Circular Motion Problems

Knowing $F = mv^2/r$ is the easy part. The marks in Paper 4 come from applying it to real arrangements: a mass swinging in a cone, a car on a banked track, an aircraft turning, a ball on a string in a vertical circle, a car going over a hump. This note works through each standard set-up with the same method, so that any new situation can be broken down the same way.

The one method behind every problem

Every circular motion problem is Newton's second law in disguise. The steps never change:

Two kinds of problem appear:

• Horizontal circles (conical pendulum, banked track, turning aircraft, turntable). The speed is constant. Resolve horizontally (towards the centre) and vertically (balanced).
• Vertical circles (ball on a string, roller coaster loop, hump-back bridge, Ferris wheel). Gravity has a component along the path, so the speed generally changes around the circle. Apply $F = mv^2/r$ at the specific point asked about, and use conservation of energy to link speeds at different heights.

The conical pendulum

A mass $m$ on a string of length $L$ moves in a horizontal circle, the string sweeping out a cone and making angle $\theta$ with the vertical.

Resolving:

• Vertically (no vertical acceleration): $T\cos\theta = mg$
• Horizontally (towards the centre): $T\sin\theta = mr\omega^2 = \dfrac{mv^2}{r}$

Dividing the second by the first eliminates $T$:

The horizontal component of the tension is the centripetal force. The vertical component holds the bob up. Faster rotation means larger $\theta$, but $\theta$ can never reach $90^\circ$, because then $T\cos\theta$ could not balance the weight.

Banked tracks and turning aircraft

On a level road, friction alone provides the centripetal force. On a banked track tilted at angle $\theta$ to the horizontal, the normal contact force $N$ is tilted towards the centre, so its horizontal component can provide the centripetal force with no need for friction.

With no friction:

• Vertically: $N\cos\theta = mg$
• Horizontally: $N\sin\theta = \dfrac{mv^{2}}{r}$

An aircraft turning in a horizontal circle is mathematically identical. The aircraft banks so that the lift force $L$ (perpendicular to the wings) tilts towards the centre: $L\cos\theta = mg$ and $L\sin\theta = mv^2/r$, so again $\tan\theta = v^2/(rg)$. Because $L = mg/\cos\theta$, the lift (and the force on the passengers from their seats) is greater than in level flight.

Vertical circles

In a vertical circle the weight acts along the radius at the top and the bottom. These are the two points examiners ask about.

At the bottom, the centre is above. Tension (or normal force) acts upwards, towards the centre; weight acts downwards, away from it:

$$T_{\text{bottom}} - mg = \frac{mv^{2}}{r} \quad\Rightarrow\quad T_{\text{bottom}} = mg + \frac{mv^{2}}{r}$$

At the top, the centre is below. Both tension and weight act downwards, towards the centre:

$$T_{\text{top}} + mg = \frac{mv^{2}}{r} \quad\Rightarrow\quad T_{\text{top}} = \frac{mv^{2}}{r} - mg$$

The tension is greatest at the bottom and least at the top.

Minimum speed to complete the circle

A string can pull but not push, so $T_{\text{top}} \ge 0$. The limiting case is $T_{\text{top}} = 0$, when gravity alone provides the centripetal force:

Below this speed the string goes slack before the top and the mass follows a projectile path inside the circle. The same condition applies to a roller coaster at the top of a loop (normal force $\ge 0$) and to water staying in a bucket swung overhead.

Linking top and bottom with energy

Ignoring air resistance, the kinetic energy gained falling from top to bottom equals the loss in gravitational potential energy over a height $2r$:

$$\tfrac{1}{2}mv_{\text{bottom}}^{2} = \tfrac{1}{2}mv_{\text{top}}^{2} + mg(2r) \quad\Rightarrow\quad v_{\text{bottom}}^{2} = v_{\text{top}}^{2} + 4gr$$

Going over a hump

A car going over the top of a hump-back bridge of radius $r$ has its centre of curvature below it. Weight acts towards the centre and the normal contact force $N$ acts away:

$$mg - N = \frac{mv^{2}}{r} \quad\Rightarrow\quad N = mg - \frac{mv^{2}}{r}$$

As $v$ increases, $N$ decreases. When $v = \sqrt{gr}$, $N = 0$ and the car loses contact with the road. Passengers feel "lighter" because the seat pushes up on them with less than their weight.

Worked examples

Practice questions

Answers

1. $T = mg/\cos 40^\circ = 0.30 \times 9.81/0.766 = 3.84\ \text{N}$. $r = 1.2\sin 40^\circ = 0.771\ \text{m}$; $v = \sqrt{rg\tan 40^\circ} = \sqrt{0.771 \times 9.81 \times 0.839} = 2.52\ \text{m s}^{-1}$. Period $= 2\pi\sqrt{1.2\cos 40^\circ/9.81} = 1.92\ \text{s}$.
2. $v = \sqrt{rg\tan\theta} = \sqrt{200 \times 9.81 \times \tan 12^\circ} = 20.4\ \text{m s}^{-1}$.
3. (a) $v = \sqrt{gr} = \sqrt{9.81 \times 8.0} = 8.86\ \text{m s}^{-1}$. (b) $v^2 = 78.5 + 4 \times 9.81 \times 8.0 = 392$, $v = 19.8\ \text{m s}^{-1}$. (c) $N = mg + mv^2/r = 589 + 60 \times 392/8.0 = 3.53 \times 10^{3}\ \text{N}$, which is $6mg$.
4. Need $r\omega^2 \ge g$: $\omega = \sqrt{9.81/0.90} = 3.30\ \text{rad s}^{-1}$, so $f = \omega/2\pi = 0.53$ revolutions per second.
5. $\omega = 2\pi/60 = 0.105\ \text{rad s}^{-1}$; $mr\omega^2 = 70 \times 20 \times 0.01097 = 15.4\ \text{N}$. Top: $N = mg - mr\omega^2 = 687 - 15.4 = 671\ \text{N}$. Bottom: $N = 687 + 15.4 = 702\ \text{N}$.
6. $N = mg - mv^2/r = 14\,715 - 1500 \times 100/30 = 9.7 \times 10^{3}\ \text{N}$. Leaves road at $v = \sqrt{gr} = \sqrt{294} = 17.2\ \text{m s}^{-1}$.
7. At the bottom $N - mg = mv^2/r$, so $N/mg = 1 + v^2/(rg) = 1 + 22\,500/4905 = 5.6$.
8. (a) $T = mg + mv^2/r = 4.905 + 0.50 \times 36/0.75 = 28.9\ \text{N}$. (b) $v_{\text{top}}^2 = 36 - 4 \times 9.81 \times 0.75 = 6.57$, $v_{\text{top}} = 2.56\ \text{m s}^{-1}$. The minimum needed is $\sqrt{gr} = \sqrt{7.36} = 2.71\ \text{m s}^{-1}$, so the string goes slack before the top: the ball does not complete the circle (equivalently, $mv^2/r - mg = -0.53\ \text{N} < 0$). (c) When the string is horizontal the weight is tangential, so the tension alone provides the centripetal force. $v^2 = 36 - 2 \times 9.81 \times 0.75 = 21.3$, $T = 0.50 \times 21.3/0.75 = 14.2\ \text{N}$.