The Central Idea
In uniform circular motion:
- the path is a circle;
- the radius remains constant;
- the speed remains constant;
- the velocity direction continually changes;
- the acceleration points inward toward the center.
Therefore, constant speed does not imply zero acceleration.
Position on a Circle
A circle of radius \(r\) centered at the origin can be parameterized by
\[ \boxed{ x=r\cos\theta, \qquad y=r\sin\theta }. \]The angle \(\theta\) identifies the object's angular position around the circle.
Angular Position
The angle
\[ \theta \]is usually measured from the positive \(x\)-axis.
Positive angular direction is conventionally counterclockwise, while clockwise rotation is represented by decreasing or negative angle.
Radians Are Natural for Circular Motion
Radian measure connects angular displacement directly with arc length.
If an object moves through angle \(\theta\) radians on a circle of radius \(r\), then the arc length is
\[ \boxed{s=r\theta}. \]This direct relationship is one reason radians are the standard unit in higher-level circular kinematics.
One Complete Revolution
One full revolution corresponds to
\[ \boxed{2\pi\text{ radians}} \]or
\[ \boxed{360^\circ}. \]Thus the circumference
\[ 2\pi r \]is exactly the arc length obtained from
\[ s=r(2\pi). \]Angular Displacement
If the angular position changes from \(\theta_i\) to \(\theta_f\), then the angular displacement is
\[ \boxed{ \Delta\theta = \theta_f-\theta_i }. \]Angular displacement can be positive or negative depending on the chosen orientation.
Angular Speed
Angular speed measures how quickly angular position changes.
For uniform circular motion,
\[ \boxed{ \omega = \frac{\Delta\theta}{\Delta t} }. \]If \(\omega\) is constant, then angular position changes linearly with time.
Angular Position as a Function of Time
Under constant angular velocity,
\[ \boxed{ \theta(t) = \theta_0+\omega t }. \]Here:
- \(\theta_0\) is the initial angle;
- \(\omega\) is the constant angular velocity;
- \(t\) is elapsed time.
Vector Position Function
Substitute
\[ \theta(t)=\theta_0+\omega t \]into the circular coordinates.
The position vector becomes
\[ \boxed{ \mathbf{r}(t) = \left\langle r\cos(\theta_0+\omega t), r\sin(\theta_0+\omega t) \right\rangle }. \]A Common Simplified Parameterization
If the object begins at
\[ (r,0) \]when \(t=0\), then
\[ \theta_0=0. \]Therefore,
\[ \boxed{ \mathbf{r}(t) = \langle r\cos(\omega t), r\sin(\omega t) \rangle }. \]Direction of Rotation
If
\[ \omega>0, \]the standard parameterization moves counterclockwise.
If
\[ \omega<0, \]it moves clockwise.
The magnitude
\[ |\omega| \]describes angular speed.
Period
The period \(T\) is the time required for one complete revolution.
Since one revolution is \(2\pi\) radians,
\[ |\omega|T=2\pi. \]Therefore,
\[ \boxed{ T=\frac{2\pi}{|\omega|} }. \]Frequency
The frequency \(f\) is the number of revolutions completed per unit time.
Frequency and period are reciprocals:
\[ \boxed{ f=\frac{1}{T} }. \]Therefore,
\[ \boxed{ |\omega|=2\pi f }. \]Frequency Units
A common frequency unit is the hertz:
\[ \boxed{ 1\text{ Hz} = 1\frac{\text{cycle}}{\text{s}} }. \]Thus a rotating object at \(5\text{ Hz}\) completes 5 revolutions each second.
Example: Period from Angular Speed
Suppose
\[ \omega=4\pi\frac{\text{rad}}{\text{s}}. \]Then
\[ T = \frac{2\pi}{4\pi} \] \[ = \boxed{\frac12\text{ s}}. \]The object completes two revolutions each second.
Example: Frequency from Period
If
\[ T=0.25\text{ s}, \]then
\[ f=\frac{1}{0.25} \] \[ = \boxed{4\text{ Hz}}. \]Tangential Speed
A point moving around a circle travels along the tangent direction.
During angular displacement \(\Delta\theta\), the arc length traveled is
\[ \Delta s=r\Delta\theta. \]Divide by time:
\[ \frac{\Delta s}{\Delta t} = r\frac{\Delta\theta}{\Delta t}. \]Therefore,
\[ \boxed{ v=r|\omega| }. \]Why Radius Affects Tangential Speed
Two points can rotate with the same angular speed while having different tangential speeds.
A point farther from the center travels a larger circumference during each revolution.
Since
\[ v=r|\omega|, \]increasing \(r\) increases tangential speed when angular speed is fixed.
Example: Tangential Speed
A point moves on a circle of radius
\[ r=3\text{ m} \]with angular speed
\[ \omega=2\frac{\text{rad}}{\text{s}}. \]Then
\[ v=r|\omega| \] \[ =3(2) \] \[ \boxed{ v=6\frac{\text{m}}{\text{s}} }. \]Tangential Speed from Circumference and Period
In one revolution, the object travels distance
\[ 2\pi r. \]One revolution takes time \(T\).
Therefore,
\[ \boxed{ v=\frac{2\pi r}{T} }. \]Since \(f=1/T\),
\[ \boxed{ v=2\pi rf }. \]Velocity Vector
Begin with
\[ \mathbf{r}(t) = \langle r\cos(\omega t), r\sin(\omega t) \rangle. \]Differentiate:
\[ \boxed{ \mathbf{v}(t) = \langle -r\omega\sin(\omega t), r\omega\cos(\omega t) \rangle }. \]Velocity Is Tangent to the Circle
The position vector points radially outward from the center.
The velocity vector points along the tangent.
Their dot product is
\[ \mathbf{r}\cdot\mathbf{v} = -r^2\omega\cos(\omega t)\sin(\omega t) + r^2\omega\sin(\omega t)\cos(\omega t). \]Therefore,
\[ \boxed{ \mathbf{r}\cdot\mathbf{v}=0 }. \]Thus position and velocity are perpendicular.
Magnitude of the Velocity Vector
Compute:
\[ |\mathbf{v}(t)| = \sqrt{ r^2\omega^2\sin^2(\omega t) + r^2\omega^2\cos^2(\omega t) }. \]Factor:
\[ |\mathbf{v}(t)| = r|\omega| \sqrt{ \sin^2(\omega t)+\cos^2(\omega t) }. \]Therefore,
\[ \boxed{ |\mathbf{v}|=r|\omega| }. \]The speed is constant.
Acceleration Vector
Differentiate velocity:
\[ \mathbf{a}(t) = \langle -r\omega^2\cos(\omega t), -r\omega^2\sin(\omega t) \rangle. \]Since
\[ \mathbf{r}(t) = \langle r\cos(\omega t), r\sin(\omega t) \rangle, \]we obtain
\[ \boxed{ \mathbf{a}(t) = -\omega^2\mathbf{r}(t) }. \]Acceleration Points Inward
The position vector points outward from the center.
The acceleration vector is
\[ -\omega^2\mathbf{r}. \]The negative sign reverses the direction.
Therefore, acceleration points toward the center of the circle.
Centripetal Acceleration
The inward acceleration in circular motion is called centripetal acceleration.
Its magnitude is
\[ |\mathbf{a}| = r\omega^2. \]Therefore,
\[ \boxed{ a_c=r\omega^2 }. \]Centripetal Acceleration in Terms of Speed
Since
\[ v=r|\omega|, \]we have
\[ \omega^2=\frac{v^2}{r^2}. \]Substitute into
\[ a_c=r\omega^2. \]Then
\[ \boxed{ a_c=\frac{v^2}{r} }. \]What the Centripetal Formula Tells Us
The formula
\[ a_c=\frac{v^2}{r} \]shows that:
- doubling the speed multiplies inward acceleration by \(4\);
- tripling the speed multiplies inward acceleration by \(9\);
- increasing the radius reduces inward acceleration when speed is fixed.
Example: Centripetal Acceleration
An object moves at
\[ 12\frac{\text{m}}{\text{s}} \]around a circle of radius
\[ 8\text{ m}. \]Then
\[ a_c=\frac{12^2}{8} \] \[ =\frac{144}{8} \] \[ \boxed{ a_c=18\frac{\text{m}}{\text{s}^2} }. \]Velocity and Acceleration Are Perpendicular
In uniform circular motion, acceleration points radially inward while velocity points tangentially.
Therefore,
\[ \boxed{ \mathbf{v}\perp\mathbf{a} }. \]This perpendicular acceleration changes the direction of velocity without changing its magnitude.
Why Speed Remains Constant
Acceleration can change speed only through a component parallel or antiparallel to velocity.
In uniform circular motion,
\[ \mathbf{v}\cdot\mathbf{a}=0. \]Therefore, acceleration changes only the direction of the velocity vector, not its magnitude.
Constant Speed Does Not Mean Constant Velocity
Velocity includes both magnitude and direction.
In uniform circular motion:
\[ |\mathbf{v}|=\text{constant}, \]but
\[ \mathbf{v}(t) \]changes continuously because its direction changes.
Therefore,
\[ \boxed{ \mathbf{a}(t)\ne\mathbf{0} }. \]Period, Frequency, Angular Speed, and Tangential Speed
The main circular-motion rate relationships are
\[ \boxed{ f=\frac1T } \] \[ \boxed{ |\omega|=\frac{2\pi}{T}=2\pi f } \] \[ \boxed{ v=r|\omega| } \] \[ \boxed{ v=\frac{2\pi r}{T}=2\pi rf }. \]Revolutions per Minute
Rotational speeds are often given in revolutions per minute, abbreviated rpm.
Since one revolution is \(2\pi\) radians,
\[ 1\frac{\text{rev}}{\text{min}} = 2\pi\frac{\text{rad}}{\text{min}}. \]Converting minutes to seconds gives angular speed in radians per second.
Example: Converting rpm to Angular Speed
Suppose a wheel rotates at
\[ 120\frac{\text{rev}}{\text{min}}. \]Convert:
\[ 120 \frac{\text{rev}}{\text{min}} \cdot 2\pi \frac{\text{rad}}{\text{rev}} \cdot \frac{1\text{ min}}{60\text{ s}}. \]Therefore,
\[ \boxed{ \omega = 4\pi\frac{\text{rad}}{\text{s}} }. \]Example: Rim Speed from rpm
If the same wheel has radius
\[ r=0.5\text{ m}, \]then
\[ v=r\omega \] \[ =0.5(4\pi) \] \[ \boxed{ v=2\pi\frac{\text{m}}{\text{s}} }. \]Number of Revolutions
If frequency is constant,
\[ \boxed{ N=ft }. \]If angular speed is known,
\[ \Delta\theta=\omega t. \]Since one revolution is \(2\pi\) radians,
\[ \boxed{ N=\frac{|\omega|t}{2\pi} }. \]Example: Number of Revolutions
An object rotates at
\[ f=3\text{ Hz} \]for
\[ t=8\text{ s}. \]Then
\[ N=ft \] \[ =3(8) \] \[ \boxed{N=24\text{ revolutions}}. \]Arc Distance Traveled
After \(N\) full revolutions on a circle of radius \(r\),
\[ d=N(2\pi r). \]Therefore,
\[ \boxed{ d=2\pi rN }. \]For constant speed,
\[ d=vt. \]Displacement after a Full Revolution
After one complete revolution, the object returns to its starting point.
Therefore,
\[ \boxed{ \Delta\mathbf{r}=\mathbf{0} }. \]But the distance traveled is
\[ \boxed{2\pi r}. \]This is another clear distinction between distance and displacement.
Average Velocity over a Full Revolution
Since the net displacement is zero,
\[ \mathbf{v}_{\text{avg}} = \frac{\Delta\mathbf{r}}{T} \] \[ = \boxed{\mathbf{0}}. \]Yet the instantaneous speed is nonzero throughout the motion.
Average Speed over a Full Revolution
The distance traveled is
\[ 2\pi r. \]Therefore,
\[ \text{average speed} = \frac{2\pi r}{T}. \]For uniform circular motion, this is simply the constant speed:
\[ \boxed{ v=\frac{2\pi r}{T} }. \]Quarter-Revolution Displacement
Suppose an object moves on a circle of radius \(r\) from
\[ (r,0) \]to
\[ (0,r). \]The displacement vector is
\[ \langle-r,r\rangle. \]Its magnitude is
\[ \sqrt{r^2+r^2} \] \[ = \boxed{r\sqrt2}. \]The arc distance traveled is instead
\[ \boxed{\frac{\pi r}{2}}. \]Acceleration in Terms of Period
Since
\[ |\omega|=\frac{2\pi}{T}, \]substitute into
\[ a_c=r\omega^2. \]Then
\[ \boxed{ a_c = \frac{4\pi^2r}{T^2} }. \]Acceleration in Terms of Frequency
Since
\[ |\omega|=2\pi f, \]we have
\[ a_c=r(2\pi f)^2. \]Therefore,
\[ \boxed{ a_c=4\pi^2rf^2 }. \]Example: Acceleration from Period
An object moves in a circle of radius
\[ r=2\text{ m} \]with period
\[ T=4\text{ s}. \]Then
\[ a_c = \frac{4\pi^2(2)}{4^2} \] \[ = \frac{8\pi^2}{16} \] \[ \boxed{ a_c=\frac{\pi^2}{2}\frac{\text{m}}{\text{s}^2} }. \]Angular Velocity versus Tangential Velocity
Angular velocity describes how rapidly the angle changes:
\[ \omega=\frac{d\theta}{dt}. \]Tangential velocity describes linear motion along the circle.
Their magnitudes are related by
\[ \boxed{ v=r|\omega| }. \]Points on a Rigid Rotating Disk
On an ideal rigid disk, all points complete each revolution in the same amount of time.
Therefore, they have the same angular speed.
But points farther from the center have greater tangential speed because
\[ v=r|\omega|. \]Example: Two Points on the Same Disk
Suppose a disk rotates with angular speed
\[ \omega=5\frac{\text{rad}}{\text{s}}. \]Point \(A\) is at radius
\[ r_A=0.2\text{ m}, \]while point \(B\) is at
\[ r_B=0.6\text{ m}. \]Their tangential speeds are
\[ v_A=0.2(5)=1\frac{\text{m}}{\text{s}} \]and
\[ v_B=0.6(5)=3\frac{\text{m}}{\text{s}}. \]Thus
\[ \boxed{ v_B=3v_A }. \]Common Error: Thinking Constant Speed Means Zero Acceleration
Acceleration measures change in velocity, not merely change in speed.
Since velocity direction changes continually in circular motion,
\[ \boxed{ \mathbf{a}\ne\mathbf{0} }. \]Common Error: Pointing Acceleration Along the Velocity
In uniform circular motion, velocity is tangent to the circle.
Acceleration points toward the center.
Therefore,
\[ \boxed{ \mathbf{v}\perp\mathbf{a} }. \]Common Error: Pointing Centripetal Acceleration Outward
The centripetal acceleration vector points inward.
The word centripetal means center-seeking.
In the standard vector model,
\[ \mathbf{a} = -\omega^2\mathbf{r}, \]and the negative sign confirms the inward direction.
Common Error: Confusing Angular and Tangential Speed
Angular speed has units such as
\[ \frac{\text{rad}}{\text{s}}, \]while tangential speed has units such as
\[ \frac{\text{m}}{\text{s}}. \]They are related by radius:
\[ \boxed{ v=r|\omega| }. \]Common Error: Using Degrees Directly in \(s=r\theta\)
The formula
\[ s=r\theta \]assumes that \(\theta\) is measured in radians.
Degree measures must first be converted to radians.
Common Error: Forgetting the Square in \(a_c=v^2/r\)
Centripetal acceleration depends on the square of speed:
\[ \boxed{ a_c=\frac{v^2}{r} }. \]Doubling speed therefore quadruples inward acceleration.
Common Error: Treating Frequency and Period as the Same Quantity
Period measures time per cycle.
Frequency measures cycles per unit time.
They are reciprocals:
\[ \boxed{ f=\frac1T }. \]Common Error: Forgetting Units When Converting rpm
Rotations per minute must usually be converted through both:
- revolutions to radians;
- minutes to seconds.
Unit cancellation makes the conversion reliable.
A Reliable Procedure for Circular-Rate Problems
- Identify the radius \(r\).
- Identify the given period, frequency, angular speed, or tangential speed.
- Convert angular quantities to radians when needed.
- Use \(f=1/T\) to move between frequency and period.
- Use \(|\omega|=2\pi/T=2\pi f\).
- Use \(v=r|\omega|\) for tangential speed.
- Use \(a_c=v^2/r=r\omega^2\) for inward acceleration.
- Check units and direction.
A Reliable Procedure from a Position Vector
- Write \(\mathbf{r}(t)\).
- Differentiate to obtain \(\mathbf{v}(t)\).
- Take \(|\mathbf{v}(t)|\) to verify constant speed.
- Differentiate again to obtain \(\mathbf{a}(t)\).
- Compare \(\mathbf{a}(t)\) with \(\mathbf{r}(t)\).
- Verify inward direction through \(\mathbf{a}=-\omega^2\mathbf{r}\).
- Compute magnitudes when needed.
Example: Full Circular-Motion Workflow
Let
\[ \mathbf{r}(t) = \langle 5\cos(2t), 5\sin(2t) \rangle. \]Radius:
\[ \boxed{r=5}. \]Angular speed:
\[ \boxed{\omega=2}. \]Velocity:
\[ \mathbf{v}(t) = \langle -10\sin(2t), 10\cos(2t) \rangle. \]Speed:
\[ |\mathbf{v}| = \sqrt{ 100\sin^2(2t)+100\cos^2(2t) } \] \[ = \boxed{10}. \]Acceleration:
\[ \mathbf{a}(t) = \langle -20\cos(2t), -20\sin(2t) \rangle. \]Therefore,
\[ \boxed{ \mathbf{a}(t) = -4\mathbf{r}(t) }. \]Its magnitude is
\[ a_c=r\omega^2 \] \[ =5(4) \] \[ \boxed{a_c=20}. \]Practice 1: Period
An object has angular speed
\[ \omega=6\pi\frac{\text{rad}}{\text{s}}. \]Find its period.
Practice 2: Frequency
A rotating object has period
\[ T=0.2\text{ s}. \]Find its frequency.
Practice 3: Tangential Speed
A point moves on a circle of radius \(4\text{ m}\) with angular speed \(3\text{ rad/s}\). Find its tangential speed.
Practice 4: Centripetal Acceleration
An object moves at \(10\text{ m/s}\) around a circle of radius \(5\text{ m}\). Find the centripetal acceleration.
Practice 5: Angular Speed from Frequency
An object rotates at
\[ f=2\text{ Hz}. \]Find its angular speed.
Practice 6: Revolutions
A disk rotates at \(4\text{ Hz}\) for 6 seconds. How many revolutions does it complete?
Practice 7: Position Vector
Write a position vector for counterclockwise uniform circular motion of radius \(3\) with angular speed \(2\), beginning at \((3,0)\).
Practice 8: Velocity and Speed
Let
\[ \mathbf{r}(t) = \langle 2\cos(3t), 2\sin(3t) \rangle. \]Find \(\mathbf{v}(t)\) and the speed.
Practice 9: Acceleration Vector
For the same motion,
\[ \mathbf{r}(t) = \langle 2\cos(3t), 2\sin(3t) \rangle, \]find \(\mathbf{a}(t)\).
Practice 10: Identify the Error
A student says:
“The object is moving at constant speed, so its acceleration must be zero.”
Explain the error for uniform circular motion.
Practice Answers
Practice 1
\[ T=\frac{2\pi}{6\pi} \] \[ = \boxed{\frac13\text{ s}}. \]Practice 2
\[ f=\frac1T \] \[ =\frac1{0.2} \] \[ \boxed{f=5\text{ Hz}}. \]Practice 3
\[ v=r\omega \] \[ =4(3) \] \[ \boxed{ v=12\frac{\text{m}}{\text{s}} }. \]Practice 4
\[ a_c=\frac{v^2}{r} \] \[ =\frac{10^2}{5} \] \[ = \boxed{ 20\frac{\text{m}}{\text{s}^2} }. \]Practice 5
\[ \omega=2\pi f \] \[ =2\pi(2) \] \[ \boxed{ \omega=4\pi\frac{\text{rad}}{\text{s}} }. \]Practice 6
\[ N=ft \] \[ =4(6) \] \[ \boxed{N=24\text{ revolutions}}. \]Practice 7
\[ \boxed{ \mathbf{r}(t) = \langle 3\cos(2t), 3\sin(2t) \rangle }. \]Practice 8
\[ \mathbf{v}(t) = \langle -6\sin(3t), 6\cos(3t) \rangle. \]Speed:
\[ |\mathbf{v}(t)| = \sqrt{ 36\sin^2(3t)+36\cos^2(3t) } \] \[ = \boxed{6}. \]Practice 9
\[ \mathbf{a}(t) = \boxed{ \langle -18\cos(3t), -18\sin(3t) \rangle }. \]Equivalently,
\[ \boxed{ \mathbf{a}(t) = -9\mathbf{r}(t) }. \]Practice 10
Acceleration measures change in velocity, not merely change in speed.
In uniform circular motion, the speed remains constant but the velocity direction changes continuously.
Therefore, the object has nonzero inward centripetal acceleration:
\[ \boxed{ a_c=\frac{v^2}{r} }. \]What to Remember
Uniform circular motion has constant speed but changing velocity.
The principal relationships are
\[ \boxed{ \theta(t)=\theta_0+\omega t } \] \[ \boxed{ f=\frac1T } \] \[ \boxed{ |\omega|=\frac{2\pi}{T}=2\pi f } \] \[ \boxed{ v=r|\omega| } \] \[ \boxed{ a_c=r\omega^2=\frac{v^2}{r} }. \]In vector form,
\[ \boxed{ \mathbf{r}(t) = \langle r\cos(\omega t), r\sin(\omega t) \rangle } \]and
\[ \boxed{ \mathbf{a}(t) = -\omega^2\mathbf{r}(t) }. \]The most important conceptual facts are:
- velocity is tangent to the circle;
- acceleration points toward the center;
- velocity and acceleration are perpendicular;
- constant speed does not mean constant velocity;
- radians connect angular motion directly with linear distance.
Where We Go Next
Uniform circular motion shows how vector direction can change even when speed remains constant. The next lesson turns to another major two-dimensional motion model in which horizontal and vertical components behave differently.
The next lesson, Projectile Motion, develops motion under approximately constant gravitational acceleration, combining horizontal motion with vertical acceleration to produce parabolic trajectories.