Circular Motion Kinematics — JEE Advanced Physics

01

Introduction to Circular and Curved Motion

Motion Along Curved Trajectories

Most real physical motions do not occur in straight lines. When an object travels along a curved path, its direction of motion changes at every point. Circular motion is the fundamental prototype of two-dimensional curved motion.

Core Principle: In two dimensions, velocity is a vector possessing both magnitude (speed) and direction. A change in either the speed or the direction produces a non-zero acceleration.

Even if an object moves at constant speed along a curve, its velocity vector changes direction continuously. Therefore, any body moving in a curved path is always accelerated.

A car moves around a circular track at a constant speed of 20 m/s. Is the car accelerating?

Decomposition of Acceleration: Tangential and Radial

At any point on a curved path, we define two perpendicular axes. One axis lies along the tangent, and the other lies along the inward normal.

The total acceleration vector on a curved path is the vector sum of the tangential acceleration and the radial acceleration:

\vec{a} = \vec{a}_t + \vec{a}_r

The roles of these two orthogonal components are completely distinct:

Tangential Acceleration ( $\vec{a}_t$ ): Acts along the tangent to the path. It is the rate of change of speed: $a_t = \dfrac{dv}{dt}$
Radial / Centripetal Acceleration ( $\vec{a}_r$ ): Acts perpendicular to the velocity, directed toward the instantaneous center of curvature. It changes the direction of motion: $a_r = \dfrac{v^2}{R}$ where $v$ is the speed and $R$ is the radius of curvature.

Resolution of the total acceleration vector into tangential and radial components on a curved path

Because the two components are mutually perpendicular, the magnitude of the total linear acceleration is given by Pythagoras' theorem:

$a = \sqrt{a_t^2 + a_r^2} = \sqrt{\left(\dfrac{dv}{dt}\right)^2 + \left(\dfrac{v^2}{R}\right)^2}$

The angle $\phi$ between the total acceleration vector and the velocity vector satisfies:

$\tan\phi = \dfrac{a_r}{a_t}$

Live simulation showing velocity and the orthogonal acceleration components along a curved path

Example: Vehicle Accelerating along a Curved Track

A car travels along a circular track of radius $R = 100\,\text{m}$ . At a given instant, its speed is $v = 20\,\text{m/s}$ and its speed is increasing at a constant rate of $3.0\,\text{m/s}^2$ . Find the magnitude and direction of the total acceleration of the car at this instant.

Show the solution

Step 1: Identify the given quantities.

The radius of curvature is $R = 100\,\text{m}$ . The instantaneous speed is $v = 20\,\text{m/s}$ . The rate of change of speed is the tangential acceleration: $a_t = \dfrac{dv}{dt} = 3.0\,\text{m/s}^2$

Step 2: Calculate the radial acceleration.

The radial (centripetal) acceleration is: $a_r = \dfrac{v^2}{R} = \dfrac{(20\,\text{m/s})^2}{100\,\text{m}} = \dfrac{400}{100} = 4.0\,\text{m/s}^2$

Step 3: Calculate the magnitude of the total acceleration.

Since $\vec{a}_t$ and $\vec{a}_r$ are perpendicular, the magnitude is: $a = \sqrt{a_t^2 + a_r^2} = \sqrt{(3.0\,\text{m/s}^2)^2 + (4.0\,\text{m/s}^2)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\,\text{m/s}^2$

Step 4: Find the angle with the direction of motion.

The angle $\phi$ between the total acceleration vector and the velocity vector is: $\tan\phi = \dfrac{a_r}{a_t} = \dfrac{4.0}{3.0} \approx 1.333 \implies \phi = \arctan(1.333) \approx 53.1^\circ$

The total acceleration of the car has a magnitude of $5.0\,\text{m/s}^2$ directed at $53.1^\circ$ inward from the tangent line.

If a particle moves along a curved path and its tangential acceleration is zero (a_t = 0), what does this indicate about its motion?

02

Uniform Circular Motion (UCM)

What is Uniform Circular Motion?

Uniform Circular Motion (UCM) is defined as the motion of an object traveling along a circular trajectory at a constant speed.

Important Distinction: Although the speed

v

is constant, the velocity

\vec{v}

is not constant. Because the direction of motion continuously rotates along the tangent, the velocity vector changes at every instant.

Because the velocity vector changes with time, an acceleration is necessarily present. In uniform circular motion, the tangential acceleration is zero ( $a_t = dv/dt = 0$ ), so the total acceleration is purely radial.

Derivation of Centripetal Acceleration

Let us derive the formula for centripetal acceleration using geometric vector subtraction.

Consider a particle moving with constant speed $v$ in a circle of radius $r$ . In a small time interval $\Delta t$ , the particle moves from position $P_1$ to position $P_2$ , sweeping out an angle $\Delta\theta$ and covering arc length $\Delta s = r\,\Delta\theta$ .

Vector subtraction showing that the change in velocity vector points toward the center of the circle

The change in velocity is the vector difference $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ . Since the speeds are equal ( $|\vec{v}_1| = |\vec{v}_2| = v$ ), the vectors $\vec{v}_1$ and $\vec{v}_2$ form an isosceles triangle with vertex angle $\Delta\theta$ .

For small angles (in radians), the chord length $|\Delta\vec{v}|$ is approximately equal to the arc length of the velocity vector: $|\Delta\vec{v}| \approx v\,\Delta\theta$

Dividing by the time interval $\Delta t$ and taking the limit as $\Delta t \to 0$ : $a_c = \lim_{\Delta t \to 0} \dfrac{|\Delta\vec{v}|}{\Delta t} = v \lim_{\Delta t \to 0} \dfrac{\Delta\theta}{\Delta t} = v\,\omega$

Using the relation between linear speed and angular speed ( $v = r\omega$ ), we obtain the standard expressions for centripetal acceleration:

The magnitude of centripetal acceleration in uniform circular motion is:

a_c = \dfrac{v^2}{r} = \omega^2 r = v\omega

$v$ is the constant linear speed of the particle in meters per second.
$r$ is the radius of the circular path in meters.
$\omega$ is the angular speed in radians per second.
The direction of $\vec{a}_c$ is always directed radially inward toward the center of the circle.

Live simulation showing velocity and centripetal acceleration vectors revolving synchronously

Time Period and Frequency

Two fundamental quantities characterize the periodicity of uniform circular motion:

Time Period ( $T$ ): The time taken by the particle to complete one full revolution ( $2\pi$ radians). $T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{\omega}$ The SI unit of time period is the second ( $\text{s}$ ).
Frequency ( $\nu$ or $f$ ): The number of complete revolutions performed by the particle per unit time. $\nu = \dfrac{1}{T} = \dfrac{\omega}{2\pi} = \dfrac{v}{2\pi r}$ The SI unit of frequency is the hertz ( $\text{Hz}$ ), where $1\,\text{Hz} = 1\,\text{s}^{-1} = 1\,\text{rev/s}$ .

From these definitions, the angular speed can be written directly in terms of frequency: $\omega = 2\pi\nu = \dfrac{2\pi}{T}$

Example: Whirling a Stone on a String

A stone of mass $m = 0.50\,\text{kg}$ is tied to a string of length $r = 0.80\,\text{m}$ and whirled in a horizontal circle at constant speed. It completes 5.0 revolutions in a time interval of 10.0 seconds. Find:

The frequency and time period of the motion.
The linear speed of the stone.
The magnitude and direction of the centripetal acceleration.

Show the solution

Step 1: Calculate the frequency and time period.

The frequency is the number of revolutions per second: $\nu = \dfrac{5.0\,\text{rev}}{10.0\,\text{s}} = 0.50\,\text{rev/s} = 0.50\,\text{Hz}$

The time period is the reciprocal of frequency: $T = \dfrac{1}{\nu} = \dfrac{1}{0.50\,\text{s}^{-1}} = 2.0\,\text{s}$

Step 2: Calculate the angular speed and linear speed.

The angular speed is: $\omega = 2\pi\nu = 2\pi (0.50\,\text{s}^{-1}) = \pi\,\text{rad/s} \approx 3.142\,\text{rad/s}$

The linear speed is: $v = r\omega = (0.80\,\text{m})(\pi\,\text{rad/s}) = 0.80\pi\,\text{m/s} \approx 2.513\,\text{m/s}$

Step 3: Calculate the centripetal acceleration.

The centripetal acceleration is: $a_c = \omega^2 r = (\pi\,\text{rad/s})^2 (0.80\,\text{m}) = 0.80\pi^2\,\text{m/s}^2 \approx 7.90\,\text{m/s}^2$

Using the alternative formula to cross-check: $a_c = \dfrac{v^2}{r} = \dfrac{(2.513\,\text{m/s})^2}{0.80\,\text{m}} = \dfrac{6.315}{0.80} \approx 7.90\,\text{m/s}^2$

The centripetal acceleration has a magnitude of $7.90\,\text{m/s}^2$ and is directed radially inward toward the center of the circle.

A particle moves in a circle of radius r with constant speed v. If the speed is doubled and the radius is halved, by what factor does the centripetal acceleration change?

A particle in uniform circular motion completes 20 revolutions in 4.0 seconds. What is its angular speed $\omega$ ?

03

Angular Variables and Rotational Kinematics

Angular Position and Angular Displacement

To analyze circular and rotational motion conveniently, we describe the position of a particle by an angle rather than Cartesian coordinates $(x, y)$ .

Radian Measure: The angular position

\theta

is measured in radians (

\text{rad}

). A radian is dimensionless because it is the ratio of arc length

s

to radius

r

:

\theta = \dfrac{s}{r}

When the arc length $s$ equals the radius $r$ , the angle subtended is exactly $1\,\text{radian} \approx 57.3^\circ$ . A full revolution corresponds to $2\pi\,\text{radians} = 360^\circ$ .

Geometric definition of angular position and the radian on a circular arc

Angular Displacement ( $\Delta\theta$ ): The change in angular position $\Delta\theta = \theta_2 - \theta_1$ . Counterclockwise angular displacement is conventionally defined as positive, while clockwise displacement is negative.

Vector Nature of Rotations: Finite angular displacements do not obey commutative vector addition (

\vec{\theta}_1 + \vec{\theta}_2 \ne \vec{\theta}_2 + \vec{\theta}_1

) and are therefore scalars. However, infinitesimal angular displacements (

d\vec{\theta}

) commute and are true axial vectors.

Angular Velocity

Average Angular Velocity ( $\omega_{avg}$ ): The total angular displacement divided by the time interval: $\omega_{avg} = \dfrac{\Delta\theta}{\Delta t}$

Instantaneous Angular Velocity ( $\omega$ ): The time derivative of angular position: $\omega = \lim_{\Delta t \to 0} \dfrac{\Delta\theta}{\Delta t} = \dfrac{d\theta}{dt}$

The SI unit of angular velocity is the radian per second ( $\text{rad/s}$ ). Common practical units include revolutions per minute ( $\text{rpm}$ ) and revolutions per second ( $\text{rps}$ ): $1\,\text{rpm} = \dfrac{2\pi\,\text{rad}}{60\,\text{s}} = \dfrac{\pi}{30}\,\text{rad/s} \approx 0.1047\,\text{rad/s}$ $1\,\text{rps} = 2\pi\,\text{rad/s} \approx 6.283\,\text{rad/s}$

Direction of Angular Velocity (Right-Hand Thumb Rule): Curl the fingers of your right hand in the direction of rotation. The outstretched thumb points along the direction of the axial vector $\vec{\omega}$ perpendicular to the plane of rotation.

The right-hand thumb rule determining the direction of the angular velocity and angular acceleration vectors

Angular Acceleration

Average Angular Acceleration ( $\alpha_{avg}$ ): The change in angular velocity divided by the time interval: $\alpha_{avg} = \dfrac{\Delta\omega}{\Delta t}$

Instantaneous Angular Acceleration ( $\alpha$ ): The time derivative of angular velocity: $\alpha = \dfrac{d\omega}{dt} = \dfrac{d^2\theta}{dt^2}$

Using the chain rule of calculus, angular acceleration can also be expressed with respect to angular position: $\alpha = \omega\dfrac{d\omega}{d\theta}$

The SI unit of angular acceleration is radians per second squared ( $\text{rad/s}^2$ ).

Direction and Signs: If the angular speed increases,

\vec{\alpha}

is in the same direction as

\vec{\omega}

(parallel). If the angular speed decreases,

\vec{\alpha}

is in the opposite direction to

\vec{\omega}

(antiparallel).

Angular Velocity About an Arbitrary Reference Point

Angular velocity is not restricted to circular paths. Any particle moving in a plane possesses an instantaneous angular velocity about any chosen reference origin $O$ .

Resolution of particle velocity to compute angular velocity relative to an arbitrary observation point

Let a particle have position vector $\vec{r}$ relative to origin $O$ and velocity $\vec{v}$ . Let $\phi$ be the angle between the extended position vector and $\vec{v}$ . We resolve the velocity into two orthogonal components:

Parallel component ( $v_\parallel = v\cos\phi$ ): Changes the distance between the particle and the origin (the radial rate of change $dr/dt$ ). It does not turn the line of sight.
Perpendicular component ( $v_\perp = v\sin\phi$ ): Rotates the line of sight joining the origin to the particle.

The angular velocity of a particle about an arbitrary origin is the ratio of its perpendicular velocity component to its distance from the origin:

\omega = \dfrac{v_\perp}{r} = \dfrac{v\sin\phi}{r} = \dfrac{|\vec{r} \times \vec{v}|}{r^2}

Example: Angular Velocity of a Particle Moving in a Straight Line

A particle moves along a straight horizontal line with a constant speed of $v = 10.0\,\text{m/s}$ . The perpendicular distance from an observation point $O$ to the line of motion is $d = 4.0\,\text{m}$ . Find the angular velocity of the particle about $O$ :

At the instant the particle is at the point of closest approach.
At the instant the position vector makes an angle of $30.0^\circ$ with the line of motion.

Show the solution

Step 1: Point of closest approach.

At the point of closest approach, the distance is $r = d = 4.0\,\text{m}$ and the velocity is perpendicular to the position vector ( $\phi = 90.0^\circ$ ). Therefore: $v_\perp = v = 10.0\,\text{m/s}$ $\omega = \dfrac{v_\perp}{r} = \dfrac{10.0\,\text{m/s}}{4.0\,\text{m}} = 2.50\,\text{rad/s}$

Step 2: When position vector is at $30.0^\circ$ to the path.

From the right-angled triangle formed by the origin, the closest approach, and the particle position: $\sin(30.0^\circ) = \dfrac{d}{r} \implies r = \dfrac{d}{\sin(30.0^\circ)} = \dfrac{4.0\,\text{m}}{0.500} = 8.0\,\text{m}$

The angle between the position vector and the velocity vector is $\phi = 30.0^\circ$ . The perpendicular velocity component is: $v_\perp = v\sin(30.0^\circ) = (10.0\,\text{m/s})(0.500) = 5.0\,\text{m/s}$

Therefore, the angular velocity at this instant is: $\omega = \dfrac{v_\perp}{r} = \dfrac{5.0\,\text{m/s}}{8.0\,\text{m}} = 0.625\,\text{rad/s}$

As the particle travels further away along the straight line, its angular velocity about $O$ continuously decreases toward zero.

Why are finite angular displacements classified as scalar quantities rather than true vectors?

A ceiling fan rotating counterclockwise begins to slow down after being switched off. What is the direction of its angular acceleration vector?

04

Relating Linear and Angular Variables

Scalar Relations Between Linear and Angular Quantities

When a particle moves along a circular arc of fixed radius $r$ , every linear quantity relates directly to its corresponding angular quantity multiplied by $r$ .

For a particle at radius

r

:

s = r\theta

v = r\omega

a_t = r\alpha

a_c = r\omega^2 = \dfrac{v^2}{r}

$s$ is the linear distance traveled along the circular arc in meters.
$\theta$ is the angular displacement in radians.
$v$ is the instantaneous linear speed in meters per second.
$\omega$ is the instantaneous angular speed in radians per second.
$a_t$ is the tangential acceleration in meters per second squared.
$\alpha$ is the angular acceleration in radians per second squared.
$a_c$ is the centripetal acceleration in meters per second squared.

Rigid Body Insight: On a rotating rigid body, all particles share the exact same angular velocity

\omega

and angular acceleration

\alpha

. However, particles situated further from the axis of rotation possess proportionally larger linear speeds (

v \propto r

).

Vector Formulations: Cross Products

In full three-dimensional vector notation, linear velocity and acceleration are expressed as vector cross products.

Three-dimensional vector cross products relating linear velocity and acceleration to angular vectors

1. Linear Velocity Vector: The linear velocity vector $\vec{v}$ is the cross product of the angular velocity vector $\vec{\omega}$ and the position vector $\vec{r}$ : $\vec{v} = \vec{\omega} \times \vec{r}$

Because $\vec{\omega}$ is perpendicular to the plane of the circle and $\vec{r}$ lies in the plane, the angle between them is $90^\circ$ . The magnitude is $|\vec{v}| = \omega r\sin(90^\circ) = \omega r$ , and the right-hand rule confirms that $\vec{v}$ is directed along the tangent.

2. Linear Acceleration Vector: Differentiating the velocity vector with respect to time using the product rule gives: $\vec{a} = \dfrac{d\vec{v}}{dt} = \dfrac{d}{dt}(\vec{\omega} \times \vec{r}) = \left(\dfrac{d\vec{\omega}}{dt} \times \vec{r}\right) + \left(\vec{\omega} \times \dfrac{d\vec{r}}{dt}\right)$ Since $d\vec{\omega}/dt = \vec{\alpha}$ and $d\vec{r}/dt = \vec{v}$ , this simplifies to:

The total linear acceleration vector is:

\vec{a} = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v}) = \vec{a}_t + \vec{a}_c

Tangential Acceleration Vector: $\vec{a}_t = \vec{\alpha} \times \vec{r}$ , acting along the tangent.
Centripetal Acceleration Vector: $\vec{a}_c = \vec{\omega} \times \vec{v} = \vec{\omega} \times (\vec{\omega} \times \vec{r})$ , pointing radially inward toward the center.

Total Linear Acceleration in Non-Uniform Circular Motion

In non-uniform circular motion, the particle speeds up or slows down while turning. Both $a_t$ and $a_c$ are simultaneously non-zero.

Orthogonal vector addition of tangential and centripetal acceleration in non-uniform circular motion

Because $\vec{a}_t$ and $\vec{a}_c$ are mutually orthogonal ( $\vec{a}_t \cdot \vec{a}_c = 0$ ), the total magnitude of acceleration is: $a = \sqrt{a_t^2 + a_c^2} = \sqrt{(r\alpha)^2 + (r\omega^2)^2} = r\sqrt{\alpha^2 + \omega^4}$

The angle $\beta$ between the total acceleration vector and the inward radial direction satisfies: $\tan\beta = \dfrac{a_t}{a_c} = \dfrac{r\alpha}{r\omega^2} = \dfrac{\alpha}{\omega^2}$

Example: Acceleration on a Rotating Flywheel

A flywheel of radius $R = 0.50\,\text{m}$ starts from rest at $t = 0$ and accelerates with constant angular acceleration $\alpha = 2.0\,\text{rad/s}^2$ . At time $t = 3.0\,\text{s}$ , find for a point on the outer rim of the flywheel:

The angular speed and linear speed.
The tangential and centripetal accelerations.
The magnitude of the total linear acceleration and its angle with the inward radius.

Show the solution

Step 1: Calculate angular speed and linear speed.

The angular speed at $t = 3.0\,\text{s}$ starting from rest is: $\omega = \omega_0 + \alpha t = 0 + (2.0\,\text{rad/s}^2)(3.0\,\text{s}) = 6.0\,\text{rad/s}$

The linear speed at the rim is: $v = R\omega = (0.50\,\text{m})(6.0\,\text{rad/s}) = 3.0\,\text{m/s}$

Step 2: Calculate tangential and centripetal acceleration components.

The tangential acceleration is: $a_t = R\alpha = (0.50\,\text{m})(2.0\,\text{rad/s}^2) = 1.0\,\text{m/s}^2$

The centripetal acceleration is: $a_c = R\omega^2 = (0.50\,\text{m})(6.0\,\text{rad/s})^2 = (0.50)(36.0) = 18.0\,\text{m/s}^2$

Step 3: Calculate the total linear acceleration.

The magnitude of the total acceleration is: $a = \sqrt{a_t^2 + a_c^2} = \sqrt{(1.0\,\text{m/s}^2)^2 + (18.0\,\text{m/s}^2)^2} = \sqrt{1.0 + 324.0} = \sqrt{325.0} \approx 18.03\,\text{m/s}^2$

The angle $\beta$ with the inward radius vector is: $\tan\beta = \dfrac{a_t}{a_c} = \dfrac{1.0}{18.0} \approx 0.0556 \implies \beta = \arctan(0.0556) \approx 3.18^\circ$

At this instant, the centripetal acceleration dominates strongly over the tangential component, so the total acceleration points almost directly toward the center.

Given position vector r = 2 i meters and angular velocity omega = 3 k rad/s, what is the linear velocity vector?

Two children sit on a rotating merry-go-round. Child A is at distance r from the center, and Child B is at distance 2r. How do their centripetal accelerations compare?

05

Equations of Motion for Constant Angular Acceleration

Kinematic Equations for Constant Angular Acceleration

Under constant angular acceleration ( $\alpha = \text{constant}$ ), the rotational equations of motion match linear 1D kinematics exactly.

The five fundamental equations of angular motion for constant

\alpha

are:

1.\quad \omega = \omega_0 + \alpha t

2.\quad \theta = \omega_0 t + \tfrac{1}{2}\alpha t^2

3.\quad \omega^2 = \omega_0^2 + 2\alpha\theta

4.\quad \theta = \left(\dfrac{\omega_0 + \omega}{2}\right)t

5.\quad \theta_n = \omega_0 + \dfrac{\alpha}{2}(2n - 1)

$\omega_0$ is the initial angular velocity in radians per second.
$\omega$ is the final angular velocity after time $t$ in radians per second.
$\theta$ is the angular displacement in radians.
$\alpha$ is the constant angular acceleration in radians per second squared.
$t$ is the time interval in seconds.
$\theta_n$ is the angular displacement during the $n^{\text{th}}$ second.

Linear to Angular Mapping: You can convert any 1D linear kinematic equation to an angular kinematic equation by replacing linear displacement

x \to \theta

, velocity

v \to \omega

, and acceleration

a \to \alpha

.

Every linear quantity and equation has a rotational twin. The bridge column converts one column into the other through the radius r. — Every linear quantity and equation has a rotational twin. The bridge column converts one column into the other through the radius $r$ .

Graphical Analysis of Angular Motion

Graphs of angular kinematics provide valuable geometric insights into rotational motion.

Angular velocity versus time graph illustrating slope as angular acceleration and area as angular displacement

Key graphical relationships include:

Slope of $\theta-t$ Graph: The instantaneous slope equals the angular velocity: $\dfrac{d\theta}{dt} = \omega$
Slope of $\omega-t$ Graph: The instantaneous slope equals the angular acceleration: $\dfrac{d\omega}{dt} = \alpha$
Area under $\omega-t$ Graph: The definite integral (area under the curve) from $t_1$ to $t_2$ equals the angular displacement: $\Delta\theta = \int_{t_1}^{t_2} \omega\,dt$
Area under $\alpha-t$ Graph: The area under the curve equals the change in angular velocity: $\Delta\omega = \int_{t_1}^{t_2} \alpha\,dt$

Variable Angular Acceleration: Calculus Methods

If the angular acceleration $\alpha$ is not constant, the standard kinematic formulas cannot be used. You must solve the differential equations using calculus:

If $\alpha = f(t)$ : $\omega(t) = \omega_0 + \int_0^t f(t')\,dt', \qquad \theta(t) = \theta_0 + \int_0^t \omega(t')\,dt'$
If $\alpha = g(\theta)$ : Use $\alpha = \omega\dfrac{d\omega}{d\theta}$ : $\int_{\omega_0}^\omega \omega'\,d\omega' = \int_0^\theta g(\theta')\,d\theta' \implies \dfrac{1}{2}(\omega^2 - \omega_0^2) = \int_0^\theta g(\theta')\,d\theta'$
If $\alpha = h(\omega)$ : Use $\alpha = \dfrac{d\omega}{dt}$ : $t = \int_{\omega_0}^\omega \dfrac{d\omega'}{h(\omega')}$

Example: Deceleration and Revolution Count of a Motor

An electric motor rotates at $1800\,\text{rpm}$ . When switched off, it comes to rest in $20.0\,\text{s}$ with constant angular deceleration. Find:

The angular acceleration of the motor.
The total number of revolutions made before coming to rest.

The situation before any algebra. The angular velocity falls along a straight line, and the slope and the shaded area are the two unknowns.

Show the solution

Step 1: Convert initial angular velocity to SI units.

The initial angular speed is: $\omega_0 = 1800\,\text{rpm} = 1800 \times \dfrac{2\pi}{60}\,\text{rad/s} = 60\pi\,\text{rad/s} \approx 188.50\,\text{rad/s}$ The final angular speed is $\omega = 0$ at $t = 20.0\,\text{s}$ .

Step 2: Calculate the angular acceleration.

Using the first equation of motion: $\omega = \omega_0 + \alpha t \implies 0 = 60\pi + \alpha(20.0)$ $\alpha = -\dfrac{60\pi}{20.0} = -3\pi\,\text{rad/s}^2 \approx -9.42\,\text{rad/s}^2$ The negative sign indicates angular deceleration (retardation).

Step 3: Calculate the total angular displacement.

Using the average angular velocity formula: $\theta = \left(\dfrac{\omega_0 + \omega}{2}\right)t = \left(\dfrac{60\pi + 0}{2}\right)(20.0) = (30\pi)(20.0) = 600\pi\,\text{rad} \approx 1884.96\,\text{rad}$

Step 4: Convert angular displacement into revolutions.

Since each full revolution equals $2\pi\,\text{radians}$ : $n = \dfrac{\theta}{2\pi} = \dfrac{600\pi\,\text{rad}}{2\pi\,\text{rad/rev}} = 300\,\text{revolutions}$ The motor makes exactly $300$ complete revolutions before coming to rest.

A wheel starts from rest with constant angular acceleration. If it turns through angle theta_1 in the first 2 seconds, what total angle does it turn through in the first 4 seconds?

The angular acceleration of a body is given by alpha = k theta, where k is a positive constant. Which method must be used to find omega as a function of theta?

06

Curved-Path Motion and Radius of Curvature

What is an Osculating Circle?

Any smooth curved path can be approximated locally at any point $P$ by a circle called the osculating circle.

Geometry of the osculating circle and radius of curvature at a point on a curve

The osculating circle is the unique circle that:

Touches the curve at point $P$ with a common tangent line.
Possesses the identical curvature (rate of turning of the tangent) as the curve at $P$ .
Has its center at the center of curvature ( $C$ ) located along the inward normal line.

The radius of this circle is defined as the radius of curvature ( $R_c$ ) of the trajectory at point $P$ .

Physical Definition of Radius of Curvature

When a particle moves along any curved path with instantaneous speed $v$ , its normal (perpendicular) acceleration $a_\perp$ is directed toward the instantaneous center of curvature:

The physical formula for the radius of curvature at any point is:

R_c = \dfrac{v^2}{a_\perp} = \dfrac{v^2}{|\vec{a} \times \hat{v}|}

$v$ is the instantaneous speed of the particle in meters per second.
$a_\perp$ is the component of total acceleration perpendicular to the velocity vector in meters per second squared.
$\hat{v} = \vec{v}/|\vec{v}|$ is the unit vector along the instantaneous velocity.

Special Geometries:

Circle of radius $R$ : The normal acceleration is everywhere $v^2/R$ , so $R_c = R = \text{constant}$ .
Straight line: The normal acceleration is zero ( $a_\perp = 0$ ), so $R_c = \infty$ . A straight line has zero curvature.

Cartesian Formula for Radius of Curvature

If the trajectory is given by a function $y = f(x)$ in Cartesian coordinates, the radius of curvature is:

In Cartesian coordinates, the radius of curvature is:

R_c = \dfrac{\left[1 + \left(\dfrac{dy}{dx}\right)^2\right]^{3/2}}{\left|\dfrac{d^2y}{dx^2}\right|}

At a local maximum or minimum (where slope $dy/dx = 0$ ), the expression reduces simply to: $R_c = \dfrac{1}{|d^2y/dx^2|}$

Radius of Curvature in Projectile Motion

Consider a projectile launched with speed $u$ at an elevation angle $\theta$ above horizontal ground under uniform gravity $g$ .

Determination of the radius of curvature at key points of a projectile trajectory

Let us determine the radius of curvature at two critical points:

At the Highest Point (Apex):
- The velocity is purely horizontal: $v = u\cos\theta$ .
- Gravity acts vertically downward, so $\vec{g}$ is strictly perpendicular to $\vec{v}$ : $a_\perp = g$ .
- The radius of curvature at the apex is: $R_{\text{apex}} = \dfrac{v^2}{a_\perp} = \dfrac{(u\cos\theta)^2}{g} = \dfrac{u^2\cos^2\theta}{g}$
At the Point of Launch ( $t = 0$ ):
- The speed is $v = u$ .
- The component of gravity perpendicular to the initial velocity is $a_\perp = g\cos\theta$ .
- The radius of curvature at launch is: $R_{\text{launch}} = \dfrac{v^2}{a_\perp} = \dfrac{u^2}{g\cos\theta}$

The ratio of the radius of curvature at launch to that at the apex is:

\dfrac{R_{\text{launch}}}{R_{\text{apex}}} = \dfrac{u^2/(g\cos\theta)}{u^2\cos^2\theta/g} = \dfrac{1}{\cos^3\theta} = \sec^3\theta

Example 1: Curvature of a Projectile Trajectory

A projectile is launched from the ground with an initial speed of $u = 20.0\,\text{m/s}$ at an angle of $\theta = 60.0^\circ$ above the horizontal. Take $g = 9.80\,\text{m/s}^2$ . Calculate:

The radius of curvature at the highest point of its trajectory.
The radius of curvature at the instant of projection.
The ratio $R_{\text{launch}} / R_{\text{apex}}$ .

Show the solution

Step 1: Radius of curvature at the apex.

At the apex, the speed is $v = u\cos(60.0^\circ) = (20.0\,\text{m/s})(0.500) = 10.0\,\text{m/s}$ . The perpendicular acceleration is $a_\perp = g = 9.80\,\text{m/s}^2$ : $R_{\text{apex}} = \dfrac{v^2}{a_\perp} = \dfrac{(10.0\,\text{m/s})^2}{9.80\,\text{m/s}^2} = \dfrac{100}{9.80} \approx 10.20\,\text{m}$

Step 2: Radius of curvature at launch.

At the launch point, the speed is $u = 20.0\,\text{m/s}$ . The perpendicular component of gravity is $a_\perp = g\cos(60.0^\circ) = (9.80)(0.500) = 4.90\,\text{m/s}^2$ : $R_{\text{launch}} = \dfrac{u^2}{a_\perp} = \dfrac{(20.0\,\text{m/s})^2}{4.90\,\text{m/s}^2} = \dfrac{400}{4.90} \approx 81.63\,\text{m}$

Step 3: Calculate the ratio.

$\dfrac{R_{\text{launch}}}{R_{\text{apex}}} = \dfrac{81.63\,\text{m}}{10.20\,\text{m}} = 8.00$ Notice that $\sec^3(60.0^\circ) = (2)^3 = 8.00$ , confirming the analytical formula exactly.

Example 2: Radius of Curvature of a Parabolic Curve

A particle moves along a parabolic wire path defined by the equation $y = 4.0x - x^2$ (where $x$ and $y$ are in meters). Find the radius of curvature of the path at its vertex.

Show the solution

Step 1: Find the vertex coordinates.

Differentiating with respect to $x$ : $\dfrac{dy}{dx} = 4.0 - 2.0x$ Setting $dy/dx = 0$ to locate the vertex: $4.0 - 2.0x = 0 \implies x = 2.0\,\text{m}$

Step 2: Calculate the second derivative.

$\dfrac{d^2y}{dx^2} = -2.0\,\text{m}^{-1} \implies \left|\dfrac{d^2y}{dx^2}\right| = 2.0\,\text{m}^{-1}$

Step 3: Apply the Cartesian curvature formula.

At $x = 2.0\,\text{m}$ , the first derivative is $dy/dx = 0$ : $R_c = \dfrac{\left[1 + (0)^2\right]^{3/2}}{2.0} = \dfrac{1.0}{2.0} = 0.50\,\text{m}$

The radius of curvature of the wire at its vertex is $0.50\,\text{m}$ .

At an inflection point on a smooth curved trajectory where d^2 y / dx^2 = 0, what is the radius of curvature?

Where is the radius of curvature minimum along a standard parabolic projectile trajectory?

07

Chapter Summary and Key Formula Reference

1. Linear and Angular Kinematic Analogies

Kinematic Quantity	Linear Motion (1D)	Rotational Motion	Connection (Radius $r$ )
Position / Displacement	$x$ or $s$ (m)	$\theta$ (rad)	$s = r\theta$
Velocity	$v = \dfrac{dx}{dt}$ (m/s)	$\omega = \dfrac{d\theta}{dt}$ (rad/s)	$v = r\omega$
Tangential Acceleration	$a = \dfrac{dv}{dt}$ (m/s $^2$ )	$\alpha = \dfrac{d\omega}{dt}$ (rad/s $^2$ )	$a_t = r\alpha$
Centripetal Acceleration	$-$	$a_c = \omega^2 r = \dfrac{v^2}{r}$	$a_c = v\omega$
Kinematic Equation 1	$v = v_0 + at$	$\omega = \omega_0 + \alpha t$	-
Kinematic Equation 2	$x = v_0 t + \tfrac{1}{2}at^2$	$\theta = \omega_0 t + \tfrac{1}{2}\alpha t^2$	-
Kinematic Equation 3	$v^2 = v_0^2 + 2ax$	$\omega^2 = \omega_0^2 + 2\alpha\theta$	-
Vector Formulation	$\vec{v}$	$\vec{\omega}$	$\vec{v} = \vec{\omega} \times \vec{r}$
Total Acceleration	$\vec{a} = \dfrac{d\vec{v}}{dt}$	$\vec{a} = \vec{a}_t + \vec{a}_c$	$\vec{a} = (\vec{\alpha} \times \vec{r}) + (\vec{\omega} \times \vec{v})$

Radius of curvature of any curved path: physical definition $R_c = \dfrac{v^2}{a_\perp}$ , Cartesian formula $R_c = \dfrac{[1 + (y')^2]^{3/2}}{|y''|}$ .

A particle moves in a circle of radius r. If its speed increases at a constant rate a_t, what is the magnitude of its total acceleration when its speed reaches v?

On a graph of angular velocity $\omega$ against time $t$ , what does the area under the curve represent?