Excellent, future engineers! Welcome to another deep dive session where we unravel the fascinating world of Physics, especially tailored for your IIT-JEE journey. Today, we're going to explore a topic that's fundamental to understanding many phenomena around us:
Uniform Circular Motion.
Grab your notebooks and let's begin our exploration, starting from the very basics and building our understanding brick by conceptual brick.
### Introduction to Circular Motion: The Dance of Constant Change
Imagine a car taking a turn, a satellite orbiting Earth, or even the hands of a clock moving. What do they all have in common? They are all exhibiting
circular motion.
At its core,
circular motion is simply the movement of an object along the circumference of a circle or a circular path.
Now, within circular motion, we have a very special and important case called
Uniform Circular Motion (UCM).
What makes it 'Uniform'?
In UCM, an object moves with a
constant speed along a circular path. This sounds simple, right? But here's the crucial point, often missed by students: *constant speed does NOT mean constant velocity*.
Think about it:
Velocity is a vector quantity, meaning it has both magnitude (speed) and direction. Even if the speed (magnitude) remains constant, the direction of motion is *continuously changing* as the object moves along the circle.
Key Takeaway (JEE Focus): In Uniform Circular Motion, the speed of the object is constant, but its velocity is continuously changing due to the change in direction. This constant change in velocity means there MUST be an acceleration acting on the object.
This concept is absolutely vital. If velocity is changing, even if only its direction, then by definition, there must be an acceleration. This acceleration is what keeps the object moving in a circle, preventing it from flying off in a straight line (which it would do if there were no acceleration, as per Newton's First Law).
### 1. Describing Motion: Angular Variables
To accurately describe circular motion, we introduce a set of
angular variables that are analogous to the linear variables we used in straight-line kinematics.
#### 1.1 Angular Displacement (Δθ)
Just as linear displacement (Δx) describes how much an object has moved in a straight line,
angular displacement (Δθ) describes how much an object has rotated around a central point.
*
Definition: When an object moves from point A to point B along a circular path, the angle swept out by the radius vector from the center to the object is its angular displacement.
*
Units: The standard unit for angular displacement is the
radian (rad). While degrees are common, radians are crucial in physics as they directly relate arc length to radius.
* Relationship: 1 revolution = 360° = 2π radians.
*
Relation to Arc Length: For a circle of radius 'r', if the angular displacement is θ (in radians), the arc length 's' covered is given by:
s = rθ
This formula only works if θ is in radians!
*
Vector Nature (JEE Advanced Concept): For infinitesimal angular displacements, Δθ can be treated as a vector pointing along the axis of rotation, following the
right-hand thumb rule. If you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of the angular displacement vector. For large angular displacements, it's not strictly a vector due to non-commutative addition, but for infinitesimal changes, it is.
#### 1.2 Angular Velocity (ω)
Just as linear velocity (v) is the rate of change of linear displacement,
angular velocity (ω) is the rate of change of angular displacement.
*
Definition: Angular velocity (ω) is the angular displacement per unit time.
*
Average Angular Velocity: ω
avg = Δθ / Δt
*
Instantaneous Angular Velocity: ω = dθ/dt
*
Units: Radians per second (rad/s). Other units like revolutions per minute (rpm) or revolutions per second (rps) are also used, but should be converted to rad/s for calculations.
* 1 rpm = (2π radians) / (60 seconds) = π/30 rad/s
*
Relation to Linear Speed (v): This is a critical connection. If an object moves a distance 's' in time 't' with angular displacement 'θ', then:
v = s/t
Since s = rθ, we substitute:
v = (rθ)/t = r(θ/t)
Therefore, the magnitude of linear velocity (speed) is:
v = rω
Physical Interpretation: This formula tells us that for a given angular velocity, points farther from the center (larger 'r') will have a greater linear speed. This is why the outer edge of a merry-go-round moves faster than the inner parts.
*
Vector Nature: Angular velocity is a vector quantity. Its direction is the same as that of angular displacement, determined by the right-hand thumb rule. For an object rotating counter-clockwise in the xy-plane, ω points along the positive z-axis.
#### 1.3 Angular Acceleration (α)
*
Definition: Angular acceleration (α) is the rate of change of angular velocity.
*
Average Angular Acceleration: α
avg = Δω / Δt
*
Instantaneous Angular Acceleration: α = dω/dt
*
Units: Radians per second squared (rad/s²).
*
In Uniform Circular Motion (UCM): Since the angular speed (magnitude of ω) is constant, the angular acceleration
α = 0.
Important Note: While α = 0 in UCM, the *linear* acceleration is definitely NOT zero. We will discuss this next.
### 2. The Mysterious Acceleration in UCM: Centripetal Acceleration
This is where UCM gets interesting and slightly counter-intuitive. An object moving at a constant speed *still* experiences acceleration. How? Because its velocity's direction is constantly changing!
This acceleration is called
centripetal acceleration (or radial acceleration).
#### 2.1 Direction of Centripetal Acceleration (a_c)
Imagine tying a ball to a string and whirling it around your head. At any instant, if you were to let go of the string, the ball would fly off *tangentially* to the circle. This tangential direction is the direction of its instantaneous linear velocity.
However, to keep the ball moving in a circle, the string must constantly pull it *inwards*, towards the center of your hand. This inward pull is what causes the change in the ball's velocity direction.
Therefore, the
centripetal acceleration (a_c) always points towards the center of the circular path. It is perpendicular to the instantaneous linear velocity vector.
Common Mistake Alert: Never confuse centripetal acceleration with tangential acceleration. Tangential acceleration changes the *speed* of the object, while centripetal acceleration changes the *direction* of the object's velocity. In UCM, tangential acceleration is zero.
#### 2.2 Derivation of Centripetal Acceleration
Let's derive the formula for centripetal acceleration using two common approaches:
Approach 1: Geometric/Vector Method (Intuitive)
Consider an object moving from point P to point Q in a small time interval Δt.
* Let its velocity at P be
v₁ and at Q be
v₂.
* Since it's UCM, |v₁| = |v₂| = v (constant speed).
* The position vectors are
r₁ and
r₂.
* The angular displacement is Δθ.
1. Draw the velocity vectors
v₁ and
v₂ (tangential to the circle).
2. Shift
v₁ parallel to itself so its tail is at the tail of
v₂.
3. The vector connecting the head of
v₁ to the head of
v₂ is
Δv = v₂ - v₁. This
Δv represents the change in velocity.
4. Notice that the triangle formed by
v₁,
v₂, and
Δv is similar to the triangle formed by the position vectors
r₁,
r₂, and
Δr (the chord connecting P and Q, which for small Δθ is approximately the arc length).
5. From similar triangles: |Δv| / v = |Δr| / r
So, |Δv| = (v/r) * |Δr|
6. Average acceleration a
avg = |Δv| / Δt = (v/r) * (|Δr| / Δt)
7. As Δt approaches 0 (for instantaneous acceleration), |Δr| / Δt approaches the linear speed v.
Therefore, instantaneous centripetal acceleration:
a_c = v²/r
Approach 2: Calculus Method (Rigorous)
Let the object move in the xy-plane in a circle of radius 'r' centered at the origin.
Its position vector at any time 't' can be written as:
r(t) = r cos(ωt) î + r sin(ωt) ĵ
1.
Linear Velocity Vector v(t): Differentiate
r(t) with respect to 't'.
v(t) = d
r/dt = -rω sin(ωt) î + rω cos(ωt) ĵ
The magnitude of this velocity is:
|v(t)| = √((-rω sin(ωt))² + (rω cos(ωt))²)
|v(t)| = √(r²ω² sin²(ωt) + r²ω² cos²(ωt))
|v(t)| = √(r²ω² (sin²(ωt) + cos²(ωt)))
|v(t)| = √(r²ω²) = rω (constant speed, as expected for UCM).
2.
Linear Acceleration Vector a(t): Differentiate
v(t) with respect to 't'.
a(t) = d
v/dt = -rω² cos(ωt) î - rω² sin(ωt) ĵ
a(t) = -ω² [r cos(ωt) î + r sin(ωt) ĵ]
Notice that the term in the square brackets is exactly the position vector
r(t).
So,
a(t) = -ω² r(t)
3.
Magnitude and Direction of a(t):
*
Magnitude: |a(t)| = ω² |r(t)| =
ω²r
*
Direction: The negative sign indicates that the acceleration vector
a(t) is always directed opposite to the position vector
r(t). Since
r(t) points outwards from the center to the object,
a(t) must point inwards, towards the center of the circle. This is indeed the centripetal acceleration.
Combining the two formulas for `a_c`:
We know v = rω, so ω = v/r.
Substitute ω into a_c = ω²r:
a_c = (v/r)² * r = v²/r
So, the magnitude of centripetal acceleration is given by:
a_c = v²/r = rω²
JEE Focus: Memorize these two forms of centripetal acceleration formula. They are used extensively. Remember its direction: always towards the center.
### 3. Period (T) and Frequency (f)
These terms are crucial for quantifying how often an object completes a revolution.
*
Period (T): The time taken for one complete revolution or cycle.
*
Units: Seconds (s).
*
Frequency (f): The number of revolutions or cycles completed per unit time.
*
Units: Hertz (Hz) or s⁻¹ (revolutions per second).
Relationship between T, f, and ω:
* Since one revolution is 2π radians, and 'T' is the time for one revolution:
ω = (Angular Displacement) / (Time) = 2π / T
So,
ω = 2π/T
* Frequency is the reciprocal of the period:
f = 1/T
* Substituting T = 1/f into the ω equation:
ω = 2πf
Summary Table of Kinematic Variables in UCM:
Variable |
Symbol |
Formula in UCM |
Units |
Direction (if vector) |
Remarks |
|---|
Angular Displacement |
Δθ |
s/r |
rad |
Along axis of rotation (RHR) |
Scalar for large, vector for infinitesimal |
Angular Velocity |
ω |
Δθ/Δt or 2π/T or 2πf |
rad/s |
Along axis of rotation (RHR) |
Constant magnitude in UCM |
Angular Acceleration |
α |
dω/dt |
rad/s² |
Along axis of rotation (RHR) |
Zero in UCM |
Linear Speed |
v |
rω |
m/s |
Not applicable (scalar) |
Constant magnitude in UCM |
Linear Velocity |
v |
Magnitude: rω |
m/s |
Tangent to circle |
Continuously changing direction |
Centripetal Acceleration |
a_c |
v²/r or rω² |
m/s² |
Towards the center |
Always present in UCM |
### Example Problems: Putting Concepts into Practice
Let's solidify our understanding with a couple of examples.
Example 1: A Car on a Circular Track
A car is negotiating a circular track of radius 200 m at a constant speed of 72 km/h.
(a) What is its angular speed?
(b) What is the magnitude of its centripetal acceleration?
(c) What is the period of its motion?
Step-by-step Solution:
Given:
Radius (r) = 200 m
Speed (v) = 72 km/h
First, convert speed to SI units (m/s):
v = 72 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 72 * (5/18) m/s = 4 * 5 = 20 m/s
(a) Angular speed (ω):
We know the relation v = rω.
So, ω = v/r
ω = 20 m/s / 200 m = 0.1 rad/s
(b) Centripetal acceleration (a_c):
Using the formula a_c = v²/r
a_c = (20 m/s)² / 200 m
a_c = 400 m²/s² / 200 m = 2 m/s²
Alternatively, using a_c = rω²:
a_c = 200 m * (0.1 rad/s)²
a_c = 200 * 0.01 m/s² = 2 m/s²
Both formulas give the same result, as expected. The acceleration is directed towards the center of the track.
(c) Period (T):
We know ω = 2π/T.
So, T = 2π/ω
T = 2π / 0.1 rad/s = 20π seconds
T ≈ 20 * 3.14159 ≈ 62.83 seconds
Example 2: An Electron in a Circular Orbit (Conceptual Application)
An electron moves in a circular orbit of radius 0.53 Å around a proton in a hydrogen atom, with a speed of 2.2 × 10⁶ m/s.
(a) What is the centripetal acceleration of the electron?
(b) What provides this acceleration?
Step-by-step Solution:
Given:
Radius (r) = 0.53 Å = 0.53 × 10⁻¹⁰ m (Note: 1 Å = 10⁻¹⁰ m)
Speed (v) = 2.2 × 10⁶ m/s
(a) Centripetal acceleration (a_c):
Using a_c = v²/r
a_c = (2.2 × 10⁶ m/s)² / (0.53 × 10⁻¹⁰ m)
a_c = (4.84 × 10¹² m²/s²) / (0.53 × 10⁻¹⁰ m)
a_c ≈ 9.13 × 10²² m/s²
This is an incredibly large acceleration, showcasing the immense forces at play at the atomic level!
(b) What provides this acceleration?
For any object to undergo circular motion, there must be a force acting towards the center, causing the centripetal acceleration. This force is called the
centripetal force.
In the case of an electron orbiting a proton, the force providing the necessary centripetal acceleration is the
electrostatic force of attraction between the negatively charged electron and the positively charged proton. This force, described by Coulomb's law, is what holds the atom together.
### Conclusion and JEE Focus
Uniform Circular Motion is a cornerstone concept in Physics. While it seems straightforward due to constant speed, the continuous change in velocity and the resulting centripetal acceleration are its defining and most tested aspects.
JEE Main & Advanced Key Points:
* Understand the difference between constant speed and changing velocity in UCM.
* Always remember that
centripetal acceleration (a_c = v²/r = rω²) is *always* directed towards the center.
* Be comfortable with converting between linear and angular variables (v = rω).
* Master the relationships between period (T), frequency (f), and angular velocity (ω = 2π/T = 2πf).
* Later, in dynamics, you will learn about the
centripetal force (F_c = m a_c = m v²/r = m rω²), which is the physical force responsible for causing this acceleration. It's crucial to understand that centripetal force is *not* a new type of force, but rather a *role* played by existing forces (like tension, gravity, friction, electrostatic force, etc.) to cause circular motion.
This detailed understanding of the basic concepts of Uniform Circular Motion will serve as a strong foundation for more advanced topics like non-uniform circular motion, banking of roads, conical pendulums, and gravitational orbits. Keep practicing, and don't hesitate to revisit these fundamentals!