CBSE Class 9 Science Advanced Physics Circular Motion Notes Set 01

Circular Motion

A. Circular Motion

(a) Definition

When a body keeps moving round a fixed point along a circle, its motion is called circular motion.

Uniform circular motion. If the body travels equal lengths of the circumference in equal times, the motion is uniform circular motion. Here the speed does not change, while the direction of the velocity changes all the time.

Examples: the moon going round the earth, and a satellite going round its planet.

(b) Why circular motion is an accelerated motion

Explanation. Picture a boy running at a steady speed along a track shaped like a regular hexagon. At each of the six corners he has to turn, so his direction changes while his speed does not. On a regular octagon track at the same speed he must turn eight times per round, and the turns come closer together in time. As we add more sides the turns become more numerous and the gaps between them shrink. A circle is the limit of a polygon with endless sides, where the turning goes on continuously without any gap.

So a body moving on a circle at constant speed has a velocity whose direction is changing at every instant. A changing velocity means acceleration, hence uniform circular motion is an accelerated motion.

Uniform linear motion compared with uniform circular motion

(c) The radian

The radian is the unit of a plane angle.

Definition. One radian is the angle made at the centre of a circle by an arc whose length is equal to the radius of that circle.

In the figure, the arc AB has length ℓ and makes the angle θ at the centre C. Then \( \theta = \frac{\ell}{r} \) radian. When ℓ = r, θ = 1 radian.

For the whole circumference the arc length is 2πr, so the angle at the centre is \( \frac{2\pi r}{r} = 2\pi \) radian. The small letter c written above a number (2π^c) is the symbol of radian, just as the small circle º is the symbol of degree.

Radian and degree. A complete circle is 2π^c = 360º, so \( 1^c = \frac{360^\circ}{2\pi} = 57.3^\circ \).

(d) Variables of circular motion

1. Angular position. To state where a point is in angular terms we must fix an origin and a reference line. The angle that the position vector (drawn from the origin to the point) makes with the reference line is the angular position. It therefore depends on the choice of both the origin and the reference line. Circular motion takes place in one plane, so it is a two-dimensional motion.

A particle P moves on a circle of radius r with centre O. Its angular position at any moment is the angle θ between OP and the line OX. In the figure the particle moves from P to P' and the angle between the two positions is Δθ.

2. Angular displacement (Δθ). The angle through which the position vector of a moving particle turns in a given interval of time is its angular displacement. It depends on the origin but not on the reference line. If the particle turns through Δθ in the time Δt, then Δθ is its angular displacement.

Points to note:

1. Angular displacement has no dimension. Its SI unit is the radian; other units are the degree and the revolution, with 2π rad = 360º = 1 rev.
2. A very small (infinitesimal) angular displacement can be treated as a vector, but a finite angular displacement cannot, because small angular displacements can be added in any order (dθ₁ + dθ₂ = dθ₂ + dθ₁), while finite ones cannot (θ₁ + θ₂ ≠ θ₂ + θ₁).
3. The direction of a small angular displacement is found with the right hand thumb rule: curl the fingers of the right hand along the motion of the point, and the thumb gives the direction of the angular displacement.

3. Angular velocity (ω). The angular displacement per unit time is the angular velocity, written ω (omega).

\[ \omega_{av} = \frac{\text{Angular displacement}}{\text{Total time taken}} = \frac{\theta_2 - \theta_1}{t_2 - t_1} = \frac{\Delta\theta}{\Delta t} \]

Here θ₁ and θ₂ are the angular positions at the times t₁ and t₂. As the book treats angular displacement as a scalar here, the average angular velocity is also treated as a scalar.

Points to note:

1. The SI unit of angular velocity is rad/s.
2. In a rigid body every point turns through the same angle in the same time, so angular velocity belongs to the body as a whole. For example, every point of the earth has the angular velocity 2π/24 rad per hour about the earth's axis.
3. If a body makes N rotations in t seconds, its average angular velocity in rad/s is \( \omega_{av} = \frac{2\pi N}{t} \).
4. If T is the time period and f the frequency of a uniform circular motion, then \( \omega_{av} = \frac{2\pi}{T} = 2\pi f \).
5. The direction of the angular velocity is along the axis of rotation and is found with the right hand thumb rule.

4. Angular acceleration (α). The rate at which angular velocity changes is the angular acceleration. If ω₁ and ω₂ are the angular velocities at the times t₁ and t₂, the average angular acceleration is

\[ \alpha_{av} = \frac{\omega_2 - \omega_1}{t_2 - t_1} = \frac{\Delta\omega}{\Delta t} \]

Points to note:

1. Average and instantaneous angular accelerations are both axial vectors (along the axis). The dimension is [T^–2] and the unit is rad/s².
2. Its direction is along the axis of rotation, given by the right hand thumb rule.
3. If α = 0, the circular motion is uniform.

(e) Relation between linear and angular quantities

A particle moves on a circle of radius r from P to Q, covering the arc x and turning through the angle θ = ωt at the centre C.

1. Linear and angular displacement. Angle = arc ÷ radius, so θ = x/r, that is x = rθ .... (i)

2. Linear and angular velocity. Dividing (i) by the time, x/t = r(θ/t), so v = rω .... (ii)

3. Linear (tangential) and angular acceleration. Dividing (ii) by the time, v/t = r(ω/t), so a = rα .... (iii). (The book's heading for this step repeats the words "angular velocity and linear velocity"; the relation is really between the accelerations.)

Centripetal acceleration and force (used in the exercises)

A body in uniform circular motion has an acceleration directed along the radius towards the centre, called the centripetal acceleration. Its size is \( a = \frac{v^2}{r} = \omega^2 r \). The inward force that causes it is the centripetal force, \( F = \frac{mv^2}{r} = m\omega^2 r \). The velocity is along the tangent, at right angles to this acceleration.

Solved Examples

Illustration 1. A fly wheel making 120 revolutions/minute. Find the angular speed of the wheel :
Answer:
120 revolutions/minute = 2 revolutions/second.
Angular speed = angle in one revolution × revolutions per second = 2π × 2 = 4π rad/s.

Illustration 2. A stone tied to the end of a string 80 cm is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 sec., what is the magnitude of the angular speed. (use \( \pi = \frac{22}{7} \))
Answer:
Angular speed = angle in one revolution × revolutions per second.
ω = 2π × \( \frac{14}{25} \) = 2 × \( \frac{22}{7} \) × \( \frac{14}{25} \) = 3.52 rad/s.
(The length of the string is not needed.)

Illustration 3. Earth revolves around the Sun in 365 days. Calculate its angular speed.
Answer:
T = 365 days = 365 × 24 × 60 × 60 s.
ω = \( \frac{2\pi}{T} = \frac{2\pi}{365 \times 24 \times 60 \times 60} \) = 1.99 × 10^–7 rad/s.

Illustration 4. Is the angular velocity of rotation of hour hand of a watch greater or smaller than the angular velocity of Earth’s rotation about its own axis ?
Answer:
The hour hand completes one rotation in 12 hours, while the Earth takes 24 hours.
Since ω = 2π/T, the angular velocity of the hour hand is double (greater than) that of the Earth’s rotation.

Illustration 5. A particle is moving with constant speed in a circular path. Find the ratio of average velocity to its instantaneous velocity when the particle describes an angle \( \theta = \frac{\pi}{2} \)
Answer:
Time to describe the angle θ: t = θ/ω = θR/v = πR/2v.
For a quarter circle the displacement is the chord = √2 R.
Average velocity = displacement ÷ time = \( \frac{\sqrt{2}R}{\pi R/2v} = \frac{2\sqrt{2}}{\pi}v \).
Instantaneous velocity = v.
Ratio = \( \frac{2\sqrt{2}}{\pi} \).

Illustration 6. Find the time period of meeting of minute hand and second hand of a clock.

Answer:
Both hands start together along the reference line.
ω_min = 2π/60 rad/min and ω_sec = 2π/1 rad/min.
They meet again when the second hand has gained one full turn: θ_sec – θ_min = 2π.
(ω_sec – ω_min)t = 2π, so 2π(1 – 1/60)t = 2π.
t = 60/59 min (about 1 min 1.02 s).