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$\omega = d\theta/dt$ and $\alpha = d\omega/dt$, integrating angular motion, $v = r\omega$, tangential and centripetal acceleration, angular velocity as a vector, and rolling without slipping.
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By the end of this lesson you will be able to describe rotation with derivatives and integrals of an angle, and find the speed and acceleration of any point on a turning or rolling body.
The first lessons of this course found velocity and acceleration by differentiating position and reversed the process by integration. From Physics 1 you know angular displacement in radians, uniform circular motion and centripetal acceleration $v^2/r$. This lesson applies the calculus of kinematics to rotation, and connects the turning of a rigid body to the motion of every point on it, from the axle out to the rim, including the special case of a wheel that rolls along the ground.
| Term | What it means |
|---|---|
| Angular position | $\theta$, in radians, the angle a reference line on the body has turned. |
| Angular velocity | $\omega = d\theta/dt$, in rad/s; its direction is along the axis by the right-hand rule. |
| Angular acceleration | $\alpha = d\omega/dt$, in rad/s². |
| Tangential acceleration | $a_t = r\alpha$, the rate of change of a point's speed along its circle. |
| Centripetal acceleration | $a_c = r\omega^2 = v^2/r$, toward the axis. |
| Rolling without slipping | Motion in which the contact point is momentarily at rest, so $v_{\text{cm}} = R\omega$. |
| Radian | The arc length divided by the radius, so $s = r\theta$; dimensionless. |
Every point of a rigid body rotating about a fixed axis turns through the same angle, so one function $\theta(t)$, in radians, describes the whole motion. The same calculus as linear kinematics gives
$$\omega = \frac{d\theta}{dt}, \qquad \alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2},$$
and integrating runs it backward: $\omega(t) = \omega_0 + \int\alpha\,dt$ and $\theta(t) = \theta_0 + \int\omega\,dt$. For constant $\alpha$, the familiar equations follow, $\omega = \omega_0 + \alpha t$ and $\theta = \theta_0 + \omega_0t + \tfrac{1}{2}\alpha t^2$.
A point at distance $r$ from the axis travels an arc $s = r\theta$, so its speed is $v = r\omega$. Its acceleration has two parts: the tangential part $a_t = r\alpha$, which changes its speed, and the centripetal part $a_c = r\omega^2$, which changes its direction and points toward the axis. Radians make these formulas work: they are a ratio of lengths, so $r\omega$ comes out in m/s.
Another way: picture
Paint a line from the center of a spinning record to its edge. The whole line turns together, so one angle describes it. A speck near the edge races around while a speck near the center barely moves, even though both turn through the same angle each second: speed grows with radius, as $v = r\omega$.
Another way: steps
The radian is defined so that an arc of length $s$ on a circle of radius $r$ subtends $\theta = s/r$. A full turn is $2\pi$ rad. Because it is a ratio of two lengths, the radian has no units, and that is what makes $v = r\omega$ correct: $r$ in meters times $\omega$ in rad/s gives m/s.
Degrees and revolutions per minute are common in practice, and every formula in this lesson needs them converted first: multiply degrees by $\pi/180$, and revolutions per minute by $2\pi/60$. An engine idling at $600$ rpm turns at $62.8$ rad/s. Forgetting the conversion is the most frequent error in rotational problems, and it produces answers wrong by a factor of $57$ or $10$.
When the angular acceleration changes, differentiate and integrate as in linear kinematics. A flywheel spun up by a motor whose torque grows might have $\alpha = 2ct$, giving $\omega = \omega_0 + ct^2$ and $\theta = \omega_0t + ct^3/3$. A turntable with $\theta = at^3 - bt$ changes its direction of spin where $\omega = 3at^2 - b$ passes through zero.
The constant-acceleration formulas hold only when $\alpha$ is constant: for a wheel switched off and slowing uniformly they give the stopping angle $\tfrac{1}{2}\omega_0T$ quickly, but for a wheel slowed by air drag, whose $\alpha$ shrinks as it slows, they do not. The question to ask first is the same as for linear motion: is the angular acceleration constant?
A point on a turning body always has a centripetal acceleration $r\omega^2$ toward the axis, because its velocity keeps changing direction. If the rotation is also speeding up or slowing down, it has a tangential acceleration $r\alpha$ along its path. The two are perpendicular, so the total is $\sqrt{(r\alpha)^2 + (r\omega^2)^2}$.
For most spinning machinery the centripetal part is by far the larger. A car tire at highway speed has $\omega$ near $90$ rad/s, and a point on the tread feels about $2700$ m/s², nearly $280g$, while its tangential acceleration during braking is only a few m/s². That enormous centripetal acceleration is why tires are balanced carefully and why stones are flung from treads.
Angular velocity has a direction: along the axis, given by the right-hand rule. Curl the fingers of the right hand the way the body turns, and the thumb points along $\vec{\omega}$. A wheel on a car moving forward has $\vec{\omega}$ pointing to the car's left. Angular acceleration is the derivative of that vector, along the axis when the axis is fixed.
With the vector, a point's velocity is $\vec{v} = \vec{\omega} \times \vec{r}$, which gives both its size $r\omega$ and its direction, tangent to its circle. This form becomes essential for rigid bodies whose axis moves, such as tops and gyroscopes, and it is the language of the torque and angular momentum lessons that follow.
A wheel rolling on a road combines translation and rotation. If it does not slip, the point touching the road is momentarily at rest, which requires the center to move at $v_{\text{cm}} = R\omega$. The top of the wheel then moves at $2v_{\text{cm}}$: the forward motion of the center plus the forward rim speed.
The rolling condition links the linear and angular motions, so one equation of motion can often replace two. A bicycle's speedometer uses it: a sensor counts wheel turns, and the known radius converts them to distance, which is why a new tire size requires recalibration. When a car's wheels lock and slide, the condition fails, and the tires skid instead of rolling.
Checking an answer. Every angle must be in radians before multiplying by a radius. The centripetal acceleration must point toward the axis and never vanish while the body turns. The contact point of a rolling wheel must have zero velocity. And the linear quantities must scale with $r$: a point twice as far out moves twice as fast.
Rotation rates are measured in many ways. A tachometer counts revolutions of an engine's crankshaft; an optical encoder on a robot joint counts thousands of marks per turn and differentiates the count for $\omega$. The gyroscopes in phones use tiny vibrating structures whose motion is deflected by rotation, and report $\omega$ directly about three axes.
From those readings, software integrates $\omega$ to track the phone's orientation and differentiates it for $\alpha$. When you turn a phone and the screen rotates, the phone has integrated its measured angular velocity to decide it has turned ninety degrees. The same calculus of this lesson, done a thousand times a second, runs inside every drone and game controller.
Every rotating machine turns rotation into linear motion or the reverse, and the link is always $s = r\theta$ and its derivatives. A winch drum of radius $0.20$ m turning at $5.0$ rad/s reels in cable at $1.0$ m/s; if the drum speeds up at $2.0$ rad/s², the load accelerates at $0.40$ m/s², provided the cable does not slip on the drum.
Gears and belts pass rotation between shafts of different radii. A belt that does not slip moves at the same speed over both pulleys, so $r_1\omega_1 = r_2\omega_2$: the smaller pulley turns faster. A bicycle's chain does the same between the chainring and the rear sprocket, which is how the choice of gear trades pedal speed against wheel speed.
In its final spin, a front-loading washer turns its drum at $1000$ to $1400$ rpm. For a drum of radius $0.25$ m at $1200$ rpm, $\omega = 2\pi \times 1200/60 = 126$ rad/s, and the centripetal acceleration at the drum wall is $r\omega^2 \approx 3900$ m/s², about $400g$. The drum wall supplies the inward force that keeps the clothes going around.
Water is different: at the holes in the drum nothing pushes it inward, so it keeps going in a straight line, tangent to the drum, and leaves. Higher spin speeds wring out more water, cutting the energy the dryer needs, and appliance makers such as those in Ohio and Iowa advertise the rpm for that reason. The same physics, at much higher rates, runs the centrifuges that separate blood in hospital labs.
A car's speedometer does not measure speed directly. A sensor on the transmission or at each wheel counts rotations, usually as pulses from a toothed ring, and the car's computer converts the rate to $\omega$. The rolling condition, $v = R\omega$, with the tire's radius, then gives the speed.
That is why fitting larger tires makes a speedometer read low: a tire $3$ percent larger in radius covers $3$ percent more ground per turn, so at an indicated $60$ mph the car is really doing about $62$. Antilock braking systems use the same wheel sensors differently: if one wheel's $\omega$ drops much faster than the car's speed allows, it is about to lock and skid, and the system releases that brake for a moment.
When a wheel spins at a steady rate, its angular acceleration is zero, and it is tempting to conclude that its points have no acceleration. But each point moves in a circle, and its velocity changes direction constantly. That change is the centripetal acceleration $r\omega^2$, directed toward the axis, and it is large for fast rotors.
A second error is to use degrees or revolutions in $v = r\omega$. The formula needs radians, because only then is the arc length $r\theta$. A wheel turning at $2$ revolutions per second has $\omega = 4\pi \approx 12.6$ rad/s, not $2$; using the revolutions directly would understate every speed and acceleration on the wheel by a factor of $2\pi$ or its square.
A disk has $\theta = 2t^3 - 6t$ (rad, s). Differentiate for the angular velocity.
$\omega = 6t^2 - 6$
Power rule.
Differentiate again for the angular acceleration.
$\alpha = 12t$
Derivative of $\omega$.
Find when the disk momentarily stops.
$6t^2 - 6 = 0 \Rightarrow t = 1.0\ \text{s}$
It reverses its direction of spin then.
Evaluate both at $t = 2.0$ s.
$\omega = 18\ \text{rad/s}, \quad \alpha = 24\ \text{rad/s}^2$
Substitute the time.
Find the rim point's accelerations at $r = 0.10$ m.
$a_t = 0.10 \times 24 = 2.4\ \text{m/s}^2, \quad a_c = 0.10 \times 18^2 = 32.4\ \text{m/s}^2$
Tangential and centripetal.
A wheel at $120$ rpm slows uniformly to rest in $10$ s. Convert the rate.
$\omega_0 = \dfrac{2\pi \times 120}{60} = 12.6\ \text{rad/s}$
Revolutions per minute to radians per second.
Find the angular acceleration.
$\alpha = \dfrac{0 - 12.6}{10} = -1.26\ \text{rad/s}^2$
Uniform.
Find the angle turned.
$\theta = \tfrac{1}{2}\omega_0T = \tfrac{1}{2} \times 12.6 \times 10 = 63\ \text{rad}$
Average angular velocity times time.
Convert to revolutions.
$\dfrac{63}{2\pi} = 10\ \text{rev}$
Half of the $20$ it would make at full speed.
Find the tangential deceleration of a rim point at $0.30$ m.
$a_t = 0.30 \times (-1.26) = -0.38\ \text{m/s}^2$
Along the rim, slowing it.
A bicycle wheel of radius $0.34$ m rolls at $8.5$ m/s. Find its angular velocity.
$\omega = \dfrac{v}{R} = \dfrac{8.5}{0.34} = 25\ \text{rad/s}$
Rolling without slipping.
Find the speed of the top of the tire.
$v_{\text{top}} = v + R\omega = 8.5 + 8.5 = 17\ \text{m/s}$
Twice the bicycle's speed.
Find the speed of the contact point.
$v_{\text{bottom}} = v - R\omega = 0$
It does not slide.
Find a rim point's acceleration relative to the axle.
$a = R\omega^2 = 0.34 \times 625 = 213\ \text{m/s}^2$
About $22g$, toward the axle.
Find how many turns it makes in $1.0$ km.
$N = \dfrac{1000}{2\pi \times 0.34} = 468$
Distance over circumference.
Find the wheel's angular acceleration if the bike speeds up at $1.7$ m/s².
$\alpha = \dfrac{a}{R} = \dfrac{1.7}{0.34} = 5.0\ \text{rad/s}^2$
The rolling condition differentiated.
Find the angular acceleration.
$\alpha = \dfrac{50 - 30}{4.0} = 5.0\ \text{rad/s}^2$
Change in rate over time.
Find the angle with the average rate.
$\theta = \dfrac{30 + 50}{2} \times 4.0$
Uniform acceleration.
Evaluate the angle.
A wheel of radius $0.3$ m spins at a steady $6$ rad/s. What is the acceleration of a point on its rim?
Complete the worked solution: a wheel of radius $0.4$ m rolls without slipping at $6$ m/s. Find its angular velocity in rad/s, the speed of the top of the wheel in m/s, and the acceleration of a rim point relative to the axle in m/s².
Divide the speed by the radius.
$\omega = \dfrac{v}{R} =$ w
No slipping: the contact point is momentarily at rest.
Add the rim speed to the axle's speed at the top.
$v_{\text{top}} = v + R\omega =$ t
At the top, rotation and translation point the same way.
Find the centripetal acceleration about the axle.
$a = R\omega^2 = \dfrac{v^2}{R} =$ a
The axle moves steadily, so this is the rim point's full acceleration.
Check the bottom of the wheel.
$v_{\text{bottom}} = v - R\omega = 0$
The contact point does not slide.
Match each rotational quantity to its expression.
| $d\theta/dt$ | $d\omega/dt$ | $r\omega$ | $r\omega^2$ | |
|---|---|---|---|---|
| angular velocity | ||||
| angular acceleration | ||||
| speed at radius r | ||||
| centripetal acceleration |
A turntable's angle is $\theta(t) = 2t^3 - 6t$ (radians, seconds). Fill in its angular velocity in rad/s and angular acceleration in rad/s² at $t = 2$ s, and the tangential acceleration then of a point $0.4$ m from the axis, in m/s².
| value | |
|---|---|
| angular velocity (rad/s) | |
| angular acceleration (rad/s²) | |
| tangential acceleration (m/s²) |
A flywheel spinning at $3$ rad/s at $t = 0$ is driven with angular acceleration $\alpha = 2t$ (rad/s², s). Write its angular velocity, in rad/s, as a formula in $t$.
Answer:
A grinding wheel spinning at $30$ rad/s is switched off and slows at a uniform rate to rest in $3$ s. Through what angle does it turn while stopping, in radians?
Answer: rad while stopping
A front-loading washer made in Ohio spins its drum, of radius $0.25$ m, at $1400$ rpm in the final spin cycle. What is the centripetal acceleration of clothes pressed against the drum wall, in m/s²?
Answer: m/s² at the drum wall
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A flywheel spinning at $8$ rad/s at $t = 0$ is driven with angular acceleration $\alpha = 2t$ (rad/s², s). Write its angular velocity, in rad/s, as a formula in $t$.
Answer:
You can use rotational kinematics. Explain to someone why the top of a rolling bicycle wheel moves twice as fast as the bicycle.
19. Your turn: a fan at $30$ rad/s speeds up uniformly to $50$ rad/s in $4.0$ s. Find its angular acceleration and the angle turned., step 3
$\theta = 160\ \text{rad}$
About $25$ revolutions.