“Same equations, new alphabet”
Rotational versions of the laws of motion
m becomes I, F becomes τ, v becomes ω, a becomes α, p becomes L. Every linear equation has an identical rotational partner.
Class 11 Physics · Gujarat Board, CBSE, GUJCET and JEE
Rotational Motion formulas
Rotational motion is linear motion with every symbol swapped: mass becomes moment of inertia, force becomes torque, velocity becomes angular velocity. Learn the translations and half the chapter is already done, because every equation you learned last month has a rotational twin.
20 formulas, 4 memory hooks, and the 5 mistakes that cost the most marks.
The translations
| What | Formula | Symbols and notes |
|---|---|---|
| Angular velocity | ω = dθ/dt = 2πf | measured in radian per second |
| Angular acceleration | α = dω/dt | the rotational twin of a |
| Equations of rotation | ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ | identical in shape to the SUVAT equations |
| Torque | τ = r × F = rF sinθ = Iα | the rotational twin of F = ma |
| Angular momentum | L = Iω = r × p | conserved when the net external torque is zero |
| Rotational kinetic energy | KE = ½Iω² | add it to ½mv² for a rolling body |
Moments of inertia
| What | Formula | Symbols and notes |
|---|---|---|
| Thin ring, about its axis | I = MR² | all the mass sits at distance R |
| Solid disc or cylinder | I = ½MR² | about the central axis |
| Solid sphere | I = ⅖MR² | about a diameter |
| Hollow sphere | I = ⅔MR² | about a diameter |
| Rod about its centre | I = ML²/12 | about a perpendicular axis |
| Rod about one end | I = ML²/3 | four times the value about the centre |
| Parallel axis theorem | I = Icm + Md² | d is the distance between the two parallel axes |
| Perpendicular axis theorem | Iz = Ix + Iy | for a flat lamina only |
Rolling
| What | Formula | Symbols and notes |
|---|---|---|
| Rolling without slipping | v = ωR | the contact point is momentarily at rest |
| Total kinetic energy | KE = ½mv²(1 + k²/R²) | k is the radius of gyration, I = mk² |
| Down an incline | a = g sinθ/(1 + k²/R²) | a ring is slowest, a solid sphere is fastest |
| Radius of gyration | k = √(I/m) | the distance at which all the mass could be placed |
Centre of mass
| What | Formula | Symbols and notes |
|---|---|---|
| Two particles | xcm = (m₁x₁ + m₂x₂)/(m₁ + m₂) | it lies nearer the heavier mass |
| Velocity of the centre of mass | vcm = (m₁v₁ + m₂v₂)/(m₁ + m₂) | unchanged by internal forces |
Taught by Darshit Bhatt, founder and lead faculty at Darshit Sir Institute. Instrumentation and Control engineer, with a master's degree and a chartered degree, teaching in Paldi, Ahmedabad since 2006.
Know the formula but not when to use it?
Darshit Sir has taught this in Paldi since 2006, in the classroom and live online.
Memory hooks
How to remember this chapter.
“Ring, disc, sphere: 1, ½, ⅖”
The three moments of inertia asked most
Ring MR², disc ½MR², solid sphere ⅖MR². The further the mass sits from the axis, the larger the value, which is why the ring is largest.
“Skater pulls arms in”
Conservation of angular momentum
Pulling the arms in reduces I, so ω rises to keep Iω constant. Kinetic energy rises too, paid for by the work the skater does.
“Fat rings lose races”
Rolling down an incline
The larger the k²/R², the slower the body. Order down a slope: solid sphere, then disc, then ring, and mass and radius make no difference at all.
Where marks go
The mistakes that cost the most.
- Using I = MR² for every round body. The shape decides the number in front, and that number decides who wins a rolling race.
- Applying the perpendicular axis theorem to a solid body. It works for flat laminas only.
- Forgetting the rotational kinetic energy for a rolling body. Its total energy is ½mv² plus ½Iω², not ½mv² alone.
- Assuming angular momentum is conserved when an external torque acts. Check the torque first, then conserve.
- Writing v = ωR when the body is slipping. That relation holds only for rolling without slipping.
Test yourself
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