Circular motion examples — cars satellites rides — Physics

Before this

This leads to

One question, three answers

Swing a bucket of water in a vertical circle, fast enough, and the water stays in, even when the bucket is upside down at the top. Drive round a racing circuit's steep, tilted corner, and the car can hold its curve with almost no help from the tyres. Look up at a communications satellite, and it seems to hang still in the sky, although it is moving at about 3 km/s.

The last lecture built one formula: an object moving at speed $v$ on a circle of radius $r$ needs a resultant force $\frac{m v ^{2}}{r}$ towards the centre. It also promised to use that formula on real designs: banked roads, rides that loop, and satellites. This lecture keeps that promise.

The formula is the same each time. What changes is the question that the earlier lectures kept asking: which real forces act, and how much of each one points towards the centre? Every problem in this lecture is solved by the same three steps.

The method for every circular-motion problem

1. Find the centre of the circle, and draw every real force on the object (weight, normal force, tension, friction).

2. Take "towards the centre" as positive. Add up the parts of the forces that point towards the centre, and subtract the parts that point away.

3. Set that total equal to $\frac{m v ^{2}}{r}$ . If anything else must balance (for example the vertical forces on a car), write that as a second equation. Then solve.

Notice what is not in step 1: a separate force called "the centripetal force". Centripetal is the name of a direction. The real forces supply it.

Banked roads: tilting the normal force

On a flat road, the only horizontal force on a car is friction from the tyres. The last lecture showed how much friction a car needs on a bend, and how quickly that need grows with speed. Engineers have a better idea. They tilt the road.

Look at the forces on a car on a banked road. The road is tilted at an angle $θ$ , sloping down towards the inside of the bend. Suppose, for now, that the road is perfectly smooth, so there is no friction at all.

A car on a road banked at angle θ (drawn steeper than usual so the parts are clear). The normal force N is at right angles to the road. Its vertical part, N cos θ, balances the weight. Its horizontal part, N sin θ, points towards the centre of the bend.

Only two forces act: the weight $W = m g$ , straight down, and the normal force $N$ from the road. The normal force always acts at right angles to the surface. On a flat road it points straight up. On a tilted road it leans towards the inside of the bend. So it has two parts.

The vertical part is $N cos θ$ . The car is not rising or sinking, so this part must balance the weight:

N cos θ = m g

The horizontal part is $N sin θ$ , and it points towards the centre of the bend. It is the only horizontal force, so it must supply the whole centripetal force:

N sin θ = \frac{m v ^{2}}{r}

Now there are two equations, and you do not know $N$ . This is the same situation as simultaneous equations in Mathematics: two equations, and a quantity that must be removed. Divide the second equation by the first. The $N$ cancels, and so does the $m$ , because $\frac{sin θ}{cos θ} = tan θ$ :

tan θ = \frac{v ^{2}}{r g}

Look at what this says. For a given bank angle and radius, there is exactly one speed at which the car needs no friction at all. It is called the design speed. The mass has cancelled, so the design speed is the same for a small car and a loaded lorry.

Stop and think

At the design speed, what provides the centripetal force: friction, the weight, or something else?

Show the answer

The normal force. More exactly, its horizontal part, $N sin θ$ . Friction is zero at the design speed. The weight cannot help, because it points straight down and has no horizontal part. This surprises many students, who think a normal force can only push "up". It pushes at right angles to the surface, and on a tilted surface that direction has a sideways part.

Faster or slower than the design speed

Real cars do not all travel at the design speed. If a car goes faster, it needs more centripetal force than $N sin θ$ can give. Then friction must act down the slope, towards the inside of the bend, to supply the rest. If a car goes slower, $N sin θ$ is more than it needs. Without friction, the car would slide down the bank, towards the inside. So friction acts up the slope to hold it.

Try this on the road below. The bend has a radius of 80 m, and the car has a mass of 1000 kg. Change the speed and the bank angle, and watch the friction arrow.

Explore — friction on a banked bend

Move the speed and the bank angle. Watch the size and direction of the friction the tyres must supply. Find the speed where no friction is needed at all.

Speed v: 20 m/s

Bank angle θ: 15°

A 1000 kg car on a bend of radius 80 m, banked at 15°. The design speed is about 14.5 m/s. The friction needed at three speeds, all values approximate (14.5 m/s is the design speed):

Speed	Friction (along the slope)	Normal force
10 m/s	1330 N up	9790 N
14.5 m/s	≈ 0	10 150 N
20 m/s	2290 N down	10 760 N

Below the design speed, friction acts up the slope to stop the car sliding inwards. Above it, friction acts down the slope to help turn the car.

What did you see? At one speed for each angle, the friction arrow disappears. Above that speed, friction points down the slope, and it grows quickly, because the force needed grows with $v^{2}$ . Below it, friction points up the slope. Even a parked car on a banked road needs friction up the slope, or it would slide down. A steeper bank raises the design speed. A flat road (0°) has a design speed of zero: every moving car on a flat bend needs friction.

Worked Example 1 — The design speed of a bend

A road is banked at 15° on a bend of radius 80 m. Find the design speed. Use $g = 9.8$ m/s².

$v^{2} = r g tan θ = 80 \times 9.8 \times tan 1 5^{\circ} \approx 80 \times 9.8 \times 0.268 \approx 210$ , so $v \approx 14.5$ m/s.

Multiply by 3.6 to convert to km/h: about 52 km/h. At this speed, a car needs no friction to hold the curve.

Check the units: $m \times m/s^{2} = m^{2} / s^{2}$ , and its square root is m/s.

Worked Example 2 — Designing the bank

An engineer wants a bend of radius 300 m to have a design speed of 25 m/s (90 km/h). What bank angle is needed?

$tan θ = \frac{v ^{2}}{r g} = \frac{2 5 ^{2}}{300 \times 9.8} = \frac{625}{2940} \approx 0.213$ , so $θ \approx 12. 0^{\circ}$ .

Check the size: a gentle bank on a wide bend at motorway speed is reasonable. Racing tracks have much steeper banks, because the cars go much faster.

Vertical circles: when gravity joins in

On a banked road, the circle is horizontal, and the weight acts at right angles to the centripetal direction. In a vertical circle, the weight sometimes points towards the centre and sometimes away from it. That is what makes vertical circles interesting.

Think of a ball on a string, swung in a vertical circle. Look at the top and the bottom, where the forces line up with the string.

Forces on a ball swung in a vertical circle, drawn to scale for Worked Example 3. At the top, the weight helps the tension pull towards the centre. At the bottom, the weight pulls away from the centre, so the tension must be larger.

At the top, the centre is below the ball. The string pulls the ball down, towards the centre, and the weight also points down, towards the centre. Both count as positive:

T + m g = \frac{m v ^{2}}{r}

T = \frac{m v ^{2}}{r} - m g

At the bottom, the centre is above the ball. The string pulls up, towards the centre. The weight still points down, which is now away from the centre, so it counts as negative:

T - m g = \frac{m v ^{2}}{r}

T = \frac{m v ^{2}}{r} + m g

Here $T$ is the tension in the string. At the top, gravity does part of the work, so the string pulls less. At the bottom, the string must hold up the weight and supply the centripetal force, so it pulls more.

Worked Example 3 — Tension at the top and the bottom

A 0.5 kg ball on a 0.8 m string moves in a vertical circle at 4 m/s. (A real ball slows down as it rises, so assume the hand keeps the speed steady.) Find the tension at the top and at the bottom.

First, the centripetal force: $\frac{m v ^{2}}{r} = \frac{0.5 \times 16}{0.8} = 10$ N. The weight is $0.5 \times 9.8 = 4.9$ N.

At the top: $T = 10 - 4.9 = 5.1$ N.

At the bottom: $T = 10 + 4.9 = 14.9$ N.

Check: the difference is $14.9 - 5.1 = 9.8$ N, which is twice the weight. That must be so, because the weight helps by 4.9 N at the top and hinders by 4.9 N at the bottom. A string is most likely to break at the bottom.

The sign at the top is where most mistakes happen. Find the mistake in this student's work.

Spot the error — the top of the circle

A 0.4 kg ball on a 0.6 m string moves at 3 m/s at the top of a vertical circle. A student finds the tension. Exactly one line contains a mistake. Tap it.

A student's working for the tension at the top. Line 3 is wrong.

Line 1. At the top, the centre is below the ball, so "towards the centre" is down. Correct.
Line 2. Weight = 0.4 × 9.8 = 3.92 N, down; tension also down, along the string. Correct.
Line 3. Towards the centre: T − 3.92 = mv²/r. Wrong: both forces point towards the centre, so it should be T + 3.92 = mv²/r.
Line 4. mv²/r = 0.4 × 9 ÷ 0.6 = 6 N. Correct.
Line 5. So T = 6 + 3.92 = 9.92 N. This carries the mistake. The right answer is T = 6 − 3.92 = 2.08 N.

The slowest speed at the top

Look again at $T = \frac{m v ^{2}}{r} - m g$ . If the ball goes more slowly at the top, the tension gets smaller. At one particular speed it becomes zero. Below that speed the formula gives a negative tension, which would mean the string has to push the ball. A string cannot push. So the string goes slack, and the ball leaves the circle and falls inwards.

The slowest speed is when $T = 0$ . Then gravity alone supplies the centripetal force:

m g = \frac{m v _{m i n}^{2}}{r}

The $m$ cancels. Multiply both sides by $r$ and take the square root:

v_{m i n} = g r

That is it. That is the whole idea behind the bucket of water. At the top, the water needs a downward force of $\frac{m v ^{2}}{r}$ to stay on the circle. If the bucket moves fast enough, gravity alone is not enough, so the bucket's base must push down on the water as well. The water is pressed into the bucket, even upside down. If the bucket moves too slowly, gravity is more than the circle needs, so the water falls out of the circle, and out of the bucket.

Worked Example 4 — A loop on a ride

A ride's loop has a radius of 8 m at the top. (Real loops are not perfect circles, but treat the top as part of a circle.) Find the minimum speed at the top. Then find the force from the seat on a 60 kg rider who passes the top at 12 m/s.

Minimum speed: $v_{m i n} = g r = 9.8 \times 8 = 78.4 \approx 8.85$ m/s.

At 12 m/s, the force needed towards the centre is $\frac{m v ^{2}}{r} = \frac{60 \times 144}{8} = 1080$ N. Gravity supplies $60 \times 9.8 = 588$ N. The seat, which is above the upside-down rider, pushes down for the rest: $1080 - 588 = 492$ N.

The rider feels pressed into the seat by 492 N, about 0.84 of their usual weight of 588 N. They feel slightly lighter than normal, but they are held firmly. Ride designers keep the speed at the top well above the minimum, to keep that margin.

Stop and think

At exactly the minimum speed, riders at the top of a loop feel weightless. Has gravity been cancelled by an outward force?

Show the answer

No. From the ground, gravity is the only force acting on the rider at that moment, and it is fully in use: it supplies exactly the $\frac{m v ^{2}}{r}$ needed. The seat pushes with zero force. We feel our weight through the push of whatever supports us, so with no push the rider feels weightless. Nothing is cancelling gravity. The rider and the seat are simply falling round the curve together. The outward "flung" feeling is not a force at all, seen from the ground.

Satellites: gravity does all the work

A satellite has no road, no string and no seat. The only force on it is gravity, pulling towards the centre of the Earth. So gravity must supply exactly $\frac{m v ^{2}}{r}$ . The last lecture used this for the International Space Station: the gravitational field there is about 8.7 m/s², so $\frac{v ^{2}}{r} = 8.7$ , which gave about 7.7 km/s.

Notice that the mass of the satellite cancels, just as the mass of the car cancelled on the banked road. Every object at a given height needs the same speed to orbit, whether it is a satellite or a spanner dropped by an astronaut. The gravitation lectures show how the field gets weaker with distance, using Newton's law of gravity, and they find the orbital speed at any height. One special orbit, below, already shows the idea.

Three classic mistakes

Trap 1 — Adding an extra "centripetal force"

A free-body diagram shows only real forces: weight, normal force, tension, friction. Drawing an extra arrow labelled "centripetal force" counts the same force twice. The centripetal force is the total of the real forces towards the centre, not one more force.

Trap 2 — The wrong sign at the top

At the top of a vertical circle, the weight points towards the centre, so it adds to the tension: $T + m g = \frac{m v ^{2}}{r}$ . At the bottom, it points away: $T - m g = \frac{m v ^{2}}{r}$ . Always find the centre first, then give each force its sign.

Trap 3 — Swapping sin and cos on the bank

The vertical part of $N$ is $N cos θ$ and the horizontal part is $N sin θ$ . Check with an extreme case: on a flat road, $θ = 0$ . Then $N cos 0 = N$ balances the weight, and $N sin 0 = 0$ , so the road gives no sideways push. That is correct: on a flat road, only friction can turn the car.

Try it yourself

Use $g = 9.8$ m/s².

A bend of radius 200 m is banked at 10°. Find its design speed in m/s and in km/h.

Hint 1
$tan θ = \frac{v ^{2}}{r g}$ , so $v^{2} = r g tan θ$ .

Hint 2
$tan 1 0^{\circ} \approx 0.176$ .

Full answer
$v^{2} = 200 \times 9.8 \times tan 1 0^{\circ} \approx 346$ , so $v \approx 18.6$ m/s. In km/h: $18.6 \times 3.6 \approx 67$ km/h. Check: the bend is wider than in Worked Example 1, so the design speed is higher even though the bank is gentler.
You swing a bucket of water in a vertical circle of radius 0.9 m (from your shoulder to the water). What is the minimum speed at the top? What is the longest time one full turn can take, if the speed were the same all the way round?

Hint 1
$v_{m i n} = g r$ .

Hint 2
One turn is a distance of $2 π r$ . Time = distance ÷ speed.

Full answer
$v_{m i n} = 9.8 \times 0.9 = 8.82 \approx 2.97$ m/s. One turn is $2 π \times 0.9 \approx 5.65$ m, so it takes at most $5.65 \div 2.97 \approx 1.9$ s. Any slower, and the water leaves the circle at the top. (In practice the bucket moves faster at the bottom, but the top is where the limit applies.)
A 0.2 kg ball on a 0.5 m string moves in a vertical circle. At the top its speed is 3 m/s, and at the bottom it is 5.3 m/s. Find the tension at the top and at the bottom.

Hint 1
At the top, $T = \frac{m v ^{2}}{r} - m g$ . At the bottom, $T = \frac{m v ^{2}}{r} + m g$ . Use the right speed for each.

Hint 2
The weight is $0.2 \times 9.8 = 1.96$ N.

Full answer
Top: $\frac{0.2 \times 9}{0.5} = 3.6$ N, so $T = 3.6 - 1.96 = 1.64$ N. Bottom: $\frac{0.2 \times 5. 3 ^{2}}{0.5} \approx 11.24$ N, so $T \approx 11.24 + 1.96 \approx 13.2$ N. Check: the top speed, 3 m/s, is above $9.8 \times 0.5 \approx 2.21$ m/s, so the string stays tight, as the positive tension shows.
A 1000 kg car takes the bend from Worked Example 1 (radius 80 m, banked at 15°) at 20 m/s, faster than the design speed. Find the friction needed and the normal force. Which way does the friction act?

Hint 1
The car is too fast, so friction acts down the slope. Resolve along the slope and at right angles to it. The centripetal acceleration, $\frac{v ^{2}}{r} = 5$ m/s², is horizontal, so it has a part $5 cos 1 5^{\circ}$ along the slope (downwards) and $5 sin 1 5^{\circ}$ into the road.

Hint 2
Along the slope: $F + m g sin θ = m \frac{v ^{2}}{r} cos θ$ . At right angles to the slope: $N - m g cos θ = m \frac{v ^{2}}{r} sin θ$ .

Full answer
Friction: $F = 1000 \times (5 cos 1 5^{\circ} - 9.8 sin 1 5^{\circ}) \approx 1000 \times (4.83 - 2.54) \approx 2290$ N, down the slope. Normal force: $N = 1000 \times (5 sin 1 5^{\circ} + 9.8 cos 1 5^{\circ}) \approx 1000 \times (1.29 + 9.47) \approx 10760$ N. The friction is about 0.21 of the normal force. Check: these are the readings in the explorer at 20 m/s and 15°. And at the design speed, 14.5 m/s, the same formula gives $F \approx 0$ , as it should.

The bigger idea

A communications satellite that seems to hang still above the equator is in a geostationary orbit. It goes round once in exactly the time the Earth takes to spin once relative to the stars: one sidereal day, 23 hours 56 minutes and 4 seconds. Only one radius works: about 42 164 km from the Earth's centre, which is about 35 786 km above the surface.

You can check that radius with this lecture's ideas. One turn in 86 164 s gives $ω = \frac{2 π}{86 164} \approx 7.29 \times 1 0^{- 5}$ rad/s, so the centripetal acceleration needed is $ω^{2} r \approx 0.224$ m/s². Gravity there is weaker than at the surface. The last lecture's Moon test supported Newton's rule that gravity weakens with the square of the distance from the centre. So it is about $9.8 \times (\frac{6.4}{42.2})^{2} \approx 0.23$ m/s². The two agree. Gravity supplies exactly what the circle needs, and nothing else holds the satellite up.

Banked roads, buckets, rides and satellites look like four different topics. They are one idea: find the real forces, and ask how much of them points to the centre.

Key concepts in this lesson

Every circular-motion problem is solved by drawing the real forces and setting their total towards the centre equal to mv²/r.
On a banked road, the horizontal part of the normal force, N sin θ, turns the car, and tan θ = v²/(rg) gives the design speed at which no friction is needed.
Above the design speed friction acts down the slope, and below it friction acts up the slope.
In a vertical circle, the tension is mv²/r − mg at the top and mv²/r + mg at the bottom.
The minimum speed at the top of a vertical circle is √(gr), when gravity alone supplies the centripetal force.
For a satellite, gravity is the only force, so it supplies the whole centripetal force, and the satellite's mass cancels.