Physics · Laws of Motion · Chapter notes
Circular Motion · Class 11 Notes
Class 11 notes on Circular Motion: banking of roads, the vertical circle, the rotor and radius of curvature, built to JEE Advanced depth with the NEET sections marked. Free to read or download.
In short
Every circular motion problem reduces to one question: which real force is pointing at the centre? Nothing supplies a new force. Gravity, normal reaction, tension or friction take that job, and the answer follows from writing mv²/r for whichever one it is. Banking, vertical circles and the rotor are all the same question asked three ways.
Contents
- ·How to Read This
- 1Angular Quantities ω, α, and the two accelerations
- 2What Supplies the Centripetal Force
- 3Banking of Roads without friction, then with it
- 4The Vertical Circle critical speeds, string against rod
- 5Held In by Friction Alone
- 6Radius of Curvature the circle a curved path is copying
- ★Circular Motion · Fact Sheet
How to Read This
- Second of a pair. The Friction set built the idea of limiting friction; this one spends it.
- Banking with friction, the turntable and the rotor all use the same ceiling μN, only now the friction points at the centre of a circle.
- Blue marks a defining fact. Red marks a trap.
- There is no such thing as a centripetal force in a free-body diagram.
- Centripetal is a job, and some ordinary force does it.
- Identify that force first, then set it equal to mv²/r.
- Every problem in this chapter is that one sentence, applied carefully.
Angular Quantities
- Describing rotation by angle rather than distance turns awkward geometry into algebra.
- Every angular quantity has a linear twin: v = ωr and a(t) = αr.
- Centripetal acceleration is a(c) = v²/r = ω²r, and it always points at the centre.
- One revolution takes T = 2π/ω, and the frequency is f = 1/T.
- If the speed is also changing, the total acceleration is √(a(c)² + a(t)²).
- Constant speed is not constant velocity.
- The direction changes every instant, so there is an acceleration of v²/r the whole time.
- Questions phrased as constant speed are inviting you to answer zero acceleration.
What Supplies the Centripetal Force
Net force toward the centre = mv²/r = mω²r
- Both halves of that claim are wrong.
- Centrifugal force is a pseudo-force, existing only in the rotating frame.
- A third-law pair must act on two different bodies.
- The reaction to the road's inward friction on the tyres is the tyres' outward friction on the road, not anything acting on the passenger.
Banking of Roads
- On a flat road only friction turns the car, which is unreliable in rain.
- Tilt the road and the normal reaction gains an inward component that does the job for free.
- With no friction at all: tanθ = v²/rg, so v(ideal) = √(rg tanθ).
- That is one exact speed, not a maximum.
- If μ ≥ tanθ the expression for v(min) goes negative, which means there is no minimum: the car can crawl round at any slow speed.
- On a flat road, θ = 0 reduces both to v(max) = √(μrg).
- Always sanity-check a banking formula by setting θ = 0.
The Vertical Circle
- Speed varies around the loop, because gravity helps going down and fights going up.
- Combine energy conservation with the centripetal equation and the whole problem opens.
- On a string: v(top) = √(gr) and v(bottom) = √(5gr), with tension 6mg at the bottom.
- On a rigid rod: it can push as well as pull, so v(top) may be zero and v(bottom) = √(4gr).
- Launched too slowly to loop but too fast to swing back, the string goes slack above the horizontal and the body becomes a projectile.
- That happens when cosφ = (v(bottom)² − 2gr) / 3gr.
Held In by Friction Alone
- Two classic setups where friction is the only thing preventing disaster.
- Turntable: friction supplies the whole centripetal force, so μmg ≥ mω²r and ω(max) = √(μg/r).
- Rotor: the wall supplies N inward, and friction on that N holds up the weight, so ω(min) = √(g/μr).
- On a turntable, spinning faster throws the coin off, so there is a maximum ω.
- In a rotor, spinning faster presses you harder into the wall and keeps you up, so there is a minimum ω.
- Read which surface the normal force acts on before reaching for a formula.
- The outermost coin on a turntable always slips first, because the required force grows with r.
Radius of Curvature
- Any curved path, even a non-circular one, is momentarily copying some circle.
- r = v² / a(perpendicular), where a(perp) is the part of the acceleration at right angles to v.
- At the top of a projectile the whole of g is perpendicular to v, so r = v²/g.
- At the launch point a(perp) is only g cosθ, so r = u²/(g cosθ).
- From vectors: a(perp) = |v × a| / |v|.
★ Circular Motion · Fact Sheet
Every rule for revision day.
THE DICTIONARY
v = ωr, a(t) = αra(c) = v²/r = ω²r
ω = 2π/T = 2πf.
TOTAL ACCELERATION
|a| = √(a(c)² + a(t)²)Uniform circular motion
is the case a(t) = 0.
CENTRIPETAL FORCE
It is a JOB, not a new force.Tension, friction, gravity or N
does it.
BANKING, NO FRICTION
tanθ = v²/rgv(ideal) = √(rg tanθ)
one exact speed, not a range.
BANKING WITH FRICTION
v(max) = √[rg(tanθ+μ)/(1−μtanθ)]v(min) flips the signs
At θ=0 both give √(μrg).
VERTICAL CIRCLE, STRING
v(top) = √(gr)v(bottom) = √(5gr)
T(bottom) = 6mg.
VERTICAL CIRCLE, ROD
A rod can push, so v(top) = 0v(bottom) = √(4gr)
Never use 5gr for a rod.
TURNTABLE vs ROTOR
Turntable: ω(max) = √(μg/r)Rotor: ω(min) = √(g/μr)
Opposite directions.
RADIUS OF CURVATURE
r = v²/a(perp)Projectile top: r = v²/g
At launch: r = u²/(g cosθ).
Before the exam
What the paper actually asks from this chapter
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