Dynamics of circular motion

In weeks four and five you saw that in uniform circular motion the acceleration is $a_c = v^2/r$ and points toward the center. This week is the dynamic side of the matter: which force provides that acceleration? The answer is different in each question. Friction on a curve, tension for a stone on a string, the component of the normal force on a banked curve, the combination of the normal force and the weight at an amusement park. The question type is always the same: draw the free-body diagram, add the forces toward the center, and set them equal to $mv^2/r$.

Golden rules:

1. The “centripetal force” is not drawn as a separate arrow in the diagram. It is the resultant of the real forces toward the center.
2. Take the direction toward the center as +. The equation: $\sum F_{center} = mv^2/r$.
3. On the perpendicular axis there is usually balance. On a flat curve $N = mg$; on a banked curve it is not.
4. Velocity is tangent, acceleration is toward the center. If the speed is changing, there is also a tangential acceleration.

Newton’s second law in uniform circular motion

The acceleration of an object turning in a circle of radius $r$ at constant speed is directed toward the center:

$a_c = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2},\qquad v = \frac{2\pi r}{T}$

The second law in the radial direction:

$\sum F_r = m\frac{v^2}{r}$

If this force disappears (the rope breaks, the road turns to ice), the object does not fly outward from the center; it goes straight in the direction of its velocity at that moment, that is, along the tangent. The first law.

Who provides the centripetal force?

The first step of the question is always to answer this question:

If the speed is given with the period, first convert it to speed with $v = 2\pi r/T$ or use $a_c = 4\pi^2 r/T^2$ directly. If revolutions per minute (rpm) is given, first convert to revolutions per second: 30 rpm = 0.5 rev/s, period 2.0 s.

Flat curve

The forces on the car: $mg$ downward, $N$ upward, static friction horizontal and toward the center. Vertical: $N = mg$. Radial:

$f_s = \frac{mv^2}{r} \le \mu_s mg \quad\Rightarrow\quad v_{max} = \sqrt{\mu_s g r}$

Example 1. $r = 50$ m, $\mu_s = 0.50$.

v_max = √(0.50 · 9.80 · 50) = √245 = 15.7 m/s  (≈ 56 km/h)

The mass cancels: a truck and a bicycle skid at the same speed (with the same $\mu_s$). On a wet road $\mu_s$ drops, and so does the safe speed.

Banked curve

If the road is banked at an angle $\theta$, the normal force is no longer vertical; its horizontal component $N\sin\theta$ points toward the center. Without friction:

$N\cos\theta = mg,\qquad N\sin\theta = \frac{mv^2}{r} \quad\Rightarrow\quad \tan\theta = \frac{v^2}{rg}$

This angle is correct for a single “design speed”. A car going faster tends to slide up, and friction turns down along the bank. A car going slower tends to slide down, and friction turns up.

Example 2. A curve of radius 150 m is to be designed frictionless for 25 m/s.

tanθ = 25² / (150 · 9.80) = 625 / 1470 = 0.425
θ = 23.0°

PITFALL · N = mg cosθ on a banked curve. On an incline that was the case, because there the acceleration was along the incline. On a curve the acceleration is horizontal (toward the center), so the vertical balance is written: $N\cos\theta = mg$, that is, $N = mg/\cos\theta$, larger than the weight. Choose the axes not according to the incline but according to the acceleration: horizontal and vertical.

Banked curve and friction together

If the vehicle goes faster than the design speed, friction is added down along the bank and the maximum safe speed increases. The axes are again horizontal (toward the center +) and vertical:

$\text{vertical: } N\cos\theta - f\sin\theta = mg,\qquad \text{horizontal: } N\sin\theta + f\cos\theta = \frac{mv^2}{r}$

Putting $f = \mu_s N$ at the threshold of sliding and dividing the two equations:

$v_{max}^2 = rg\,\frac{\sin\theta + \mu_s\cos\theta}{\cos\theta - \mu_s\sin\theta}$

For a slower vehicle, friction turns upward, the signs of $\mu_s$ in the formula change and the minimum speed comes out: $v_{min}^2 = rg(\sin\theta - \mu_s\cos\theta)/(\cos\theta + \mu_s\sin\theta)$. If the numerator is negative, that is, if $\mu_s > \tan\theta$, there is no lower limit: even if the vehicle stops and waits on the curve, it does not slide down. For example, for a $15^\circ$ bank and $\mu_s = 0.30$, $\tan 15^\circ = 0.268 < 0.30$, so there is only an upper limit. Instead of memorizing the two formulas, it is safer to write the two equations each time; this is where sign errors happen most in the exam.

Vertical circle

Here the speed is usually not constant, but it is enough to write the centripetal component of the acceleration at the very top and very bottom points.

Example 3. A 2.0 kg bucket in a vertical circle of radius 1.0 m, 5.0 m/s both at the bottom and at the top.

v²/r = 25 / 1.0 = 25 m/s²
Bottom:  T = m(g + v²/r) = 2.0 (9.80 + 25) = 69.6 N
Top:     T = m(v²/r - g) = 2.0 (25 - 9.80) = 30.4 N
So the rope does not go slack at the top: T ≥ 0  →  v ≥ √(g r) = √9.80 = 3.13 m/s

The largest tension is at the very bottom; ropes always break there. At the minimum speed at the top, gravity alone provides the centripetal force and the water does not spill from the bucket.

Example 4: Bump. A 1200 kg car passes over the top of a bump of radius 30 m at 12 m/s.

N = m(g - v²/r) = 1200 (9.80 - 144/30) = 1200 (9.80 - 4.80) = 6000 N
The weight is 11 760 N. The car "gets lighter".
The speed at which N = 0: v = √(g r) = √294 = 17.1 m/s. If faster, the car becomes airborne.

Conical pendulum

A string of length $L$ turns in a horizontal circle making an angle $\theta$ with the vertical. The radius is not the string length, but $r = L\sin\theta$. The vertical component of the tension balances the weight, and the horizontal component provides the centripetal force:

$T\cos\theta = mg,\qquad T\sin\theta = \frac{mv^2}{r}\quad\Rightarrow\quad v^2 = rg\tan\theta$

Example 6. A string of length 1.0 m makes 40° with the vertical. What are the speed, period and tension?

r = L sinθ = 1.0 · sin40° = 0.643 m
v² = r g tanθ = 0.643 · 9.80 · 0.839 = 5.29  →  v = 2.30 m/s
Period: T_per = 2πr / v = 2π · 0.643 / 2.30 = 1.76 s
Tension: T_string = mg / cos40° = m · 12.8 m/s²  (1.31 times the weight)

As the angle grows, the tension rises quickly; the string can never be exactly horizontal, because then there would be no vertical component left to balance the weight.

Non-uniform circular motion

If the speed is changing, the acceleration has two components:

$a_r = \frac{v^2}{r}\ (\text{toward the center}),\qquad a_t = \frac{dv}{dt}\ (\text{tangent}),\qquad a = \sqrt{a_r^2 + a_t^2}$

The net force also has two components: $\sum F_r = ma_r$, $\sum F_t = ma_t$. In this case the net force no longer points toward the center.

Example 5. A car going 8.0 m/s on a track of radius 40 m is speeding up at 2.0 m/s².

a_r = 64 / 40 = 1.6 m/s²   (toward the center)
a_t = 2.0 m/s²             (tangent, in the direction of the velocity)
a = √(1.6² + 2.0²) = 2.56 m/s²

TIP · unit check. If speed is asked, the answer should contain a square root ($\sqrt{\mu g r}$, $\sqrt{gr}$). If you see a value like 441 m/s among the options, it is the result of someone who forgot to take the root.

Common mistakes