OlympIQ
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No External Torque Means Iω Stays Constant: Track What Happens to I

When no external torque acts, angular momentum L = I ω is conserved. Any change in the moment of inertia, from a person moving or a string being pulled, changes the angular speed in the opposite way, and the kinetic energy changes too.

NSEJS · Rotational Motion · 11 linked questions · 100% coverage
Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction
Study deeper: NCERT Science · Rotational Motion

Changing I changes ω

If the Earth lost its atmosphere suddenly its moment of inertia would drop, so ω would rise and the day would become shorter. An ant walking along a diameter of a rotating disc from one edge to the other first reduces I (so ω rises) while approaching the centre, and then raises I again (so ω falls). A person of 50 kg who steps from the centre of a turntable (I = 100 kg m², one turn in 2 s) to 2 m away makes I = 100 + 50 x 4 = 300, so ω falls from π to π/3 rad/s. A sphere whirled on a string, with the string pulled so that the radius halves, has I reduced to a quarter, so ω becomes four times, 3ω going to 12ω.

Energy and angular momentum relations

K = L²/(2I), so with L fixed, doubling the radius of a mass on a string multiplies I by 4 and cuts the kinetic energy to one quarter

A sphere whose radius grows freely in space keeps its angular momentum, though ω, I and the rotational energy all change.

L = sqrt(2 I K), so for equal kinetic energies the body with the larger moment of inertia has the larger angular momentum

A man walking around the edge of a freely turning platform makes the platform turn the opposite way so that the total angular momentum stays zero.

Bodies joining together

When a non-rotating disc of moment of inertia I2 is placed on a spinning disc of I1, ω' = I1 ω / (I1 + I2). For two discs of equal I spinning at ω1 and ω2 and then joined, ω' = (ω1 + ω2)/2, and the kinetic energy lost to the interaction is (1/4) I (ω1 - ω2)². Angular momentum is conserved in such collisions but kinetic energy is not.

OlympIQ exam lens

  • ✓No external torque: I1 ω1 = I2 ω2
  • ✓K = L²/(2I): at fixed L, a larger I means a smaller K; L = sqrt(2IK)
  • ✓Pulling a string through the centre keeps L constant and increases ω as r decreases
  • ✓Joining discs: ω' = (I1ω1 + I2ω2)/(I1 + I2); energy is lost
  • ✓A person walking on a free platform makes it counter-rotate
OlympIQ trap

Do not assume kinetic energy is conserved with angular momentum. When I changes, energy changes; when two discs join, energy is lost, so only L can be equated before and after.

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