O and the line of motion, given by the right-hand rule.
Why is it constant?
NCERT's Example 6.6 proves it. As a particle with constant velocity moves, r and θ both change, but r sin θ = d stays fixed because the line of motion does not move. The direction of r × v also stays the same. So L is constant in size and direction. This matches the rule that L changes only when there is a torque; with no force, there is no torque about any point.
Worked example 1: a particle on a horizontal line
A 2.0 kg particle moves in the +x direction at 3.0 m/s along the line y = 4.0 m. Find its angular momentum about the origin.
Perpendicular distance from O to the line y = 4.0 m is d = 4.0 m.
L = 2.0 × 3.0 × 4.0 = 24 kg m²/s.
Direction: with r = (x, 4, 0) and p = (6, 0, 0), the z-component of r × p is x(0) − 4(6) = −24. So L = −24 k̂ kg m²/s, clockwise when viewed from +z.
The answer does not depend on x, so it is the same at every instant.
Change the reference point. About (0, 1.0 m), d = 3.0 m, so L = 18 kg m²/s. About any point on the line y = 4.0 m, d = 0 and L = 0. Angular momentum always belongs to a chosen point, so state which one.
Worked example 2: a projectile, where L is not constant
A 0.20 kg ball is launched from the origin at 20 m/s and 60° above the horizontal. Take g = 10 m/s². Find its angular momentum about the launch point at t = 1.0 s and t = 2.0 s.
Horizontal velocity is constant: vₓ = 20 cos 60° = 10 m/s.
Working out r × p for x = vₓt and y = v_y t − ½gt² gives L = ½ m g vₓ t², directed into the page for a ball moving left to right.
½ × 0.20 × 10 × 10 = 10, so L = 10t² kg m²/s.
At t = 1.0 s, L = 10 kg m²/s. At t = 2.0 s, L = 40 kg m²/s.
Check with torque. Gravity mg acts at horizontal distance x = vₓt from O, so τ = m g vₓ t = 0.20 × 10 × 10 × t = 20t N m. Differentiating 10t² gives 20t. The two agree. The ball is still in flight at 2.0 s, since its time of flight is 2 × 20 sin 60° / 10 ≈ 3.5 s.
Which mistakes cost marks?
"It is not rotating, so L = 0." Rotation of the body is not needed; only a lever arm about the reference point.
Using r instead of d. The full distance r changes with time; the perpendicular distance does not.
Forgetting the reference point. L about the origin and L about the centre of a rod are different numbers.
Assuming p and L are conserved together. A particle tied to a string and moving in a circle about the fixed point has changing p but constant L about that point.
These ideas set up the collision problems in our post on conservation of angular momentum, where a particle strikes a rod or a disc.
A practice routine
For five straight-line motions, find L about three different points each, with directions.
Redo the projectile example with a 45° launch and check it using torque.
Solve two problems where a particle hits a rod pivoted at one end, writing L of the particle about the pivot first.
Explain to a friend, in two sentences, why a ball moving straight past you has angular momentum about you.
If you get stuck on the vector product or the right-hand rule, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Try this tonight: find L about the origin for a 0.5 kg particle moving at 4 m/s along the line x = 3 m.
