Circular Motion in Action: Vertical Circles, Conical Pendulum, Banked Roads and Spin Dryers – Projectile and Uniform Circular Motion (Senior 4 Physics) | Ganzaa

Projectile and Uniform Circular Motion

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Senior 4 · Physics

Projectile and Uniform Circular Motion

5 lessons

Lesson 5 of 5

Circular Motion in Action: Vertical Circles, Conical Pendulum, Banked Roads and Spin Dryers

25 min

Upside down, and still not spilling

Half-fill a bucket with water, stand in an open space with nobody near, and swing it quickly in a vertical circle at arm's length. At the top of the circle the bucket is upside down, yet the water stays in. Swing too slowly and it pours on your head. In Lesson 4 you learned to find the real forces that point towards the centre of a circle. Here you use that one idea to explain vertical circles, a bob on a string swinging round in a cone, why fast bends are tilted, and how a spin dryer gets clothes dry.

Motion in a vertical circle

Consider a small body of mass $m$ whirled on a string of length $r$ in a vertical circle (Figure 5.1). Its weight now sometimes points towards the centre and sometimes away, so the tension changes round the circle.

At the top, the tension $T$ and the weight $m g$ both point down, towards the centre:

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

A string can pull but not push, so $T_{\text{top}} \ge 0$ . The slowest speed at the top that keeps the string taut is when $T_{\text{top}} = 0$ : the weight alone provides the centripetal force, and

v_{\text{top,min}} = \sqrt{g r}

At the bottom, the tension points up (towards the centre) and the weight down:

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

If no energy is lost, the body speeds up as it falls through the height $2 r$ from top to bottom: $v_{\text{bottom}}^{2} = v_{\text{top}}^{2} + 4gr$ . So to complete the circle it needs at least $v_{\text{bottom}} = \sqrt{5gr}$ at the bottom, and the tension there is then $6 m g$ .

Figure 5.1: Forces on a body whirled in a vertical circle, at the top and at the bottom.

Definition

In a vertical circle the resultant of the forces along the radius must equal $\frac{mv^{2}}{r}$ at every point. The critical speed at the top is the smallest speed with which a body on a string (or water in a bucket, or a car in a loop) can pass the top and stay on the circle: $v_c = \sqrt{gr}$ , when the weight alone provides the centripetal force. Example: for a circle of radius 0.40 m, $v_c = \sqrt{9.8 \times 0.40} = 2.0\,\mathrm{m\,s^{-1}}$ .

Worked example 1: the bucket of water

The centre of mass of the water in a bucket moves in a vertical circle of radius 0.90 m. The water has mass 2.0 kg. Find (a) the smallest speed at the top for the water to stay in, (b) the speed at the bottom if the bucket has this speed at the top, (c) the force of the bucket on the water at the bottom.

(a) $v_{\text{top}} = \sqrt{gr} = \sqrt{9.8 \times 0.90} = 2.97\,\mathrm{m\,s^{-1}}$

(b) $v_{\text{bottom}} = \sqrt{v_{\text{top}}^{2} + 4gr} = \sqrt{5gr} = \sqrt{5 \times 9.8 \times 0.90} = 6.64\,\mathrm{m\,s^{-1}}$

(c) $N - mg = \frac{mv^{2}}{r} \Rightarrow N = 2.0 \times \left(\frac{6.64^{2}}{0.90} + 9.8\right) = 118\,\mathrm{N}$ , which is $6 m g$ .

At the top, the bucket's base does not need to push on the water at all; the water's weight is just enough to make it follow the circle. Slower than this, gravity pulls the water into a tighter curve than the bucket follows, and it falls out.

Worked example 2: over a hump-backed bridge

A car of mass 1000 kg drives over a bridge whose top is an arc of radius 20 m. Find the normal force on the car at the top when its speed is $10\,\mathrm{m\,s^{-1}}$ , and the speed at which it would just lose contact.

At the top the centre of the arc is below the car, so $mg - N = \frac{mv^{2}}{r}$ :

N = m\left(g - \frac{v^{2}}{r}\right) = 1000 \times \left(9.8 - \frac{10^{2}}{20}\right) = 4.8 \times 10^{3}\,\mathrm{N}

The road pushes up with less than half the car's weight of 9800 N: passengers feel lighter. Contact is lost when $N = 0$ : $v = \sqrt{gr} = \sqrt{9.8 \times 20} = 14\,\mathrm{m\,s^{-1}}$ , about 50 km/h. With little grip at the top, steering and braking there are dangerous.

The conical pendulum

A small bob on a string of length $L$ , set moving in a horizontal circle, sweeps the string round the surface of a cone: a conical pendulum (Figure 5.2, left). With the string at angle $\theta$ to the vertical, the circle has radius $r = L\sin\theta$ . The bob does not move up or down, so the forces balance vertically, and the horizontal component of the tension provides the centripetal force:

T\cos\theta = mg, \qquad T\sin\theta = \frac{mv^{2}}{r} = m\omega^{2} r

Dividing, $\tan\theta = \frac{v^{2}}{rg}$ . Using $r = L\sin\theta$ in the second equation gives $T = m\omega^{2}L$ , and then $\cos\theta = \frac{g}{\omega^{2}L}$ , so the period of one revolution is

T_{\text{p}} = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{L\cos\theta}{g}}

The faster the bob goes round, the larger the angle; the string can never become horizontal, because then nothing would balance the weight.

Figure 5.2: Left: the forces on the bob of a conical pendulum. Right: the forces on a car on a road banked at angle θ.

Banked roads, leaning riders and spin dryers

Banked roads. On a flat bend the centripetal force must come from friction, which may fail on a wet road. If the road is tilted (banked) at angle $\theta$ , the normal reaction $N$ leans towards the centre of the bend (Figure 5.2, right), and its horizontal component can provide the centripetal force with no friction at all:

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

This fixes the design speed of the bend. Faster cars need friction acting down the slope as well, slower cars need friction up the slope; with friction coefficient $μ \mu$ , the greatest safe speed rises to $v_{\max} = \sqrt{rg\,\frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta}}$ . Cycle tracks (velodromes) are banked very steeply because riders go fast round tight curves.

Leaning riders and aircraft. A cyclist or motorcyclist leans into a bend at the angle given by the same formula, $\tan\theta = \frac{v^{2}}{rg}$ , so that the resultant of the road's upward push and sideways friction passes through the centre of mass. An aircraft banks so that the horizontal component of the lift on its wings provides the centripetal force.

Spin dryers and centrifuges. The drum of a spin dryer turns very fast. Its wall pushes the clothes inwards with the large force $m\omega^{2}r$ needed to keep them on the circle. Water drops in the cloth are pushed too, but where a drop meets a hole in the wall there is nothing to push it inwards, so it carries on along the tangent, out through the hole. A laboratory centrifuge uses the same effect to separate the parts of a liquid sample: the denser parts need a larger inward force than the liquid around them can supply, so they move outwards and settle at the bottom of the tube.

Worked example 3: a conical pendulum

A bob of mass 0.20 kg hangs from a string 1.0 m long and moves in a horizontal circle with the string at $30^\circ$ to the vertical. Find the tension, the speed of the bob and the period.

T = \frac{mg}{\cos 30^\circ} = \frac{0.20 \times 9.8}{0.866} = 2.26\,\mathrm{N}, \qquad r = L\sin 30^\circ = 0.50\,\mathrm{m}

v = \sqrt{rg\tan 30^\circ} = \sqrt{0.50 \times 9.8 \times 0.577} = 1.68\,\mathrm{m\,s^{-1}}, \qquad T_{\text{p}} = 2\pi\sqrt{\frac{1.0 \times 0.866}{9.8}} = 1.87\,\mathrm{s}

Worked example 4: designing a bend, and a spin dryer

(a) At what angle should a bend of radius 80 m be banked so that cars at $15\,\mathrm{m\,s^{-1}}$ need no friction? (b) A spin-dryer drum of radius 0.25 m turns at 1200 revolutions per minute. Find the acceleration of clothes pressed against its wall, as a multiple of $g$ .

(a) $\tan\theta = \frac{v^{2}}{rg} = \frac{15^{2}}{80 \times 9.8} = 0.287 \Rightarrow \theta = 16.0^\circ$

(b) $\omega = 1200 \times \frac{2\pi}{60} = 126\,\mathrm{rad\,s^{-1}}$ ; $a = \omega^{2}r = 126^{2} \times 0.25 = 3.9 \times 10^{3}\,\mathrm{m\,s^{-2}}$ , about 400 times $g$ . Each kilogram of wet cloth needs an inward push of nearly 4 kN from the drum.

Track cyclists racing in the Samen Velodrome, Mashhad, Iran, in 2013. Notice how steeply the track is banked and how the riders lean with it: the push of the tilted track has a horizontal part towards the centre of the bend, so they can corner fast without relying on friction alone. Photo: Mehdi Ghorbani, CC BY 4.0, via Wikimedia Commons.

A motorcyclist leaning into a bend at the Brands Hatch circuit, England, in July 2023. Notice the large angle of lean: the faster the bike and the tighter the bend, the larger the angle, as tan θ = v² ÷ rg predicts, and the more sideways friction the tyres must supply. Photo: Alan Shearman, CC BY 2.0, via Wikimedia Commons.

Warning

Common mistakes. (1) Writing $T - m g$ at the top of a vertical circle: there both forces point towards the centre, so they add. (2) Forgetting that the speed changes round a vertical circle: use energy conservation to link the top and the bottom. (3) In a conical pendulum, using $L$ instead of $r = L\sin\theta$ for the radius. (4) On a banked road, resolving the weight along the slope (as for a block at rest) instead of resolving $N$ horizontally and vertically: the acceleration is horizontal, towards the centre, not along the slope.

Tip

Resolve along the acceleration. The acceleration in all these problems is horizontal or along the radius, towards the centre. Resolve forces in that direction (sum equals $\frac{mv^{2}}{r}$ ) and at right angles to it (sum equals zero). Then divide the two equations: the mass often cancels and a $\tan\theta$ appears.

Demonstrating Why Water Stays in a Bucket Revolving in a Vertical Circle (Flipping Physics). Watch the water stay in the bucket at the top, and how the explanation uses the forces on the water there, as in Figure 5.1 and Worked example 1.

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Practice

A ball of mass 0.40 kg is whirled on a string 0.60 m long in a vertical circle. Find the smallest speed it can have at the top, and the tension in the string at the bottom when its speed there is $6.0\,\mathrm{m\,s^{-1}}$ .
A car of mass 1000 kg crosses a hump of radius 20 m at $10\,\mathrm{m\,s^{-1}}$ . Find the normal reaction at the top, and explain why passengers feel lighter there.
A bob of mass 0.10 kg on a string 0.80 m long moves as a conical pendulum with the string at $40^\circ$ to the vertical. Find the tension, the radius of the circle, the speed and the period.
A bend of radius 120 m is to be banked so that cars at $20\,\mathrm{m\,s^{-1}}$ need no friction. Find the angle of banking.
A cyclist rides round a bend of radius 20 m at $10\,\mathrm{m\,s^{-1}}$ . At what angle to the vertical must she lean?
A spin-dryer drum of radius 0.20 m turns at 900 revolutions per minute. Find the acceleration of the clothes against the wall and the inward force on a wet towel of mass 0.50 kg. Explain why water leaves through the holes but the towel does not.
For a body on a string in a vertical circle, use energy conservation to show that $T_{\text{bottom}} - T_{\text{top}} = 6mg$ .
Evaluate the claim: "A banked road is always safer than a flat one." Consider fast cars, slow lorries and an icy road surface.

Show answersHide answers

$v_{\min} = \sqrt{9.8 \times 0.60} = 2.4\,\mathrm{m\,s^{-1}}$ ; $T = 0.40 \times \left(\frac{6.0^{2}}{0.60} + 9.8\right) = 27.9\,\mathrm{N}$ .
$N = 1000 \times (9.8 - 5.0) = 4.8 \times 10^{3}\,\mathrm{N}$ . The seat pushes up on each passenger with less than their weight, because part of the weight is used to give the centripetal acceleration; the push of the seat is what we feel as weight.
$T = \frac{0.10 \times 9.8}{\cos 40^\circ} = 1.28\,\mathrm{N}$ ; $r = 0.80\sin 40^\circ = 0.514\,\mathrm{m}$ ; $v = \sqrt{rg\tan 40^\circ} = 2.06\,\mathrm{m\,s^{-1}}$ ; $T_{\text{p}} = 2\pi\sqrt{\frac{0.80\cos 40^\circ}{9.8}} = 1.57\,\mathrm{s}$ .
$\tan\theta = \frac{20^{2}}{120 \times 9.8} = 0.340 \Rightarrow \theta = 18.8^\circ$ .
$\tan\theta = \frac{10^{2}}{20 \times 9.8} = 0.510 \Rightarrow \theta = 27^\circ$ .
$\omega = 900 \times \frac{2\pi}{60} = 94.2\,\mathrm{rad\,s^{-1}}$ ; $a = 94.2^{2} \times 0.20 = 1.78 \times 10^{3}\,\mathrm{m\,s^{-2}}$ (about 180 g); $F = 0.50 \times 1776 = 888\,\mathrm{N}$ . The wall pushes the towel inwards everywhere; at a hole nothing pushes the water drop inwards, so it moves off along the tangent, out of the drum.
Top: $T_{\text{t}} = \frac{mv_{\text{t}}^{2}}{r} - mg$ ; bottom: $T_{\text{b}} = \frac{mv_{\text{b}}^{2}}{r} + mg$ . Energy: $v_{\text{b}}^{2} = v_{\text{t}}^{2} + 4gr$ . So $T_{\text{b}} - T_{\text{t}} = \frac{m(v_{\text{b}}^{2} - v_{\text{t}}^{2})}{r} + 2mg = 4mg + 2mg = 6mg$ .
Banking helps vehicles near the design speed, which need no friction, and raises the greatest safe speed. But a slow or stopped vehicle on a steep bank needs friction up the slope to avoid sliding down, and on ice it may slide inwards; tall, slow lorries can also overturn towards the inside. So banking is chosen for the typical speed of the traffic, and is safer for it, but not for everyone in every condition.

Quick check

A ball on a string passes the top of a vertical circle at the smallest speed that keeps the string taut. The tension there is:

Key points

Vertical circle on a string: top $T + mg = \frac{mv^{2}}{r}$ , bottom $T - mg = \frac{mv^{2}}{r}$ ; least speed at the top $\sqrt{gr}$ ; link top and bottom by energy.
Over a hump $N = m\left(g - \frac{v^{2}}{r}\right)$ ; contact is lost at $v = \sqrt{gr}$ .
Conical pendulum: $T\cos\theta = mg$ , $T\sin\theta = \frac{mv^{2}}{r}$ , $\tan\theta = \frac{v^{2}}{rg}$ , period $2\pi\sqrt{\frac{L\cos\theta}{g}}$ .
Banked road (no friction) and leaning rider: $\tan\theta = \frac{v^{2}}{rg}$ . Spin dryers and centrifuges work because nothing gives free water or dense particles the inward force they need.

1Upside down, and still not spilling

2Motion in a vertical circle

3The conical pendulum

4Banked roads, leaning riders and spin dryers

Upside down, and still not spilling

Motion in a vertical circle

The conical pendulum

Banked roads, leaning riders and spin dryers