Classical-Mechanics · Unit 11 · Video 5 · Interactive Practice

The Two-Direction Rule: Circular Motion on a Cone and a Turntable

IKey Formulas

Formula	Name	What you need
$N\cos\theta = \dfrac{m v_0^2}{r}, \qquad N\sin\theta = mg$	Newton's law in two directions (cone)	Half-angle $\theta$ , speed $v_0$
$r = \dfrac{v_0^2}{g}\tan\theta$	Orbit radius inside the cone	$v_0$ , $\theta$
$T = \dfrac{2\pi r}{v_0} = \dfrac{2\pi v_0 \tan\theta}{g}$	Period of one orbit	$v_0$ , $\theta$
$\omega_{\max} = \sqrt{\dfrac{\mu g}{R}}$	Turntable slipping limit	$\mu$ , $R$

Key Insight: The contact force always splits into two jobs — one component cancels $mg$ , the other supplies $mv^2/r$ . Dividing the vertical equation by the radial one kills both $N$ and $m$ , which is why neither the radius nor the slipping speed depends on how heavy the object is.

IIInside the Cone: One Push, Two Jobs

A single wall push must cancel gravity and bend the path — what speed does a given radius demand?

Half-angle θ 40° Radius r (m) 2.5

As $\theta \to 90\degree$ the wall flattens out and $N\cos\theta \to 0$ : a horizontal frictionless surface has no inward push left to give, so no radius can hold a moving object at all.

IIIThe Turntable: Friction Has a Ceiling

Static friction alone bends the coin's path, and it cannot grow past $\mu N$ — how fast is too fast?

Angular speed ω (rad/s) 3.0 Coefficient μ 0.30

Nothing caps the cone's normal force — it simply tilts and grows to match any speed — but friction is capped at $\mu N$ , and that ceiling is what fails first on the turntable.

IVTwo Equations, Every Time

One direction balances gravity, the other supplies $mv^2/r$ — and dividing them cancels $N$ and $m$ .

Problem Step 1

1 — Radial, inward negative

-N\cos\theta = -\frac{m v_0^2}{r} \;\Longrightarrow\; N\cos\theta = \frac{m v_0^2}{r}

2 — Vertical, no acceleration

N\sin\theta - mg = 0 \;\Longrightarrow\; N\sin\theta = mg

3 — Divide vertical by radial

\frac{N\sin\theta}{N\cos\theta} = \frac{mg}{m v_0^2 / r} \;\Longrightarrow\; \tan\theta = \frac{rg}{v_0^2}

4 — Solve for the radius

r = \frac{v_0^2}{g}\tan\theta

5 — One circumference at constant speed

T = \frac{2\pi r}{v_0} = \frac{2\pi v_0 \tan\theta}{g}

1 — Radial, inward negative

-f_s = -m R \omega^2 \;\Longrightarrow\; f_s = m R \omega^2

2 — Vertical, no acceleration

N - mg = 0 \;\Longrightarrow\; N = mg

3 — Friction has a ceiling

(f_s)_{\max} = \mu N = \mu m g

4 — On the verge of slipping

\mu m g = m R \omega_{\max}^2

5 — The mass cancels

\omega_{\max} = \sqrt{\frac{\mu g}{R}}

VQuiz Questions

Problem 1 · Radius Inside the Cone

Given: an object slides without friction inside a cone of apex half-angle $\theta = 60\degree$ at constant speed $v_0 = 3.0$ m/s, with $g = 9.8$ m/s² — find the orbit radius $r$ .

✅ Correct!

r = \dfrac{v_0^2}{g}\tan\theta = \dfrac{9.0}{9.8}(1.732) = 1.59

m — a wide, shallow cone (

\theta

near

90\degree

) pushes the orbit far from the axis.

❌ The ratio is upside down. Dividing the vertical equation by the radial one gives

\tan\theta = rg/v_0^2

, so

r

carries a factor of

\tan\theta

, not

\cot\theta

. You computed

\dfrac{v_0^2}{g}\cot 60\degree = 0.53

❌ The geometry is missing.

0.92

m is just

v_0^2/g

— that would be the answer only for

\theta = 45\degree

, where

\tan\theta = 1

. The wall's tilt must enter through

\tan\theta

❌ Not quite. Check the units:

v_0^2\tan\theta

has units of m²/s², so it must be divided by

g

to leave a length.

Show solution

Radial direction (inward negative), where the inward part of the normal force is $N\cos\theta$ :

-N\cos\theta = -\frac{m v_0^2}{r} \;\Longrightarrow\; N\cos\theta = \frac{m v_0^2}{r}

Vertical direction (the object neither rises nor falls):

N\sin\theta = mg

Divide the vertical equation by the radial one — both $N$ and $m$ cancel:

\tan\theta = \frac{rg}{v_0^2} \;\Longrightarrow\; r = \frac{v_0^2}{g}\tan\theta

Substituting $v_0 = 3.0$ m/s, $\theta = 60\degree$ , $g = 9.8$ m/s²:

r = \frac{(3.0)^2}{9.8}\tan 60\degree = (0.918)(1.732) = 1.59\ \text{m}

Problem 2 · Does the Mass Matter?

Given: two pucks circle inside the same frictionless cone at the same speed $v_0$ . Puck A has twice the mass of puck B — compare their orbit radii and the normal forces they feel.

✅ Correct!

r = \dfrac{v_0^2}{g}\tan\theta

contains no

m

, while

N = \dfrac{mg}{\sin\theta}

is proportional to

m

— the wall pushes a heavier puck harder in exactly the right proportion.

❌ Mass does not set the radius.

m

appears on both sides of

N\cos\theta = mv_0^2/r

and

N\sin\theta = mg

, so it cancels when you divide. Heavier means a bigger

N

, not a bigger circle.

❌ Half right. The radius really is mass-independent, but

N

is not: the vertical balance

N\sin\theta = mg

forces

N = mg/\sin\theta

, which doubles when the mass doubles.

❌ Not quite. Solve each equation separately:

r

comes from the ratio (mass cancels), but

N

comes from the vertical balance alone (mass survives).

Show solution

The two equations of motion are

N\cos\theta = \frac{m v_0^2}{r}, \qquad N\sin\theta = mg

Radius: dividing removes $N$ and $m$ :

\tan\theta = \frac{rg}{v_0^2} \;\Longrightarrow\; r = \frac{v_0^2}{g}\tan\theta

Same $v_0$ and same $\theta$ means the same radius, whatever the mass.

Normal force: from the vertical equation alone,

N = \frac{mg}{\sin\theta} \;\Longrightarrow\; \frac{N_A}{N_B} = \frac{m_A}{m_B} = 2

Puck A feels twice the normal force. This is exactly why the mass drops out: doubling $m$ doubles both the required centripetal force and the wall's push, leaving the geometry untouched.

Problem 3 · When Does the Coin Fly Off?

Given: a coin rests at the rim of a turntable, $R = 0.15$ m from the axis, with coefficient of static friction $\mu = 0.30$ and $g = 9.8$ m/s² — find $\omega_{\max}$ , then the turntable's rotation period $T$ at that instant.

What is the maximum angular speed?

What is the rotation period at that speed?

✅ Correct!

\omega_{\max} = \sqrt{\mu g/R} = \sqrt{19.6} = 4.43

rad/s, and

T = 2\pi/\omega_{\max} = 1.42

s — about 42 revolutions per minute.

❌ You stopped one step early.

\mu g/R = 19.6

\omega_{\max}^2

(units s

^{-2}

), not

\omega_{\max}

. Take the square root:

\sqrt{19.6} = 4.43

rad/s.

❌ Check which symbols belong inside the root. Setting

\mu m g = m R\omega_{\max}^2

and cancelling

m

leaves

\omega_{\max} = \sqrt{\mu g/R}

— dropping

\mu

gives

8.08

rad/s, dropping

R

gives

1.71

rad/s.

❌ Check the period relation.

T = 2\pi/\omega

: dividing by

2\pi

instead gives the frequency

0.70

rev/s, forgetting

2\pi

altogether gives

0.23

s, and multiplying by

2\pi

gives

27.8

Show solution

Step 1 — radial: static friction is the only inward force, so it supplies the centripetal force:

f_s = m R \omega^2

Step 2 — vertical: the coin stays flat on the table, so $N = mg$ , and friction's ceiling is

(f_s)_{\max} = \mu N = \mu m g

Step 3 — on the verge of slipping, required equals maximum; the mass cancels:

\mu m g = m R \omega_{\max}^2 \;\Longrightarrow\; \omega_{\max} = \sqrt{\frac{\mu g}{R}} = \sqrt{\frac{(0.30)(9.8)}{0.15}} = \sqrt{19.6} = 4.43\ \text{rad/s}

Step 4 — period:

T = \frac{2\pi}{\omega_{\max}} = \frac{6.283}{4.43} = 1.42\ \text{s}

Problem 4 · Two Coins, One Table

Given: two identical coins sit on the same turntable — one at the rim (radius $R$ ), one at radius $R/4$ . The table is slowly spun faster and faster — which slips first, and at what angular speed does the other one go?

✅ Correct!

\omega_{\max}(r) = \sqrt{\mu g/r}

, so quartering the radius multiplies the limit by

\sqrt{4} = 2

— the outermost coin always goes first.

❌ $\mu$ sets the ceiling, not the demand. Both coins share the same ceiling

\mu g

, but the required centripetal acceleration

r\omega^2

grows with

r

— so the rim coin reaches its ceiling at a lower

\omega

❌ The demand shrinks with radius. The required inward acceleration is

r\omega^2

, which is smaller at smaller

r

for the same

\omega

. A tighter circle at a fixed angular speed means a slower linear speed, hence less friction needed.

❌ Watch the square root.

\omega_{\max}\propto 1/\sqrt{r}

, not

1/r

— quartering the radius doubles the limit, it does not quadruple it.

Show solution

For a coin at radius $r$ the two-direction rule gives $f_s = m r\omega^2$ and $N = mg$ , so slipping begins when

\mu m g = m r \omega^2 \;\Longrightarrow\; \omega_{\max}(r) = \sqrt{\frac{\mu g}{r}}

The limit falls off as $1/\sqrt{r}$ , so the rim coin slips first. For the inner coin at $r = R/4$ :

\omega_{\max}(R/4) = \sqrt{\frac{\mu g}{R/4}} = \sqrt{\frac{4\mu g}{R}} = 2\sqrt{\frac{\mu g}{R}} = 2\,\omega_{\max}(R)

Equivalently, at a common $\omega$ the required inward acceleration is $r\omega^2$ : the rim coin demands four times as much friction as the coin at $R/4$ , while both are held by the same ceiling $\mu g$ .

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