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Inertial frames, the third law and momentum, and the second law in polar coordinates: $a_r = \ddot{r} - r\dot{\phi}^2$ and $a_\phi = r\ddot{\phi} + 2\dot{r}\dot{\phi}$.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to write Newton's second law in polar coordinates, compute the centripetal and Coriolis terms, and use the two equations to find a motion and the constraint force that produces it.
From Physics C you can write Newton's second law as a differential equation in Cartesian components and solve it for constant and velocity-dependent forces. You know uniform circular motion, with centripetal acceleration $v^2/r$, and you can differentiate products and compositions. This lesson rewrites the second law in polar coordinates, where circular and orbital motion become simple, and shows where the centripetal acceleration comes from.
| Term | What it means |
|---|---|
| Inertial frame | A frame in which Newton's first law holds: a free particle moves in a straight line at constant speed. |
| Polar coordinates | $(r, \phi)$: the distance from the origin and the angle from the $x$ axis. |
| Unit vectors $\hat{r}$ and $\hat{\phi}$ | Directions outward and around, which turn as the particle moves. |
| Radial component | The part along $\hat{r}$: $a_r = \ddot{r} - r\dot{\phi}^2$. |
| Transverse component | The part along $\hat{\phi}$: $a_\phi = r\ddot{\phi} + 2\dot{r}\dot{\phi}$. |
| Centripetal term | $-r\dot{\phi}^2$, the inward acceleration of anything that turns. |
| Coriolis term | $2\dot{r}\dot{\phi}$, the sideways acceleration of something moving outward while it turns. |
Newton's second law, $\vec{F} = m\ddot{\vec{r}}$, is one vector equation. In Cartesian coordinates it splits at once into $F_x = m\ddot{x}$ and $F_y = m\ddot{y}$, because the unit vectors $\hat{x}$ and $\hat{y}$ never change. In polar coordinates the position is $\vec{r} = r\hat{r}$, and the unit vector $\hat{r}$ points wherever the particle is, so it turns as the particle moves. Differentiating it gives
$$\frac{d\hat{r}}{dt} = \dot{\phi}\,\hat{\phi}, \qquad \frac{d\hat{\phi}}{dt} = -\dot{\phi}\,\hat{r},$$
and differentiating $\vec{r} = r\hat{r}$ twice with these rules gives the velocity and the acceleration:
$$\vec{v} = \dot{r}\,\hat{r} + r\dot{\phi}\,\hat{\phi}, \qquad \vec{a} = (\ddot{r} - r\dot{\phi}^2)\,\hat{r} + (r\ddot{\phi} + 2\dot{r}\dot{\phi})\,\hat{\phi}.$$
The second law becomes two equations, $F_r = m(\ddot{r} - r\dot{\phi}^2)$ and $F_\phi = m(r\ddot{\phi} + 2\dot{r}\dot{\phi})$. The extra terms are the centripetal term $-r\dot{\phi}^2$ and the Coriolis term $2\dot{r}\dot{\phi}$. Neither is a force: each is part of the acceleration, produced by the turning of the directions it is measured along.
Another way: picture
Stand at the origin and watch a particle. The arrow from you to it is $\hat{r}$, and the arrow at right angles, pointing the way it goes around, is $\hat{\phi}$. When the particle moves around, both arrows swing with it. A particle on a circle at steady speed has constant $r$ and constant $\dot{\phi}$, yet its velocity keeps changing direction; the centripetal term is that change, seen in turning axes.
Another way: steps
Write the unit vectors in Cartesian form: $\hat{r} = \cos\phi\,\hat{x} + \sin\phi\,\hat{y}$ and $\hat{\phi} = -\sin\phi\,\hat{x} + \cos\phi\,\hat{y}$. Differentiating the first with the chain rule gives $\dot{\phi}(-\sin\phi\,\hat{x} + \cos\phi\,\hat{y}) = \dot{\phi}\,\hat{\phi}$, and the second gives $-\dot{\phi}\,\hat{r}$ the same way. A unit vector cannot change length, so its derivative is always at right angles to it, and the rate is the rate of turning.
Now apply the product rule to $\vec{v} = \dot{r}\hat{r} + r\dot{\phi}\hat{\phi}$. The first term gives $\ddot{r}\hat{r} + \dot{r}\dot{\phi}\hat{\phi}$. The second gives $\dot{r}\dot{\phi}\hat{\phi} + r\ddot{\phi}\hat{\phi} - r\dot{\phi}^2\hat{r}$. Collecting terms, the two copies of $\dot{r}\dot{\phi}\hat{\phi}$ add to make the factor of two in the Coriolis term: one copy comes from the radial velocity turning, and the other from the transverse velocity $r\dot{\phi}$ growing as $r$ grows.
Checking an answer. Three limits catch most slips. On a circle at steady speed, $a_r$ must reduce to $-v^2/r$ with $v = r\dot{\phi}$. For motion along a fixed ray, $\dot{\phi} = 0$ and both components must reduce to the one-dimensional $\ddot{r}$ and zero. And every term must have units of m/s²: $r\dot{\phi}^2$ is m times (rad/s)², and the radian is dimensionless.
The first law does more than restate the second with zero force. It defines the frames in which the second holds. An inertial frame is one in which a particle with no net force moves in a straight line at constant speed. A frame attached to a merry-go-round is not inertial: a puck sliding freely on the platform curves as seen by the riders.
This is the reason the Coriolis and centripetal terms appear in the acceleration and not as forces. The polar coordinates here are measured in an inertial frame; only the directions $\hat{r}$ and $\hat{\phi}$ turn. A later lesson writes the laws in a rotating frame, where the same terms move to the other side of the equation and reappear as the centrifugal and Coriolis forces. The Earth's surface is nearly inertial for a baseball, and noticeably not for a Foucault pendulum or a hurricane.
For a system of particles, the second law applied to each and summed gives $\dot{\vec{P}} = \vec{F}^{\text{ext}} + \sum \vec{F}_{\text{internal}}$. The third law says the internal forces come in equal and opposite pairs, so they cancel, and the total momentum changes only through external forces. With no external force, momentum is conserved.
The third law is not universal. Two moving charges exert magnetic forces on each other that are not equal and opposite, and the missing momentum is carried by the electromagnetic field. For contact forces, gravity and electrostatics the third law holds, and the stronger form, that the pair acts along the line joining the particles, also gives conservation of angular momentum. Everything in this course assumes that form unless a lesson says otherwise.
The transverse component can be written $a_\phi = \frac{1}{r}\frac{d}{dt}(r^2\dot{\phi})$, which you can check by differentiating the product. When the force always points toward the origin, a central force, $F_\phi = 0$ and so $r^2\dot{\phi}$ is constant.
That constant is the angular momentum per unit mass, $\ell/m$. It also measures the rate at which the line from the origin sweeps out area, $\frac{dA}{dt} = \frac{1}{2}r^2\dot{\phi}$. So a planet under the Sun's gravity sweeps equal areas in equal times, which is Kepler's second law, obtained in one line from the transverse equation. The same equation explains why a skater spins faster when pulling in her arms, and why the puck on the string speeds up as the string is pulled in: $\dot{\phi}$ must grow as $1/r^2$.
A skateboard rolls without friction inside a half-pipe of radius $R$. Put the origin at the center of the circle, so $r = R$ always and the angle $\phi$ is measured from the bottom. The normal force $N$ points toward the center, along $-\hat{r}$. Gravity has a radial part $mg\cos\phi$ outward and a transverse part $-mg\sin\phi$ back toward the bottom.
The transverse equation is $mR\ddot{\phi} = -mg\sin\phi$: the equation of a pendulum. For small angles, $\sin\phi \approx \phi$, and $\ddot{\phi} = -(g/R)\phi$, simple harmonic motion with $\omega = \sqrt{g/R}$. The radial equation, $-mR\dot{\phi}^2 = mg\cos\phi - N$, then gives the normal force: $N = mg\cos\phi + mR\dot{\phi}^2$. One equation found the motion and the other found the constraint force, which is the pattern of every constrained problem in this course.
Polar coordinates are not always an improvement. For a projectile under uniform gravity, the force $-mg\hat{y}$ has both radial and transverse parts that change as the projectile moves, and the polar equations are coupled and hard; in Cartesian coordinates they separate into $\ddot{x} = 0$ and $\ddot{y} = -g$. The rule is to choose coordinates that match the symmetry of the forces and constraints.
The next two lessons, on air resistance, stay in Cartesian coordinates, because a drag force points along the velocity and the equations separate for linear drag. Polar coordinates return for central forces in unit 4, and generalized coordinates in unit 3 remove the need to find these acceleration terms by hand at all: the Lagrangian produces them automatically.
A child walks outward along a radius of a playground merry-go-round turning at $0.8$ rad/s, at $1.0$ m/s relative to the platform. In the ground's frame the child has $\dot{r} = 1.0$ m/s and $\dot{\phi} = 0.8$ rad/s, so the Coriolis term gives a transverse acceleration $2\dot{r}\dot{\phi} = 1.6$ m/s². For a $30$ kg child the platform must push sideways with $48$ N through friction on the feet, and the child feels this as a sideways shove, against the direction of rotation, that must be resisted.
At the same time the centripetal term grows with the radius: at $2$ m out it is $r\dot{\phi}^2 = 1.28$ m/s², or $38$ N inward. The two needs add as perpendicular vectors, and when their total exceeds the friction the shoes can supply, about $\mu_s mg$, the child slides. That is why walking toward the rim feels much harder than standing still at the same radius, and why playground safety guidance asks children to sit down before the ride spins up.
Skate parks across California build half-pipes of radius $3$ to $4$ m. The skateboard example gives a period of about $4$ s for small swings, and a normal force at the bottom of $mg + mv^2/R$: a rider reaching $6$ m/s in a $4$ m pipe feels nearly twice their weight, which is why riders crouch at the bottom and extend at the top to pump energy into the motion.
The same radial equation sets the minimum speed at the top of a vertical loop, where the normal force must stay positive: $mv^2/R \ge mg$ at the top, so $v \ge \sqrt{gR}$. Roller coaster designers at parks such as Cedar Point in Ohio avoid circular loops for this reason. A circular loop that is safe at the top gives riders six times their weight at the bottom, so modern loops are clothoids whose radius is small at the top and large at the bottom, keeping the peak load near four times weight.
It is tempting to guess that the acceleration in polar coordinates is $\ddot{r}$ outward and $r\ddot{\phi}$ around, by analogy with $\ddot{x}$ and $\ddot{y}$. That guess says a particle on a circle at steady speed has no acceleration, which contradicts everything known about circular motion. The analogy fails because $\hat{r}$ and $\hat{\phi}$ turn, and their turning adds $-r\dot{\phi}^2$ and $2\dot{r}\dot{\phi}$.
A second error is to treat those terms as forces and add a centrifugal force to the free-body diagram. In an inertial frame there is no such force: the terms belong on the acceleration side. Put them on the force side only when working in a rotating frame, and then with the opposite sign.
A particle circles at $r = 2.0$ m with $\dot{\phi} = 3.0$ rad/s. Write the rates of change of $r$.
$\dot{r} = 0, \quad \ddot{r} = 0$
The radius is fixed on a circle.
Write the rate of change of $\dot{\phi}$.
$\ddot{\phi} = 0$
The angular velocity is steady.
Evaluate the radial component.
$a_r = 0 - 2.0 \times 3.0^2 = -18\ \text{m/s}^2$
Only the centripetal term is left.
Evaluate the transverse component.
$a_\phi = 2.0 \times 0 + 2 \times 0 \times 3.0 = 0$
No speeding up around the circle.
Compare with the familiar formula.
$v = r\dot{\phi} = 6.0\ \text{m/s}, \quad \dfrac{v^2}{r} = \dfrac{36}{2.0} = 18\ \text{m/s}^2$
The polar formula reproduces $v^2/r$, pointing inward.
A particle moves with $r = 2t$ m and $\phi = 3t$ rad. Differentiate the radius.
$\dot{r} = 2\ \text{m/s}, \quad \ddot{r} = 0$
The radius grows steadily.
Differentiate the angle.
$\dot{\phi} = 3\ \text{rad/s}, \quad \ddot{\phi} = 0$
It turns steadily.
Write the radial component at $t = 1$ s, where $r = 2$ m.
$a_r = 0 - 2 \times 3^2 = -18\ \text{m/s}^2$
Inward, although $r$ is increasing.
Write the transverse component.
$a_\phi = 2 \times 0 + 2 \times 2 \times 3 = 12\ \text{m/s}^2$
Only the Coriolis term.
Interpret the transverse part.
$r\dot{\phi} = 6t \quad\Rightarrow\quad \text{the speed around grows}$
Moving outward at fixed $\dot{\phi}$ means covering more ground each turn, so something must push the particle forward.
A $60$ kg skateboarder rolls in a frictionless half-pipe of radius $R = 4.0$ m. Write the transverse equation.
$mR\ddot{\phi} = -mg\sin\phi$
Gravity's component along $\hat{\phi}$, with $r = R$ fixed so $\dot{r} = 0$.
Make the small-angle approximation.
$\ddot{\phi} = -\dfrac{g}{R}\phi$
$\sin\phi \approx \phi$ for small $\phi$ in radians.
Read off the angular frequency.
$\omega = \sqrt{\dfrac{g}{R}} = \sqrt{\dfrac{9.8}{4.0}} = 1.57\ \text{rad/s}$
The equation of simple harmonic motion.
Find the period.
$T = \dfrac{2\pi}{\omega} = \dfrac{6.283}{1.57} = 4.0\ \text{s}$
Independent of the amplitude while it stays small.
Write the radial equation.
$-mR\dot{\phi}^2 = mg\cos\phi - N$
Gravity's radial part points outward; the normal force points in.
Solve for the normal force at the bottom at $6.0$ m/s.
$N = mg + \dfrac{mv^2}{R} = 60 \times 9.8 + \dfrac{60 \times 36}{4.0} = 588 + 540 = 1128\ \text{N}$
At the bottom $\phi = 0$ and $R\dot{\phi} = v$.
Write the radial component.
$a_r = 0 - 3 \times 2^2 = -12\ \text{m/s}^2$
Fixed radius, so only the centripetal term.
Write the transverse component.
$a_\phi = 3 \times 1 + 2 \times 0 \times 2$
No radial motion, so no Coriolis term.
Evaluate the transverse component.
A particle moves on a circle of radius $5$ m with constant angular velocity $\dot{\phi} = 5$ rad/s. What is its radial acceleration?
Complete the worked solution: a bead on a turning wire has $r = 3$ m, $\dot{r} = 1$ m/s, $\ddot{r} = 1$ m/s² and $\dot{\phi} = 2$ rad/s, with $\ddot{\phi} = 0$. Find the centripetal term, the radial acceleration and the transverse acceleration, in m/s².
Multiply the radius by the square of the angular velocity.
$r\dot{\phi}^2 =$ c
The centripetal term, present whenever the bead turns.
Subtract the centripetal term from the radial acceleration term.
$a_r = \ddot{r} - r\dot{\phi}^2 =$ a
The radial component in polar coordinates.
Double the product of the radial speed and the angular velocity.
$a_\phi = 0 + 2\dot{r}\dot{\phi} =$ k
With no angular acceleration only the Coriolis term is left.
Check the signs against the motion.
$\text{outward and turning} \Rightarrow a_\phi > 0$
The wire must push the bead forward to keep it turning at the same rate as it moves out.
Match each term of the polar acceleration to what it describes.
| $\ddot{r}$ | $-r\dot{\phi}^2$ | $r\ddot{\phi}$ | $2\dot{r}\dot{\phi}$ | |
|---|---|---|---|---|
| change of radial speed | ||||
| centripetal term | ||||
| angular acceleration term | ||||
| Coriolis term |
At one instant a particle has $r = 2$ m, $\dot{r} = 2$ m/s, $\ddot{r} = 6$ m/s², $\dot{\phi} = 2$ rad/s and $\ddot{\phi} = 3$ rad/s². Fill in the radial and transverse components of its acceleration, in m/s².
| acceleration (m/s²) | |
|---|---|
| radial component | |
| transverse component |
A particle moves with $r = 4t^2$ (meters) and $\phi = 5t$ (radians). Find its radial acceleration $a_r$ as a formula in $t$, in m/s².
Answer:
A puck of mass $0.2$ kg slides on a frictionless table, tied to a string that passes through a hole in the table. At the moment the puck is $0.2$ m from the hole it circles at $\dot{\phi} = 2$ rad/s while the string is pulled in at a steady speed. What is the tension in the string, in newtons?
Answer: N of string tension
At a playground in Ohio, a child of mass $21$ kg walks straight outward along a painted radius of a merry-go-round at a steady $0.5$ m/s, relative to the platform. The merry-go-round turns at a steady $0.8$ rad/s. What sideways (transverse) force must the platform exert on the child's feet, in newtons?
Answer: N sideways
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle moves with $r = 4t^2$ (meters) and $\phi = 4t$ (radians). Find its radial acceleration $a_r$ as a formula in $t$, in m/s².
Answer:
You can use Newton's laws in polar coordinates. Explain to someone why a particle moving on a circle at steady speed is accelerating even though $r$ never changes.
18. Your turn: a particle on a circle of radius $3$ m has $\dot{\phi} = 2$ rad/s and $\ddot{\phi} = 1$ rad/s². Find both acceleration components., step 3
$a_\phi = 3\ \text{m/s}^2$
The particle is speeding up around the circle.