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Angular Kinematics
Angular Position and Displacement
When an object rotates, every point moves through a curved path called an arc. The angle swept by this motion, measured in radians, tells us how far the object has rotated regardless of its size.
$s = r\theta$
The arc length equals radius times angle in radians, linking linear distance to angular rotation.
$s$=arc length (distance along curved path)(meters (m))
$r$=radius of circular path(meters (m))
$\theta$=angular displacement in radians(radians (rad))
$\theta = 2\pi$ rad
→Complete revolution, arc length = circumference ($2\pi r$)
$\theta = 0^\circ$
→No rotation, arc length = 0
Radian Advantage: One radian is the angle where arc length equals radius, making $s = r\theta$ elegantly simple
Conversion Check: 1 revolution = $2\pi$ rad = 360°, so convert revolutions to radians before using formulas
Angular Displacement: $\Delta\theta = \theta_f - \theta_i$, measured counterclockwise as positive
Full Rotation Significance: Points closer to center travel shorter arcs but sweep same angle
Proportionality: $s \propto r$ at fixed $\theta$, and $s \propto \theta$ at fixed $r$ — doubling either doubles arc length
Angular Velocity (The Rate of Spin)
Angular velocity measures how quickly an object rotates, just like linear velocity measures how fast something moves in a straight line. The faster the spin, the greater the angular velocity.
$\omega = \frac{\Delta\theta}{\Delta t}$
Angular velocity equals angular displacement divided by time, showing how rotation rate connects to geometry.
$\omega$=angular velocity (rate of rotation)(radians per second (rad/s))
$\Delta\theta$=change in angular position(radians (rad))
$\Delta t$=time interval(seconds (s))
$\Delta\theta = 2\pi$ rad in 1s
→$\omega = 2\pi$ rad/s (about 6.28 rad/s)
$\omega > 0$
→Counterclockwise rotation (conventionally positive)
Sign Convention: Positive ω = counterclockwise, negative ω = clockwise by convention
RPM to Rad/s: Multiply revolutions per minute by $\frac{2\pi}{60}$ to convert to rad/s
Period Connection: $\omega = \frac{2\pi}{T}$ where T is the time for one complete rotation
Frequency Link: $\omega = 2\pi f$ where f is frequency in hertz (revolutions per second)
The linear velocity of any point on a rotating object depends on both how fast it spins AND how far from the center it is. Points farther out move faster even though they share the same angular velocity.
$v = r\omega$
Tangential speed equals radius times angular velocity, linking linear and rotational motion.
$v$=linear (tangential) velocity(meters per second (m/s))
$r$=distance from rotation axis(meters (m))
$\omega$=angular velocity(radians per second (rad/s))
$r = 0$ (at center)
→$v = 0$ despite any ω - the center doesn't move
$\omega$ doubles
→$v$ doubles - speed is directly proportional to spin rate
Tangential Direction: Velocity is always perpendicular to radius, tangent to the circular path
CD Example: Outer edge of CD (r=6cm) moves 4× faster than inner point (r=1.5cm)
Practical Insight: This is why ceiling fan blades move air faster at their tips than near center
Proportionality: $v \propto r$ at constant $\omega$ — double the radius, double the speed
The direction of angular velocity follows the right-hand rule, creating a vector perpendicular to the plane of rotation. This points along the axis of spin.
$\vec{\omega} = \omega \hat{n}$
Angular velocity as a vector, with direction given by the right-hand rule along the rotation axis.
$\vec{\omega}$=angular velocity vector(rad/s with direction)
$\hat{n}$=unit vector along rotation axis(dimensionless)
$\omega$=magnitude of angular velocity(rad/s)
Counterclockwise rotation
→Vector points toward you (out of page)
Clockwise rotation
→Vector points away from you (into page)
Right-Hand Rule: Curl fingers in rotation direction, thumb points along ω vector
Axis Alignment: ω always points along rotation axis, not in direction of motion
Why This Matters: Essential for torque, angular momentum, and rotational dynamics
Angular Acceleration
When an object's rotation speeds up or slows down, it experiences angular acceleration. This tells us how quickly the angular velocity changes over time.
$\alpha = \frac{\Delta\omega}{\Delta t}$
Angular acceleration equals change in angular velocity divided by time interval.
$\alpha$=angular acceleration(radians per second squared (rad/s²))
$\Delta\omega$=change in angular velocity(radians per second (rad/s))
$\Delta t$=time interval(seconds (s))
$\alpha > 0$
→Angular velocity increasing (speeding up in rotation)
$\alpha < 0$
→Angular velocity decreasing (slowing down)
Speeding Up: ω and α have same sign (both positive or both negative)
Slowing Down: ω and α have opposite signs
Sign Meaning: Positive α means counterclockwise increase OR clockwise decrease (more negative)
Angular acceleration causes tangential acceleration, which changes the speed of points moving along their circular paths. But rotating objects also have centripetal acceleration even at constant speed.
$a_t = r\alpha, \quad a_c = \omega^2 r$
Tangential acceleration from speeding rotation, centripetal acceleration from curved path direction change.
$a_t$=tangential acceleration (speed change)(m/s²)
$a_c$=centripetal acceleration (direction change)(m/s²)
$\alpha$=angular acceleration(rad/s²)
$\omega$=angular velocity(rad/s)
$\alpha = 0$ (constant ω)
→$a_t = 0$ but $a_c \neq 0$ - still has centripetal acceleration
$\omega = 0$ (starting from rest)
→$a_c = 0$ - only tangential acceleration exists
Tangential Direction: Points along tangent, either same direction as v (speeding up) or opposite (slowing)
Centripetal Direction: Always points toward center of circle, perpendicular to tangential
Total Acceleration: Since $\vec{a}_t \perp \vec{a}_c$, the magnitude is $a = \sqrt{a_t^2 + a_c^2}$
Proportionality: $a_c \propto \omega^2$ — doubling angular velocity quadruples centripetal acceleration
Car Turn Example: Steering wheel angle gives centripetal, gas/brake gives tangential
In uniform circular motion, the angular velocity remains constant. While the object continuously accelerates toward the center, it doesn't speed up or slow down in its rotation.
$\alpha = 0, \quad \omega = \text{constant}$
Uniform circular motion has zero angular acceleration but non-zero centripetal acceleration.
$\alpha = 0$=no change in rotation rate(rad/s²)
$a_c = \omega^2 r$=constant centripetal acceleration(m/s²)
Satellite in orbit
→Uniform circular motion with constant ω
CD spinning at constant speed
→$\alpha = 0$ but edge points have $a_c$
Key Insight: Speed doesn't change, but direction does - that's why there's acceleration
Force Requirement: Centripetal force needed to maintain circular path
Energy Conservation: In uniform circular motion, kinetic energy stays constant
The Equations of Rotational Motion
The kinematic analogy between linear and angular motion is perfect - every linear equation has an angular counterpart. When angular acceleration is constant, we can predict rotational motion precisely.
$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$
The 'timeless' rotational kinematic equation, useful when time isn't given or needed.
$\omega_f$=final angular velocity(rad/s)
$\omega_i$=initial angular velocity(rad/s)
$\alpha$=constant angular acceleration(rad/s²)
$\Delta\theta$=angular displacement(rad)
$\omega_i = 0$ (from rest)
→$\omega_f = \sqrt{2\alpha\Delta\theta}$ - simpler form
$\omega_f = 0$ (to rest)
→$\Delta\theta = -\frac{\omega_i^2}{2\alpha}$ - stopping distance
Linear-Angular Map: x→θ, v→ω, a→α, t→t (time stays the same!)
Other Equations: $\omega_f = \omega_i + \alpha t$ and $\Delta\theta = \omega_i t + \frac{1}{2}\alpha t^2$
When to Use: This equation is perfect when you know ω values, α, and need Δθ (or vice versa)
The complete set of rotational kinematic equations mirrors the linear ones perfectly. All require constant angular acceleration to be valid.
$$\begin{aligned}\omega_f &= \omega_i + \alpha t \\\\ \Delta\theta &= \omega_i t + \frac{1}{2}\alpha t^2 \\\\ \Delta\theta &= \left(\frac{\omega_i + \omega_f}{2}\right)t \\\\ \omega_f^2 &= \omega_i^2 + 2\alpha\Delta\theta\end{aligned}$$
Four rotational kinematic equations for constant angular acceleration, analogous to linear kinematics.
$t$=time interval(seconds (s))
$\frac{\omega_i + \omega_f}{2}$=average angular velocity(rad/s)
$\alpha = 0$
→Reduces to $\omega_f = \omega_i$ and $\Delta\theta = \omega t$
Starting from rest
→$\omega_i = 0$, simplifying all equations
Equation 1: Relates velocities, acceleration, and time (no θ needed)
Equation 2: Relates displacement, time, and acceleration (no ω_f needed)
Equation 3: Relates displacement, average velocity, and time (no α needed)
Equation 4: Relates velocities, acceleration, and displacement (no t needed)
Strategy: List knowns, identify unknown, choose equation with all knowns + unknown
Application Strategy
Worked Examples
Problem Statement
A laboratory centrifuge rotates at 120 revolutions per minute (RPM). What is its angular velocity in radians per second?
unit-conversionangular-velocityrpm-to-rads
Strategic Toolkit
$\omega = \frac{\text{revolutions}}{\text{minute}} \times \frac{2\pi\text{ rad}}{1\text{ rev}} \times \frac{1\text{ min}}{60\text{ s}}$
Step 1
Identify Given and Target
We need to convert 120 RPM to rad/s. RPM means revolutions per minute, and we want radians per second.
$\text{Given: } 120\text{ rev/min} \quad \text{Find: } \omega \text{ in rad/s}$
Step 2
Set Up Conversion Factors
Convert revolutions to radians using 2π rad/rev, and minutes to seconds using 1 min/60 s.
$\omega = 120\frac{\text{rev}}{\text{min}} \times \frac{2\pi\text{ rad}}{1\text{ rev}} \times \frac{1\text{ min}}{60\text{ s}}$
Step 3
Perform the Calculation
The revolutions and minutes cancel out, leaving rad/s. Multiply and divide.
$\omega = \frac{120 \times 2\pi}{60}\text{ rad/s} = \frac{240\pi}{60}\text{ rad/s} = 4\pi\text{ rad/s}$
Step 4
Calculate Numerical Value
Evaluate the final expression to get a numerical answer.
$\omega = 4\pi\text{ rad/s} \approx 12.57\text{ rad/s}$
Step 5
Interpret the Result
The centrifuge rotates at 12.57 radians per second, completing about 2 full rotations every second (since 2π rad ≈ 1 rev).
$\omega \approx 12.6\text{ rad/s}$
Problem Statement
A centrifuge starts from rest and accelerates uniformly at 50 rad/s² until it reaches an angular velocity of 500 rad/s. How many radians does it rotate through during this acceleration?
kinematic-equationsconstant-angular-accelerationangular-displacement
Strategic Toolkit
$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta \quad \Rightarrow \quad \Delta\theta = \frac{\omega_f^2 - \omega_i^2}{2\alpha}$
Step 1
List Known Quantities
Extract all given information from the problem. 'From rest' tells us initial angular velocity is zero.
$\omega_i = 0\text{ rad/s}, \quad \omega_f = 500\text{ rad/s}, \quad \alpha = 50\text{ rad/s}^2$
Step 2
Identify Unknown
We need to find the angular displacement Δθ. Time is not asked for and not given.
$\text{Find: } \Delta\theta \text{ (angular displacement in radians)}$
Step 3
Select Appropriate Equation
We have ω_i, ω_f, and α, and need Δθ. The 'timeless' equation fits perfectly.
$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$
Step 4
Rearrange for Δθ
Isolate the angular displacement on one side.
$\Delta\theta = \frac{\omega_f^2 - \omega_i^2}{2\alpha}$
Step 5
Substitute Known Values
Plug in $\omega_i = 0$, $\omega_f = 500$ rad/s, $\alpha = 50$ rad/s².
$\Delta\theta = \frac{(500)^2 - (0)^2}{2(50)} = \frac{250000}{100}$
Step 6
Calculate
Evaluate the expression.
$\Delta\theta = 2500\text{ rad}$
Step 7
Convert to Revolutions (Optional)
For intuition, convert radians to revolutions by dividing by 2π.
$\text{Revolutions} = \frac{2500\text{ rad}}{2\pi\text{ rad/rev}} \approx 398\text{ revolutions}$
Step 8
Interpret the Result
The centrifuge rotates through 2500 radians (about 398 complete rotations) while accelerating to 500 rad/s.
$\Delta\theta = 2500\text{ rad} \approx 398\text{ revolutions}$
Problem Statement
An electric fan rotating at 3 rev/s is switched off. It comes to rest in 18.0 s. Assuming uniform deceleration, find the angular acceleration and the number of revolutions before stopping.
unit-awarenessdecelerationkinematic-equations
Strategic Toolkit
$\alpha = \frac{\omega_f - \omega_i}{t}, \quad \theta = \omega_i t + \frac{1}{2}\alpha t^2$
Step 1
Identify Given (Watch the Units!)
The fan starts at 3 rev/s and stops. Since the problem uses rev/s throughout, we can stay in revolutions — no need to convert to rad/s here.
$\omega_i = 3.0\text{ rev/s}, \quad \omega_f = 0, \quad t = 18.0\text{ s}$
Step 2
Find Angular Acceleration
Use the first kinematic equation rearranged for α.
$\alpha = \frac{\omega_f - \omega_i}{t} = \frac{0 - 3.0}{18.0} = -0.167\text{ rev/s}^2$
Step 3
Find Angular Displacement
Use the second kinematic equation with the values already in rev/s.
$\theta = (3.0)(18.0) + \frac{1}{2}(-0.167)(18.0)^2 = 54 - 27 = 27\text{ rev}$
Step 4
Key Takeaway
When all quantities share the same angular unit (rev/s, rev/s²), you can solve entirely in revolutions. Convert to radians only when mixing with linear quantities (v = rω, a_t = rα).
$\alpha = -0.167\text{ rev/s}^2 \approx -1.05\text{ rad/s}^2, \quad \theta = 27\text{ rev} \approx 170\text{ rad}$
Memorize formulas
Arc-Angle Bridge
Arc length directly connects to angle through radius - the foundation of all angular-linear relationships.
Formula
$s = r\theta$
Linear-Angular Velocity
Tangential speed grows with both spin rate AND distance from center - outer points move faster.
Formula
$v = r\omega$
Angular Velocity Definition
Angular velocity is simply how much angle is swept per unit time.
Formula
$\omega = \frac{d\theta}{dt}$
Angular Acceleration Definition
Angular acceleration is how quickly spin rate changes - the rotational analog of linear acceleration.
Formula
$\alpha = \frac{d\omega}{dt}$
Rotational 'Timeless' Equation
Final velocity from initial velocity, acceleration, and displacement - time not needed!
Formula
$\omega_f^2 = \omega_i^2 + 2\alpha\theta$
Tangential Acceleration
Angular acceleration creates linear speed change along the circular path.
Formula
$a_t = r\alpha$
Centripetal Acceleration
Even at constant speed, rotating objects always accelerate toward the center.
Formula
$a_c = \omega^2 r$
RPM Conversion
Convert revolutions per minute to rad/s — multiply by 2π and divide by 60.
Formula
$\omega = \frac{2\pi N}{60}$
Review
1
A centrifuge rotor of radius 0.12 m accelerates uniformly from rest to 6000 RPM in 8 seconds. What is the tangential acceleration of a point on the rotor rim?
Ans
9.42 m/s²
Review
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Convert the RPM and find angular acceleration
First convert 6000 RPM into rad/s, then use constant-angular-acceleration motion from rest to get $\alpha$.
$$\omega_f = \frac{2\pi(6000)}{60} = 200\pi\ \text{rad/s},\qquad \alpha = \frac{200\pi}{8} = 25\pi\ \text{rad/s}^2$$
Use tangential acceleration at the rim
Tangential acceleration is the angular acceleration multiplied by radius.
$$a_t = \alpha r = (25\pi)(0.12) = 3\pi \approx 9.42\ \text{m/s}^2$$
2
A flywheel rotating at 30 rad/s undergoes constant angular deceleration of 2 rad/s². What is the angular displacement during the time interval from t = 5 s to t = 10 s?
Ans
75 rad
Review
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Find the angular speed at the start of the interval
Use $\omega = \omega_0 + \alpha t$ at $t = 5\,\text{s}$ to get the speed that actually applies over the interval from 5 s to 10 s.
$$\omega_5 = 30 + (-2)(5) = 20\ \text{rad/s}$$
Compute displacement over the next 5 seconds
Now treat the 5-to-10 second interval as a new constant-acceleration segment with initial speed 20 rad/s.
$$\Delta\theta = \omega_5\Delta t + \tfrac12\alpha(\Delta t)^2 = 20(5) + \tfrac12(-2)(5^2) = 75\ \text{rad}$$
3
Which of the following is true for an object in uniform circular motion (constant angular velocity)?
Ans
Angular acceleration is zero, but centripetal acceleration exists
Review
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Apply the definition of uniform circular motion
Constant angular velocity means the rate of rotation is not changing, so the angular acceleration is zero.
$$\alpha = 0$$
Check the acceleration that still remains
Even with constant speed, the direction of velocity changes continuously, so centripetal acceleration is still present.
$$a_c = \omega^2 r \neq 0$$
4
A fan slows down from 5 rad/s to 1 rad/s while rotating through 12 radians. Assuming constant angular acceleration, what is α?
Ans
-1.0 rad/s²
Review
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Use the angular kinematics relation
Since angular acceleration is constant, relate the initial and final speeds through the angular displacement.
$$\omega_f^2 = \omega_i^2 + 2\alpha\Delta\theta$$
Substitute the values and solve
Insert the given speeds and 12 radians of rotation, then isolate $\alpha$.
$$1^2 = 5^2 + 2\alpha(12) \Rightarrow 1 = 25 + 24\alpha \Rightarrow \alpha = -1.0\ \text{rad/s}^2$$
5
A wheel rotates counter-clockwise when viewed from above. Using the right-hand rule, the angular velocity vector points:
Ans
Toward you (out of the page)
Review
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Apply the right-hand rule
Curl the fingers of your right hand in the direction of rotation; the thumb gives the direction of the angular velocity vector.
Counter-clockwise rotation means the thumb points toward you, out of the page.
6
A wheel starts from rest with α = 3 rad/s². After turning through 24 radians, a second identical wheel starts from rest with α = 12 rad/s². Both wheels reach the same final angular velocity at the same instant. What angular displacement does the second wheel cover?
Ans
6 rad
Review
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Use the from-rest kinematics form
For each wheel starting from rest, the final angular speed satisfies $\omega_f^2 = 2\alpha\Delta\theta$, so displacement is inversely proportional to angular acceleration when $\omega_f$ is the same.
$$\Delta\theta \propto \frac{1}{\alpha}$$
Scale the displacement by the acceleration ratio
The second wheel has four times the angular acceleration, so it needs one quarter the angle to reach the same final speed.
$$\Delta\theta_2 = \frac{24}{4} = 6\ \text{rad}$$
7
Two gears are meshed together. Gear A has radius 10 cm and gear B has radius 30 cm. If gear A rotates at 90 rad/s, what is the angular velocity of gear B?
Ans
30 rad/s
Review
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Match tangential speeds at the contact point
Meshed gears do not slip, so the linear speed of the teeth where they touch must be the same for both gears.
$$r_A\omega_A = r_B\omega_B$$
Solve for the larger gear's angular speed
Substitute the radii and the known angular speed of gear A, then divide by the larger radius.
$$10(90) = 30\omega_B \Rightarrow \omega_B = 30\ \text{rad/s}$$
8
A point on a spinning disc has centripetal acceleration 12 m/s² at radius 0.3 m. If the radius is doubled (r = 0.6 m) while keeping the same angular velocity, the centripetal acceleration becomes:
Ans
24 m/s²
Review
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Use the radius dependence of centripetal acceleration
At fixed angular velocity, centripetal acceleration is directly proportional to radius.
$$a_c = \omega^2 r \;\Rightarrow\; a_c \propto r$$
Double the acceleration with the radius
Since the radius doubles from 0.3 m to 0.6 m, the centripetal acceleration also doubles from 12 m/s².
$$a_c' = 2(12) = 24\ \text{m/s}^2$$
9
A disc starts from rest and rotates through θ₁ radians in the first 2 seconds under constant angular acceleration. How many radians does it rotate through in the next 2 seconds (from t = 2s to t = 4s)?
Ans
3θ₁
Review
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Write the angle formula for motion from rest
With constant angular acceleration and zero initial angular speed, the angle grows as the square of time.
$$\theta = \tfrac12\alpha t^2$$
Compare the first and second 2-second intervals
At 2 s the disc has rotated by $\theta_1$, and at 4 s the total angle is four times that value, so the next 2 seconds add the difference.
$$\theta(4) = 4\theta_1 \Rightarrow \theta_{2\to4} = 4\theta_1 - \theta_1 = 3\theta_1$$
10
A wheel decelerates from 10 rad/s to 4 rad/s while completing 7 revolutions. What is the angular acceleration?
Ans
−0.955 rad/s²
Review
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Convert revolutions to radians
The kinematic equation uses angular displacement in radians, so convert the 7 revolutions before substituting.
$$\Delta\theta = 7(2\pi) = 14\pi\ \text{rad}$$
Apply the angular kinematics equation
Use the initial and final angular speeds with the converted displacement, then solve for the constant angular acceleration.
$$4^2 = 10^2 + 2\alpha(14\pi) \Rightarrow \alpha = -\frac{3}{\pi} \approx -0.955\ \text{rad/s}^2$$
Question 1 of 10
0:00
1
A centrifuge rotor of radius 0.12 m accelerates uniformly from rest to 6000 RPM in 8 seconds. What is the tangential acceleration of a point on the rotor rim?