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Chapter 06 · Part I · Mechanics
Revised on ____________________

Chapter 06 · Circular Motion

The force that turns you, and never speeds you up

sheets
diagrams
results

Formulas and conditions only — no derivations, no solved numbers. The diagrams are the reader's own.

Angular kinematics

RADIANS THROUGHOUT

I.Arc s = r θ

II.Angular velocity ω = dθdt = 2πT = 2π f

III.Speed v = r ω

IV.Angular acceleration α = dωdt

V.Centripetal acceleration a_c = v²r = ω² r always toward the centre

VI.Tangential acceleration a_t = r α only this one changes the speed

VII.Total acceleration a = √(a_c² + a_t²)

VIII.Uniform circular motion constant ω, constant a_t = 0 speed constant, velocity not

The centripetal force is a requirement, not a new force. Something real — tension, friction, gravity, a normal force — must supply mv²/r, or the body leaves the circle.

Horizontal circles

WHAT SUPPLIES mv²/r

I.Centripetal force F_c = m v²r = m ω² r

II.Car on a flat road v_max = √(μ_s g r) friction is the only supply

III.Banked, frictionless tan θ = v²r g

IV.Banked with friction v_max = √( r g tan θ + μ1 − μ tan θ )

V.Conical pendulum tan θ = v²r g , T = 2π √( L cos θg )

VI.Its tension T_string = m gcos θ

III

The vertical circle

STRING AND ROD ARE DIFFERENT PROBLEMS

I.At the top, string v_top ≥ √(g L) tension may not be negative

II.At the bottom, then v_bot ≥ √(5 g L)

III.Tension difference T_bot − T_top = 6 m g independent of speed

IV.Rod or tube v_top ≥ 0 a rod can push

V.Leaves the circle when T = 0 or N = 0 between 90° and 180° for a string

VI.Over a hill crest N = m g − m v²r airborne at v = √(g r)

Rotating frames

ONLY IF YOU RIDE THE TURNTABLE

I.Centrifugal force F = m ω² r outward, pseudo, rotating frame only

II.Coriolis force F = 2 m (v × ω) only for motion within the frame

III.In the ground frame there is no centrifugal force at all

IV.Banking of a river / rail outer edge rises by h = v² dr g d = track gauge

1 rev

2π rad = 360°

rpm → rad s⁻¹

× π/30

Earth's spin

ω = 7.3×10⁻⁵ s⁻¹

at the equator

a_c = 0.034 m s⁻²

Loop, string

v_bot = √(5gL)

Flat-road limit

v = √(μgr)

Non-uniform circular motion

WHEN THE SPEED CHANGES TOO

I.Radial equation ΣF_r = m v²r

II.Tangential equation ΣF_t = m r α

III.Speed from energy v² = v₀² − 2 g L (1 − cos θ) a swinging bob

IV.Angle of the net a tan φ = a_ta_c

Where marks are lost in this chapter

1Drawing a separate F_c arrow in the FBD. It is the resultant of the real forces, not an extra one.

2Writing T = mg for a conical pendulum. It is T = mg/cos θ — twice the weight at 60°.

3Using v_top = √(gL) for a rod. A rod can push, so the condition is only v ≥ 0.

4Reading constant ω as zero acceleration. a_t is zero; a_c = ω²r is not.