Circular motion: motion along a circle of radius r. Uniform circular motion (UCM): constant speed, variable velocity direction. Non-uniform circular motion (non-UCM): speed also changes. Examples: satellite orbit (UCM), vertically swinging ball (non-UCM).

• In UCM: acceleration is non-zero (centripetal) even though speed is constant — NEET conceptual trap.
• Period T = time for one complete revolution. Frequency n = 1/T = number of revolutions per second (Hz). ω = 2πn = 2π/T rad/s.
• NEET question: 'Which of the following is constant in UCM?' → Speed (yes), velocity (no), acceleration magnitude (yes, = v²/r), acceleration direction (no).

Angular displacement θ (rad). Angular velocity ω = dθ/dt (rad/s) — axial vector. Angular acceleration α = dω/dt (rad/s²). Linear-angular: v = ωr; a = αr; s = rθ.

• Angular velocity is an axial vector — it points along the axis of rotation (by right-hand rule: curl fingers in direction of rotation, thumb points along ω).
• ω = 2π/T = 2πn. Relation to linear speed: v = ωr → v ∝ r for constant ω (outer edge of rotating disc moves faster).
• Time period of second's hand of watch = 60 second. Then ω = 2π/60 = π/30 rad/s.

Centripetal acceleration a_c = v²/r = ω²r, directed radially inward (toward center). In UCM, speed is constant → no tangential acceleration. In non-UCM, both centripetal (a_c) and tangential (a_t = rα) accelerations exist.

• In UCM: a = v²/r (centripetal only). Net acceleration = v²/r directed toward center.
• In non-UCM: a_net = √(a_c² + a_t²) where a_c = v²/r and a_t = dv/dt. The acceleration is not toward the center.
• NEET trap: 'In UCM, the particle is in equilibrium since its speed is constant.' FALSE — the particle is NOT in equilibrium because it has centripetal acceleration. Equilibrium requires zero acceleration.

Centripetal force F_c = mv²/r = mω²r, always directed toward the center. It is provided by different real forces depending on the situation: tension (string), friction (road), normal force (bowl), gravity (satellite), etc. Centripetal force is NOT a separate force — it is the resultant toward the center.

• Table of centripetal force providers: (a) Car turning on flat road: static friction. (b) Ball on string in horizontal circle: tension. (c) Earth orbiting Sun: gravitational force. (d) Ball in vertical circle: combination of tension and gravity.
• NEET: 'What provides centripetal force for a car on a flat turn?' → Static friction between tyres and road. Not tension, not normal force.
• F_c depends on speed: F ∝ v². Doubling speed quadruples the required centripetal force — explains why fast cars are more prone to skidding on turns.

Centrifugal force is a fictitious (pseudo) force that appears in the rotating (non-inertial) reference frame. Magnitude = mv²/r = mω²r, directed radially outward. Does NOT exist in an inertial (ground) reference frame.

• In the ground frame: centripetal force (real, inward) causes circular motion. In the rotating frame: centrifugal force (fictitious, outward) + centripetal force (real, inward) = 0 → equilibrium in the rotating frame.
• NEET assertion-reason: 'Centrifugal force is a real force acting on the body in circular motion.' FALSE in inertial frame. TRUE only from the perspective of the rotating frame observer.
• Application: A person in a merry-go-round feels pushed outward — this is the centrifugal pseudo-force they perceive. Water in a spinning bucket surface becomes parabolic due to centrifugal force in the rotating frame.

A cyclist taking a circular turn must lean inward at angle θ with the vertical: tan θ = v²/(rg). The leaning creates a horizontal component of the normal force N that provides centripetal force. Vertical: N cosθ = mg; Horizontal: N sinθ = mv²/r → tanθ = v²/(rg).

• The formula tan θ = v²/(rg) applies to both a cyclist leaning inward AND the banking angle of a road.
• For higher speed v at same radius r, the required lean angle θ increases (tanθ increases). For larger radius r, same speed requires less lean.
• NEET: 'A cyclist increases speed on a turn. The lean angle must' → increase (tanθ = v²/(rg), so θ increases with v).

A banked road is inclined at angle θ to the horizontal. For ideal speed (no friction needed): tan θ = v²/(rg). Safe speed range with friction coefficient μ: v_min = √[rg(tanθ−μ)/(1+μtanθ)] to v_max = √[rg(tanθ+μ)/(1−μtanθ)].

• NEET basic question: 'Given r and v, find banking angle θ.' Use tanθ = v²/(rg) → θ = arctan(v²/rg).
• At the ideal speed: no friction is needed. Below ideal speed: friction acts up the slope (toward center). Above ideal speed: friction acts down the slope (away from center).
• Ideal speed for no friction: v = √(rg tanθ). This is the most-tested NEET formula from this subtopic.

A vehicle (half-width a, center of mass height h) overturns when inner wheel's normal force = 0. Critical speed: v_overturn = √(gra/h). At v < v_overturn: vehicle stable. At v > v_overturn: vehicle overturns outward.

• Derivation: Taking moments about the outer wheel — mg × a = (mv²/r) × h → v² = gra/h → v = √(gra/h). When mg × a < centrifugal torque, vehicle overturns.
• The overturning speed increases with wider wheelbase (larger a) and lower center of mass (smaller h) — explains why SUVs overturn more easily than low sports cars.
• NEET may ask: 'A vehicle with half-width 1.5 m and CM height 1 m on r = 50 m turn. Maximum speed before overturning?' v = √(10×50×1.5/1) = √750 ≈ 27.4 m/s.

At the top: T + mg = mv²_top/R → T = mv²_top/R − mg. For minimum tension T = 0: v_top_min = √(gR). Energy conservation bottom to top: (1/2)mv²_bottom = (1/2)mv²_top + 2mgR (height = 2R) → v²_bottom = v²_top + 4gR. Minimum v_bottom = √(5gR).

• At the top, both tension T and weight mg act downward (toward center). So T + mg = mv²/R → T = m(v²/R − g). Minimum v when T = 0: v = √(gR).
• At the bottom, weight acts downward but centripetal force is upward: T − mg = mv²/R → T = m(v²/R + g). Maximum tension is at the bottom.
• NEET question: 'What is the minimum speed at the bottom of a vertical circle of radius R for the block to complete the circle?' → v = √(5gR). The '5' comes from energy equation: v_bottom² = v_top² + 4gR → 5gR.

In non-uniform circular motion, the speed changes along with direction. Two accelerations act simultaneously: (1) centripetal acceleration a_c = v²/r (directed toward center, changes direction); (2) tangential acceleration a_t = αr = dv/dt (along tangent, changes speed). Net acceleration = √(a_c² + a_t²).

• Centripetal acceleration a_c = v²/r acts radially inward (maintains circular path). Tangential acceleration a_t = αr acts along the tangent (speeds up or slows down the body).
• Angle of net acceleration with radius: tanφ = a_t/a_c. When a_t = 0 (uniform circular motion), net acceleration = centripetal only, directed radially.
• NEET: 'A particle moves in a circle of radius r with increasing speed. The net acceleration is...' → at an angle to the radius (neither radially inward nor tangential — it is the vector sum).

A conical pendulum is a mass suspended by a string that rotates in a horizontal circle, with the string tracing a cone. The tension T provides both the centripetal force (horizontal) and supports the weight (vertical). Two equations: T cosθ = mg (vertical); T sinθ = mv²/r (horizontal). Dividing: tanθ = v²/(rg).

• Period: T_period = 2π√(L cosθ/g) where L is string length and θ is half-angle of cone. As θ increases (faster rotation), T_period decreases.
• For small θ: cos θ ≈ 1 → T_period ≈ 2π√(L/g), similar to simple pendulum. This is the limiting case.
• NEET: 'A conical pendulum rotates faster. The angle θ increases because...' → faster rotation means more centripetal force needed → larger sinθ component required → θ increases. Height h = L cosθ decreases.