Uniform Circular Motion - Physics Class 11 - NEET PDF Download

Circular Motion

Circular motion is the motion of an object along the circumference of a circle. Circular motion may be uniform or non-uniform. In uniform circular motion the speed (magnitude of velocity) is constant while the direction of the velocity changes continuously; in non-uniform circular motion the speed also changes with time.

Common examples of circular motion include a man-made satellite orbiting the Earth, a rotating ceiling fan, a car wheel in motion, the blades of a windmill, and gears in turbines. Because the direction of motion changes continuously, it is useful to describe circular motion using angular variables as well as linear variables.

Uniform Circular Motion

Uniform circular motion is the motion of a particle around a circle with constant speed. Although the speed is constant, the velocity changes because its direction changes continuously. Therefore the particle is accelerating even when its speed is constant.

There must be an acceleration directed towards the centre of the circle that continuously changes the direction of the velocity vector. This acceleration is called centripetal (radial) acceleration. The force that produces this acceleration is called centripetal force; it may be due to tension, gravity, friction or normal force depending on the situation.

The magnitude of the centripetal acceleration is given by the relations

a_c = v²/r = rω²

Using Newton's second law, the magnitude of the centripetal force for a particle of mass m is

F_c = m a_c = m v²/r = m r ω²

Examples of forces that can act as centripetal force are tension in a string, friction between tyres and road when a vehicle turns, gravitational force for satellites, electrostatic force for electrons around a nucleus (in simplified classical picture), etc.

Key properties (uniform circular motion)

• Speed is constant.
• Velocity is continuously changing (direction changes).
• There is no tangential acceleration (a_t = 0).
• Radial (centripetal) acceleration: a_c = v²/r = r ω².
• Relation between linear speed and angular speed: v = r ω.

Derivation of a_c = v²/r (geometric argument)

Consider a particle moving with constant speed v along a circle of radius r. Let its velocity vectors at times t and t + Δt be v and v + Δv. The change in velocity Δv is due only to change in direction. For a small central angle Δθ the magnitudes satisfy

Δs = r Δθ and Δv ≈ v Δθ.

Therefore

a = lim(Δv/Δt) = lim(v Δθ/Δt) = v (dθ/dt) = v ω.

Using ω = v/r gives

a = v × (v/r) = v²/r.

Variables in Circular Motion

• Angular displacement (Δθ): the angle swept by the radius vector; Δθ = Δs / r. Unit: radian (rad).
• Angular velocity (ω): rate of change of angular displacement; ω = dθ/dt = Δθ/Δt. Unit: rad s^-1. Relation: v = r ω.
• Angular acceleration (α): rate of change of angular velocity; α = dω/dt = d²θ/dt². Unit: rad s^-2. Relation with tangential acceleration: a_t = r α.
• Period (T) and frequency (f): T is the time for one complete revolution; f = 1/T. Angular velocity is related by ω = 2π f = 2π / T.
• Centripetal (radial) acceleration: a_c = v²/r = r ω². Direction: always towards the centre of the circle.

Non-uniform circular motion

In non-uniform circular motion the speed of the particle changes with time. There is a tangential acceleration component a_t = dv/dt (or a_t = r α), which changes the speed. The resultant acceleration vector has two perpendicular components: radial (centripetal) acceleration a_r = v²/r directed towards the centre, and tangential acceleration a_t tangent to the path. The magnitude of the resultant acceleration is

|a| = √(a_r² + a_t²).

Common examples of uniform circular motion

• Motion of artificial satellites in circular orbits (approximate).
• Motion of electrons around the nucleus (classical model approximation).
• Blades of windmills and fans (points on blades move in (approximately) uniform circular paths).
• The tip of the second hand of a watch.

Solved Questions 

Q.1. During the course of a turn, an automobile doubles its speed. How much must additional frictional force the tires provide if the car safely makes around the curve? Since Fc varies with v2, an increase in velocity by a factor of two must be accompanied by an increase in centripetal force by a factor of four.

A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×10⁷ meters from the earth's center. What is the magnitude of the acceleration felt by a geosynchronous satellite?

Solution:

The first part (automobile):
If initial speed is v and centripetal force required is F ∝ v², then when speed doubles (2v) the required force becomes (2v)² = 4 v², i.e., four times the original centripetal force.
Therefore the additional frictional force required = 4F - F = 3F (three times the original centripetal force).

The satellite part:
Radius of orbit r = 4.23 × 10⁷ m.
Orbital period T = 1 day = 86400 s.
Centripetal acceleration a = 4π² r / T².

Substitute values:
4π² ≈ 39.478.
Numerator = 39.478 × 4.23 × 10⁷ ≈ 1.67 × 10⁹.
Denominator = (86400)² ≈ 7.465 × 10⁹.
a ≈ 1.67 × 10⁹ / 7.465 × 10⁹ ≈ 0.224 m/s².

Q.2. A cyclist moves around a circular path of a radius of 50 m with a speed of 10 m/s.
(a) Find the acceleration of the cyclist?
(b) Considering the combined mass of the cyclist and cycle to be 120 kg, what is the net force applied on them?
Solution: (a)
The centripetal acceleration is given by a_c = v²/r.
v = 10 m/s.
r = 50 m.
a_c = 10²/50 = 100/50 = 2.0 m/s².

(b)
The net (centripetal) force is F_net = m a_c.
m = 120 kg.
F_net = 120 × 2 = 240 N directed towards the centre of the circular path.

Q.3. A race car travelling around a circular path of a radius of 400 m with a speed of 50 m/s. Find the centripetal acceleration of the car.
Solution:
a_c = v²/r.
v = 50 m/s, r = 400 m.
a_c = 50²/400 = 2500/400 = 6.25 m/s².

This acceleration is directed towards the centre of the circle.

Q.4. A racer is moving with a constant tangential speed of 50 m/s, takes one lap around a circular track in 40 seconds. Calculate the magnitude of the acceleration of the car.
Solution:
Given v = 50 m/s and time for one lap T = 40 s.
Radius r = Tv/(2π).
But acceleration a = v²/r.
Substitute r to eliminate it: a = v² / (T v / 2π) = (v × 2π) / T = 2π v / T.
Compute 2π/T = 2π / 40 ≈ 6.2832 / 40 = 0.15708 s^-1.
a = 50 × 0.15708 ≈ 7.85 m/s².

Q.5. An object moving in a circular motion has a centripetal acceleration of 20 m/s². If the radius of the motion is 0.5 m, calculate the frequency of the motion.
Solution:
Given a_c = 20 m/s², r = 0.5 m.
a_c = v²/r, therefore v² = a_c r = 20 × 0.5 = 10.
v = √10 ≈ 3.1623 m/s.
Relation between v and period T: v = 2π r / T, therefore T = 2π r / v.
Compute 2π r = 2π × 0.5 = π ≈ 3.1416.
T = 3.1416 / 3.1623 ≈ 0.995 s.
Frequency f = 1/T ≈ 1.005 Hz.

Summary

• Circular motion requires a centripetal acceleration pointing toward the centre; its magnitude is a_c = v²/r = r ω².
• In uniform circular motion speed is constant and tangential acceleration is zero; in non-uniform motion tangential acceleration changes the speed.
• Angular and linear quantities are related by v = r ω, ω = 2π f = 2π / T, and a_t = r α.