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Circular Motion revision notes

Every idea in the Circular Motion chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 33 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Angular variables and centripetal acceleration

Angular variables
11× asked, last 2025
θ=sr,ω=dθdt=2πT=2πf,v=ωr\theta=\frac{s}{r},\qquad \omega=\frac{d\theta}{dt}=\frac{2\pi}{T}=2\pi f,\qquad v=\omega r
θ\theta in radians. NN rpm =2πN/60=2\pi N/60 rad/s. Every point of a rigid body has the same ω\omega.
  • Clock hands: second hand ω=2π/60\omega=2\pi/60 rad/s, minute hand 2π/36002\pi/3600 rad/s, hour hand 2π/432002\pi/43200 rad/s. Distance moved by a tip =rθ=r\theta, with θ=2π×\theta=2\pi\times turns.
  • Angles in degrees or arc seconds must be turned into radians before s=rθs=r\theta: 1∘=π/1801^\circ=\pi/180 rad and 1′′1'' is 1/36001/3600 of a degree.
  • Not rpm in place of rad/s. Multiply rpm by 2π/602\pi/60 (or rev/s by 2π2\pi) before using ω2r\omega^2 r. Leaving out 2π2\pi makes ω2r\omega^2 r about 40 times too small.
Angular velocity as a vector
v⃗=ω⃗×r⃗\vec v=\vec\omega\times\vec r
ω⃗\vec\omega lies along the axis by the right-hand rule; r⃗\vec r is drawn from any point on the axis.
Constant angular acceleration
2× asked, last 2021
ω=ω0+αt,θ=ω0t+12αt2,ω2=ω02+2αθ\omega=\omega_0+\alpha t,\qquad \theta=\omega_0t+\frac12\alpha t^2,\qquad \omega^2=\omega_0^2+2\alpha\theta
Only when α\alpha is constant. Turns made =θ/2π=\theta/2\pi; tangential acceleration at=αra_t=\alpha r.
  • Quick method. Turns while speeding up uniformly = average rate × time: n=(f1+f2)t/2n=(f_1+f_2)t/2. With rates in rpm, put tt in minutes. No need to find α\alpha.
Variable angular acceleration
1× asked, last 2026
ω=dθdt,α=dωdt=ω dωdθ\omega=\frac{d\theta}{dt},\qquad \alpha=\frac{d\omega}{dt}=\omega\,\frac{d\omega}{d\theta}
Use when α\alpha changes with time or angle. Going back: θ=∫ω dt\theta=\int\omega\,dt with the initial values.
  • Not the leading term alone. Differentiate every term of θ(t)\theta(t) with its sign, then put in the time. A negative term in θ\theta stays negative in ω\omega and α\alpha.
Displacement and average velocity
2× asked, last 2023
∣Δr⃗∣=2Rsin⁡θ2,∣v⃗avg∣v=sin⁡(θ/2)θ/2|\Delta\vec r|=2R\sin\frac{\theta}{2},\qquad \frac{|\vec v_{\text{avg}}|}{v}=\frac{\sin(\theta/2)}{\theta/2}
Constant speed vv, turned through θ\theta in radians. Distance travelled =Rθ=R\theta, the arc.
  • Not the arc length. Displacement is the straight chord 2Rsin⁡(θ/2)2R\sin(\theta/2), and average velocity is chord/time. Arc length/time is the average speed.
Change in velocity
2× asked, last 2023
∣Δv⃗∣=v12+v22−2v1v2cos⁡θ,∣Δv⃗∣=2vsin⁡θ2  (v1=v2)|\Delta\vec v|=\sqrt{v_1^2+v_2^2-2v_1v_2\cos\theta},\qquad |\Delta\vec v|=2v\sin\frac{\theta}{2}\;(v_1=v_2)
θ\theta = angle between the two velocities = angle turned. At 90∘90^\circ: v12+v22\sqrt{v_1^2+v_2^2}.
ABθ\thetav⃗1\vec v_1v⃗2\vec v_2v⃗1\vec v_1v⃗2\vec v_2Δv⃗\Delta\vec vθ\theta
Tail to tail, the velocities make the same angle θ\theta: ∣Δv⃗∣=2vsin⁡θ2|\Delta\vec v|=2v\sin\frac{\theta}{2}
  • Not v2−v1v_2-v_1. Velocity change is a vector difference. Get each speed from energy, find the angle between the velocities, then combine. A quarter turn means perpendicular velocities.
Centripetal acceleration
7× asked, last 2025
ac=v2r=ω2r=ωv,a⃗=−ω2r⃗a_c=\frac{v^2}{r}=\omega^2 r=\omega v,\qquad \vec a=-\omega^2\vec r
Points to the centre, perpendicular to v⃗\vec v. Uniform motion: size constant, direction keeps turning.
v⃗\vec va⃗c\vec a_c
Uniform motion: v⃗\vec v along the tangent, a⃗\vec a always to the centre
  • For r⃗=a(cos⁡ωt i^+sin⁡ωt j^)\vec r=a(\cos\omega t\,\hat i+\sin\omega t\,\hat j): v⃗⊥r⃗\vec v\perp\vec r and a⃗=−ω2r⃗\vec a=-\omega^2\vec r, so the net force is opposite to r⃗\vec r. L⃗=mr⃗×v⃗\vec L=m\vec r\times\vec v stays constant along the axis.
  • Not constant velocity. Only the speed is constant. At the opposite end of a diameter both v⃗\vec v and a⃗\vec a reverse; v⃗\vec v stays tangent and a⃗\vec a points to the centre.
  • Not one component reversed. a⃗\vec a is along −r⃗-\vec r, so at P(R,θ)P(R,\theta), a⃗=−(v2/R)(cos⁡θ i^+sin⁡θ j^)\vec a=-(v^2/R)(\cos\theta\,\hat i+\sin\theta\,\hat j). Check at θ=90∘\theta=90^\circ: it must be −(v2/R)j^-(v^2/R)\hat j.

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Practise Circular Motion

26 more concepts in 6 topics

Non-uniform circular motion and tangential acceleration, Dynamics of uniform circular motion, Banking of roads and conical pendulum, Rotating frame and centrifugal force and Vertical circular motion, plus ideas from angular momentum. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Non-uniform circular motion and tangential acceleration

  • Tangential and net acceleration2×
  • Equal tangential and centripetal parts2×

Dynamics of uniform circular motion

  • Centripetal force6×
  • Stone whirled on a string5×
  • Mass on a rotating spring5×
  • Rotating rod or liquid column2×
  • Coin on a turntable3×
  • Car on a level curve4×
  • Over a hump or through a dip

Banking of roads and conical pendulum

  • Banking without friction2×
  • Banking with friction4×
  • Normal force on a banked turn1×
  • Conical pendulum2×
  • Pendulum in a turning vehicle2×
  • Bead on a rotating wire1×

Rotating frame and centrifugal force

  • Centrifugal force5×
  • Ball in a smooth radial groove1×
  • Rotor and death well2×
  • Spinning Earth: weight at the equator1×

Vertical circular motion

  • Tension at any point4×
  • Speeds from energy conservation7×
  • Completing the circle on a string3×
  • Rod or tube: completing the circle
  • Where the string slackens2×
  • Loop from a height1×
  • Leaving a smooth sphere1×

Also asked here: angular momentum

  • Key points1

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Practise Circular Motion

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