Classical-Mechanics · Unit 11 · Video 1 · Interactive Practice

The Invisible Tether: Newton's Second Law for Circular Motion

IKey Formulas

Formula	Name	What it governs
$\vec{a} = -r\dot{\theta}^2\,\hat{r} + r\ddot{\theta}\,\hat{\theta}$	Acceleration in polar coordinates	Splits any circular motion into an inward piece and an along-path piece
$F_r = -m r \dot{\theta}^2$	Radial equation ( $\hat{r}$ components)	How hard the object is turned toward the center
$F_\theta = m r \ddot{\theta}$	Tangential equation ( $\hat{\theta}$ components)	How the speed along the circle changes
$a_r = -\dfrac{4\pi^2 R}{T^2}$	Uniform orbit of period $T$	Centripetal acceleration, from $\omega = 2\pi/T$

Key Insight: Constant speed kills the tangential term ( $\ddot{\theta} = 0 \Rightarrow F_\theta = 0$ ) but never the radial one. The minus sign in $-r\dot{\theta}^2\,\hat{r}$ says anything on a circle is forever accelerating inward — so a real inward force must exist.

IIThe Frame That Rides Along

The unit vectors $\hat{r}$ and $\hat{\theta}$ are not fixed in space — they turn with the object.

IIIWhich Piece of the Force Survives?

The inward piece is always there; a force along the path appears only when $\ddot{\theta} \neq 0$ .

Angular acceleration α 0.0 Time t 0.00

IVWhat the Moon's Orbit Demands

One period and one distance are enough to pin down the Moon's inward acceleration.

Step 1

Step 1 — One turn per period

\omega = \dot{\theta} = \frac{2\pi}{T}

Step 2 — Keep the radial term only

\ddot{\theta} = 0 \;\Rightarrow\; \vec{a} = -r\dot{\theta}^2\,\hat{r} = -r\omega^2\,\hat{r}

Step 3 — Substitute $r = R_{e,m}$ and $\omega = 2\pi/T$

a_r = -R_{e,m}\left(\frac{2\pi}{T}\right)^{2}

Step 4 — Simplify and evaluate

a_r = -\frac{4\pi^2 R_{e,m}}{T^2} \approx -2.72\times10^{-3}\ \mathrm{m/s^2}

💡 Newton's second law proves an inward force must exist and fixes its size, but not its identity — naming it is the next step.

VQuiz Questions

Problem 1 · Uniform Circular Motion

Given: A stone on a string is whirled around a circle of radius $r$ at constant speed. Using $F_r = -m r \dot{\theta}^2$ and $F_\theta = m r \ddot{\theta}$ , which pair of statements is correct?

✅ Correct! Constant speed means

\dot{\theta}

is constant, so

\ddot{\theta} = 0

and the tangential force vanishes — the entire net force is the inward centripetal force.

❌ Reversed. Constant speed removes the tangential piece, not the radial one:

F_\theta = m r \ddot{\theta} = 0

while

F_r = -m r \dot{\theta}^2

survives.

❌ Not quite. Constant speed is not constant velocity — the direction keeps turning, so the inward acceleration

-r\dot{\theta}^2

never vanishes.

❌ The components are swapped.

\hat{r}

pairs with

-m r \dot{\theta}^2

and

\hat{\theta}

pairs with

m r \ddot{\theta}

, never the other way round.

❌ Not quite. Set

\ddot{\theta} = 0

in the two scalar equations and see which one survives.

Show solution

Constant speed on a fixed radius means the angular velocity never changes:

\dot{\theta} = \text{constant} \;\Rightarrow\; \ddot{\theta} = 0

Feed that into the tangential equation:

F_\theta = m r \ddot{\theta} = m r (0) = 0

The radial equation is untouched, because it contains $\dot{\theta}^2$ , not $\ddot{\theta}$ :

F_r = -m r \dot{\theta}^2 \neq 0

So the whole net force points along $-\hat{r}$ , straight at the center. That surviving inward force is the centripetal force.

Problem 2 · Halving the Period

Given: A satellite circles at constant speed with $a_r = -\dfrac{4\pi^2 r}{T^2}$ . Its period $T$ is halved while the radius $r$ stays the same. By what factor does the magnitude of its centripetal acceleration change?

✅ Correct!

|a_r| \propto 1/T^2

, so halving

T

multiplies the acceleration by

(1/\tfrac{1}{2})^2 = 4

❌ Close. The period enters squared:

|a_r| \propto 1/T^2

, so the factor is

2^2 = 4

, not

2

❌ Inverted. A shorter period means a faster orbit and a larger inward acceleration — the factor is

4

, not

\tfrac{1}{4}

❌ Not quite. With

r

fixed,

|a_r| = 4\pi^2 r / T^2

depends only on

T

, and it does so through

T^2

Show solution

Hold $r$ fixed and read off the dependence on $T$ :

|a_r| = \frac{4\pi^2 r}{T^2} \;\Rightarrow\; |a_r| \propto \frac{1}{T^2}

Replace $T$ by $T/2$ :

|a_r|_{\text{new}} = \frac{4\pi^2 r}{(T/2)^2} = \frac{4\pi^2 r}{T^2/4} = 4\,|a_r|_{\text{old}}

The acceleration grows four-fold. Equivalently, $\omega = 2\pi/T$ doubles, and $|a_r| = r\omega^2$ scales with the square of $\omega$ .

Problem 3 · The Moon's Inward Acceleration

Given: The Moon's orbit is very nearly a circle of radius $R_{e,m} = 3.84\times10^{8}\ \mathrm{m}$ , travelled once every $T = 27.3$ days, with $1\ \mathrm{day} = 86400\ \mathrm{s}$ .

What is the angular velocity $\omega = 2\pi/T$ ?

What is the magnitude of the centripetal acceleration?

✅ Correct!

\omega = 2.66\times10^{-6}\ \mathrm{rad/s}

and

|a_r| = 4\pi^2 R_{e,m}/T^2 = 2.72\times10^{-3}\ \mathrm{m/s^2}

— tiny, but relentlessly inward.

❌ You dropped the $2\pi$ . One revolution is

2\pi

radians, so

\omega = 2\pi/T

, not

1/T

❌ Wrong units.

T

must be in seconds:

27.3\ \mathrm{d} \times 86400\ \mathrm{s/d} = 2.36\times10^{6}\ \mathrm{s}

❌ Recompute $\omega$ . Convert the period to seconds first, then divide

2\pi

by it.

❌ That is a speed, not an acceleration.

R_{e,m}\,\omega \approx 1.0\times10^{3}\ \mathrm{m/s}

is the Moon's orbital speed; the acceleration needs

\omega

squared.

❌ That is $g$ at the Earth's surface. The Moon is

60

Earth radii away, and its inward acceleration is thousands of times smaller.

❌ Recompute. Use

|a_r| = R_{e,m}\,\omega^2

with

\omega = 2.66\times10^{-6}\ \mathrm{rad/s}

, and keep track of the powers of ten.

Show solution

Step 1 — Period in seconds.

T = 27.3 \times 86400 = 2.36\times10^{6}\ \mathrm{s}

Step 2 — Angular velocity. One revolution is $2\pi$ radians per period:

\omega = \frac{2\pi}{T} = \frac{6.283}{2.36\times10^{6}} = 2.66\times10^{-6}\ \mathrm{rad/s}

Step 3 — Radial term of the acceleration. The orbit is uniform, so $\ddot{\theta} = 0$ and only $-r\dot{\theta}^2$ survives:

|a_r| = R_{e,m}\,\omega^2 = (3.84\times10^{8})(2.66\times10^{-6})^2

= (3.84\times10^{8})(7.10\times10^{-12}) = 2.72\times10^{-3}\ \mathrm{m/s^2}

Equivalently, in one line:

a_r = -\frac{4\pi^2 R_{e,m}}{T^2} = -\frac{(39.48)(3.84\times10^{8})}{(2.36\times10^{6})^2} = -2.72\times10^{-3}\ \mathrm{m/s^2}

The minus sign confirms the acceleration points at the Earth's center.

Problem 4 · A Car Braking Through a Curve (Transfer)

Given: A car drives around a circular track of radius $r$ and is slowing down as it goes. Which description of the net force on the car is correct?

✅ Correct! The car still turns, so

F_r = -m r \dot{\theta}^2 \neq 0

; and it is slowing, so

\ddot{\theta}

opposes

\dot{\theta}

and

F_\theta = m r \ddot{\theta}

points backwards along the path.

❌ That is the uniform case. A purely centripetal force requires

\ddot{\theta} = 0

, i.e. constant speed — but this car is braking.

❌ Not quite. Braking supplies the tangential piece, but the car is still being turned:

F_r = -m r \dot{\theta}^2

is nonzero for any

\dot{\theta} \neq 0

❌ No outward force. The radial equation carries a minus sign,

F_r = -m r \dot{\theta}^2

— circular motion always demands an inward radial force.

❌ Not quite. Check each component separately: is

\dot{\theta}

zero? is

\ddot{\theta}

zero?

Show solution

Take the two scalar equations one at a time.

Radial. The car is moving around the circle, so $\dot{\theta} \neq 0$ and

F_r = -m r \dot{\theta}^2 < 0,

a force pointing along $-\hat{r}$ , toward the center. This is true whether the car speeds up, slows down, or holds its speed.

Tangential. Slowing down means the angular speed is decreasing, so $\ddot{\theta}$ has the opposite sign to $\dot{\theta}$ , and

F_\theta = m r \ddot{\theta}

points backwards along the path — the braking force.

The net force is therefore the sum of both pieces: inward and backwards, tilted behind the radius. Only when $\ddot{\theta} = 0$ does it collapse to the purely centripetal case.

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