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Stand a metre stick on its end, keep the bottom from slipping, and let it topple. It looks like simple falling, but the stick is rotating about a fixed point, and parts of it end up accelerating faster than g. This problem appears in many forms in JEE Advanced practice: angular speed at an angle, speed of the tip, and the force on the hinge. This post solves all three with one set of numbers, explains the result that surprises students, and lists the usual mistakes.
Which syllabus covers this?
The JEE (Advanced) 2026 brochure lists "Dynamics of rigid bodies with fixed axis of rotation" and moments of inertia of uniform bodies. The JEE (Main) 2026 syllabus lists rigid body rotation, equations of rotational motion and the axis theorems in Unit 5. Check your exam year's bulletin.
Which moment of inertia do you need?
The rod turns about its end, not its centre. NCERT's Table 6.1 gives ML²/12 about the midpoint. The parallel axis theorem adds M(L/2)², giving ML²/3 about the end. OpenStax University Physics derives the same ML²/3 directly by integration. Our posts on the axis theorems and on moment of inertia by integration cover both routes.
How fast is the rod turning at angle θ?
Let θ be the angle from the vertical. The centre of mass drops by (L/2)(1 − cos θ). The hinge does no work, so energy is conserved:
Mg(L/2)(1 − cos θ) = ½(ML²/3)ω², which gives ω² = 3g(1 − cos θ)/L.
For angular acceleration, take torques about the hinge: Mg(L/2) sin θ = (ML²/3)α, so α = 3g sin θ/(2L).
Worked example. A uniform rod of mass 2.0 kg and length 1.2 m is hinged at its base and released from upright. Take g = 10 m/s². Find its motion when it reaches horizontal (θ = 90°).
Angular speed: ω² = 3 × 10 × 1/1.2 = 25, so ω = 5.0 rad/s.
Tip speed: ωL = 5.0 × 1.2 = 6.0 m/s. A ball dropped from 1.2 m lands at √(2 × 10 × 1.2) ≈ 4.9 m/s, so the tip arrives faster.
Angular acceleration: α = 3 × 10/(2 × 1.2) = 12.5 rad/s².
Tip tangential acceleration: αL = 12.5 × 1.2 = 15 m/s², which is 1.5g.
Why does the tip fall faster than g?
Gravity acts on the whole rod, but the rod is rigid, so the inner parts drag the outer parts along. At horizontal, a point at distance r from the hinge has tangential acceleration (3g/2L) × r. This equals g at r = 2L/3, which is 0.8 m here. Every point beyond it accelerates faster than a freely falling object. A coin resting near the tip would be left behind as the rod nears horizontal.
What force does the hinge exert?
The centre of mass moves on a circle of radius L/2, so it has two acceleration components. At horizontal:
Centripetal, towards the hinge: ω²(L/2) = 25 × 0.6 = 15 m/s².
Tangential, downward: α(L/2) = 12.5 × 0.6 = 7.5 m/s².
Vertical: Mg − N = M × 7.5, so N = 20 − 15 = 5 N upward, which is Mg/4. Horizontal: H = M × 15 = 30 N towards the hinge. The total hinge force is √(5² + 30²) ≈ 30.4 N, about 1.5 times the rod's weight, and mostly sideways.
The same steps at a general angle give a hinge force along the rod of Mg(3/2 − (5/2) cos θ), positive meaning a pull towards the hinge. It is zero when cos θ = 3/5, at about 53°: before that the hinge pushes the rod outward, after it the hinge pulls. The perpendicular component is Mg sin θ/4. At 60°, these give a 5.0 N pull and 4.33 N, a useful check.
Forces inside a falling rod matter too: tall chimneys can break in mid-air as they topple. Varieschi and Kamiya (2003) reviewed the theory of the falling chimney and tested it with toy models. They found ruptures can come from shear forces near the base or from bending, with bending breaks most likely between one-third and one-half of the height.
Which mistakes cost marks?
Using ML²/12 about the hinge. That is the value about the centre.
Using v = √(2gh) for the tip. The tip is not in free fall.
Setting the hinge force equal to Mg. It changes with angle and is not vertical.
Forgetting the centripetal part of the centre-of-mass acceleration.
A practice routine
Derive ω² = 3g(1 − cos θ)/L and α = 3g sin θ/2L from scratch.
Repeat the worked example at θ = 60° and check the hinge components above.
Find the angle at which the hinge force along the rod is zero, without looking.
Try the same rod with a small mass fixed to its tip and see how each answer changes.
When you know the answer but cannot see why a step follows, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Topple a ruler on your desk tonight, then calculate its tip speed at horizontal from its length.
