Circular motion for JEE Main

Circular motion joins kinematics to Newton's laws. The kinematics tells you a body moving in a circle must accelerate towards the centre; Newton's laws tell you some real force (tension, friction, gravity or a normal force) must supply that acceleration. Most JEE Main questions in this chapter ask you to identify that force and set it equal to mv²/r.

The chapter also feeds rotation, gravitation (orbits) and charged particles in magnetic fields, so it is worth learning properly.

Angular quantities

For constant α, the rotational equations mirror the linear ones: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.

Centripetal acceleration

A body moving in a circle at constant speed still accelerates, because its direction keeps changing. That acceleration points towards the centre:

a_c = v²/r = ω²r

The net force towards the centre must be F = mv²/r. "Centripetal force" is not a new kind of force. It is the name for whichever real force, or combination of forces, points to the centre.

Non-uniform circular motion. If the speed also changes, there is a tangential acceleration a_t = dv/dt as well. The two are perpendicular, so the total acceleration is √(a_c² + a_t²).

Curves on roads

Flat curve. Only friction can push the car towards the centre: mv²/r ≤ μmg, so v_max = √(μrg).

Banked curve, no friction. The road is tilted at angle θ. The normal force has a horizontal component that provides the centripetal force: N sin θ = mv²/r and N cos θ = mg. Dividing, tan θ = v²/(rg). At this speed no friction is needed.

Banked curve with friction. The allowed speeds lie between a minimum and a maximum:

If tan θ ≤ μ, the car can go as slowly as it likes without sliding down.

Conical pendulum

A bob on a string of length L moves in a horizontal circle, with the string at angle θ to the vertical. The radius is r = L sin θ.

• Vertical: T cos θ = mg.
• Horizontal: T sin θ = mv²/r.
• So tan θ = v²/(rg), and the period is 2π √(L cos θ / g).

Vertical circle

A body tied to a string of length r moves in a vertical circle. Its speed changes, because gravity does work.

At the top: T + mg = mv_top²/r. The string stays taut only if T ≥ 0, so v_top ≥ √(gr).

From top to bottom (energy): v_bottom² = v_top² + 4gr. So the minimum speed at the bottom is √(5gr).

Tension difference: T_bottom = mg + mv_b²/r and T_top = mv_t²/r − mg. Subtracting and using the energy relation: T_bottom − T_top = 6mg.

A rod can push as well as pull, so the body can pass the top at almost zero speed. A string can only pull.

Hump and dip. A car on a convex bridge of radius r: mg − N = mv²/r, so N falls as speed rises and the car leaves the road at v = √(gr). In a dip, N = mg + mv²/r, which is why you feel heavier.

Worked problems

Problem 1: banking. A curve of radius 100 m is banked at 45°. At what speed is no friction needed? (g = 10 m/s²)

v² = rg tan θ = 100 × 10 × 1 = 1000, so v ≈ 31.6 m/s (about 114 km/h).

Problem 2 (numerical answer): flat curve. A car takes an unbanked curve of radius 80 m. μ between the tyres and the road is 0.5 (g = 10 m/s²). What is the maximum safe speed in km/h?

v_max = √(0.5 × 80 × 10) = √400 = 20 m/s = 20 × 3.6 = 72 km/h.

Problem 3: vertical circle. A 0.5 kg stone on a 2 m string is whirled in a vertical circle (g = 10 m/s²). Find the minimum speed at the bottom and the tension there at that speed.

v_bottom = √(5 × 10 × 2) = 10 m/s.
T_bottom = mg + mv²/r = 5 + 0.5 × 100/2 = 5 + 25 = 30 N, which is 6mg, since the tension at the top is zero.

Problem 4: speeding up on a track. A car on a circular track of radius 50 m has a speed of 10 m/s and is speeding up at 2 m/s². Find its total acceleration.

a_c = 100/50 = 2 m/s²; a_t = 2 m/s².
Total = √(4 + 4) = 2√2 ≈ 2.83 m/s², at 45° to the velocity.

Practice set

1. Find the centripetal acceleration of a body moving at 10 m/s in a circle of radius 5 m.
2. What is the minimum speed at the top of a vertical circle of radius 0.9 m on a string (g = 10 m/s²)?
3. A particle moves in a circle at constant speed. Which of these stay constant: speed, velocity, kinetic energy, acceleration?
4. A wheel accelerates uniformly from rest to 600 rpm in 10 s. Find its angular acceleration and the number of revolutions made.
5. A 1000 kg car crosses the top of a bridge of radius 20 m at 10 m/s (g = 10 m/s²). What is the normal force on it?
6. A bob on a light rigid rod of length 0.4 m is to complete a vertical circle. What is the minimum speed at the bottom (g = 10 m/s²)?
7. A road of radius 10 m is to be banked for a speed of 36 km/h (g = 10 m/s²). Find the banking angle.
8. A conical pendulum has a string of length 1 m at 60° to the vertical (g = 10 m/s²). Find its period and the tension in terms of mg.

Answers

1. a = v²/r = 100/5 = 20 m/s².
2. √(gr) = √9 = 3 m/s.
3. Speed and kinetic energy are constant. Velocity and acceleration change direction; only the magnitude of acceleration stays the same.
4. ω = 600 × 2π / 60 = 20π rad/s, so α = 20π/10 = 2π rad/s². θ = ½ × 2π × 100 = 100π rad = 50 revolutions.
5. N = m(g − v²/r) = 1000 × (10 − 5) = 5000 N.
6. 2√(gr) = 2√4 = 4 m/s.
7. 36 km/h = 10 m/s; tan θ = 100/(10 × 10) = 1, so θ = 45°.
8. Period = 2π √(1 × 0.5 / 10) = 2π √0.05 ≈ 1.4 s. T = mg / cos 60° = 2mg.

What to do next

• Derive the vertical-circle results (√(gr), √(5gr), 6mg) from scratch once.
• Solve 25 problems mixing banking, conical pendulums and vertical circles.
• For each, write the "towards the centre" equation before anything else.
• Revise Newton's laws and friction if the force diagrams slow you down, then move on to work, energy and power.