Chapter 1 · Mechanics

1-11 Circular Motion: Horizontal and Vertical

❓ Lesson question

Why can water remain inside a bucket at the top of a vertical circle?

You will learn to

1Identify the forces in horizontal and vertical circular motion.
2Analyse horizontal and vertical circular motion.
3Determine the tension and minimum speed in vertical circular motion.

Tie a small bucket containing water to a rope and move it rapidly in a vertical circle above your head. At the highest point of the circle, the bucket is upside down, yet the water remains in the bucket and does not spill out. Now gradually reduce the speed of the bucket. Below a certain speed, the water begins to spill out near the top of the circle.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) At the highest point, what keeps the water in the bucket: the pull of the rope, the push of the bucket, gravity, or nothing at all?

(b) Is there a minimum speed at the highest point below which the water spills out? If so, what determines this minimum speed?

Write a line for each before this opens.

🔬 Explore — Moving a bucket of water in a vertical circle

What you’ll do: You will find how fast a bucket of water must be moved in a vertical circle to keep the water in, and where the water first leaves the bucket when the speed is too slow.

⏱️ 30 min🧰 A small bucket with a handle, or a plastic cup tied securely with strong string. · Water, enough to part-fill the bucket. · A phone or camera with a slow-motion setting. · A clear, open space outdoors.

1Part-fill the bucket with water. Hold the handle and move the bucket in a vertical circle at a steady speed at your side, fast enough that the water stays in throughout the full circle.
2Ask a partner to film the circular motion in slow motion, with the whole circle in view.
3Gradually reduce the speed, on each revolution, until the water first begins to leave the bucket.
4From the slow-motion clip, identify where on the circle the water first leaves — at the bottom, at the side, or near the top.
5Compare the bucket’s speed at the top in the trial where the water first spills with its speed there in the faster trial that keeps the water in.

A safety note

Carry out the activity in a clear, open space outdoors, well away from other people and from windows. Use only a small amount of water, and check that the handle or string and its attachment are secure before each trial.

Analyse what you collected. (a) At the first spill, where on the circle does the water begin to leave the bucket — at the bottom, at the side, or near the top? (b) Explain why the bucket must be moved fast enough for the water to stay in at the top. What provides the force that keeps the water moving in a circle there, and what happens when the rotation is too slow?

In Explore, the water remained inside the bucket throughout its motion in a vertical circle when the bucket's speed was sufficiently high. As the speed was gradually reduced, the water first began to leave the bucket near the top of the circle.

At first, it might seem that an upward force is needed to keep the water in the bucket at the top. The slow-motion video shows otherwise: the required resultant force is directed toward the centre of the circle, which at the top is vertically downward. The evidence from the experiment supports the model of vertical circular motion, in which the speed changes around the circle.

🎛️ Simulation — tension around a vertical circle

Mass m0.5 kg

Radius r0.8 m

Speed at the bottom4 m/s

At 4.00 m/s the object never reaches the top at all — it runs out of speed on the way up and leaves the circular path.

At the top the critical speed is v_min = √(gr) = 2.80 m/s. Because the object must first climb a height 2r to get there, that corresponds to a speed at the bottom of √(5gr) = 6.26 m/s — noticeably more than most people guess. Set the slider to exactly that and the tension arrow at the top disappears: the weight alone is then precisely the force needed to bend the path into a circle. Slower and there is no way to make up the shortfall, because gravity is fixed and a string can only pull.

At the top

T = mv²/r − mg

speed 0.00 m/s

T = -4.90 N

Weight points at the centre here, so it does part of the job and the string does less.

At the side

T = mv²/r

speed 0.57 m/s

T = 0.20 N

Weight is at right angles to the centre, so none of it counts toward the centre.

At the bottom

T = mv²/r + mg

speed 4.00 m/s

T = 14.90 N

Weight points away from the centre, so the string must supply the whole of it and more.

Press Exam-style p.68: the bottom tension reads 14.90 N, which is what the book's answer works out to — and notice the verdict box, which points out that this particular ball never gets to the top at all. Then press √(5gr) and watch the green arrow at the top vanish while the red one does not: that single frame is the whole of "why the water stays in the bucket". Notice also that the bottom tension is always the largest of the three and the top the smallest — the string is nearest to breaking at the bottom, not at the top where it looks most alarming.

Consider an object of mass $m$ , attached to a string of length $r$ , being moved in a vertical circle. Its speed now changes around the circle — greatest at the bottom, least at the top. At each position, resolve toward the centre and write $F = m a_{c}$ . The weight is added, subtracted or ignored according to which way it points relative to the centre:

At the side

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

Weight is perpendicular to the radius, so it has no component toward the centre.

At the top

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

Weight points toward the centre, so it helps — and the string does less.

At the bottom

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

Weight points away from the centre, so the string must overcome it as well.

The tension is least at the top and greatest at the bottom — which is worth remembering, because the string is nearest to breaking at the bottom, where it looks safest.

Minimum speed at the top

At the top of the circle, both the tension and the weight act toward the centre, so:

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

A string can only pull, so its tension cannot be negative; the limiting case is when it just falls to zero. Setting $T = 0$ :

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

This is the minimum speed needed so that the object just maintains circular motion without the string going slack at the top of the circle. Note that $m$ cancels: a heavy bucket and a light one need the same speed.

Consider an object of mass $m$ attached to a string $P Q$ of length $l$ . $P$ is fixed to a support, and the object moves in a horizontal circle of radius $r$ . When the object moves at constant speed, the string makes a constant angle $θ$ with the vertical.

Two forces act: the weight $W = m g$ and the tension $T$ , which resolves into a horizontal component $T sin θ$ and a vertical component $T cos θ$ .

Since the object moves in a circular path, a centripetal force must act toward the centre of the circle, $O$ . This is provided by the horizontal component of the tension:

T sin θ = \frac{m v ^{2}}{r}

(1) toward the centre

The object does not move vertically, so the vertical forces are balanced:

T cos θ = m g

(2) vertically

Dividing (1) by (2):

tan θ = \frac{v ^{2}}{r g}

Why the vertical equation is an equilibrium and the horizontal one is not

This is the step students most often get backwards. The bob circles at a constant height, so it has no vertical acceleration and the vertical forces genuinely balance. It does not circle at a constant velocity, though — the direction changes — so horizontally there is an acceleration and the forces must not balance. One equation is set to zero and the other to

m a_{c}

, and which is which is decided by the motion, not by preference.

Point

• For a ball moving on the smooth inner surface of a conical container, the resultant of the normal force and the weight points toward the centre of the circle, and for a circle of radius $r$ whose slant makes an angle $θ$ with the horizontal, $v^{2} = g r tan θ$ .
• Motion in a vertical circle is not uniform circular motion: the speed is greatest at the bottom of the circle and least at the top, and the tension follows it. Set up the equation of motion toward the centre at each moment.
• The object only just passes the highest point when the tension falls to $T = 0$ and the weight alone supplies the centripetal force. That gives $v_{m i n}^{2} = g r$ .

📐 Equation box

Minimum tension T (top)

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

Maximum tension T (bottom)

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

Minimum speed at the top

v_{m i n} = g r

Conical pendulum

tan θ = \frac{v ^{2}}{r g}

🛠️ Worked example (book p.66)

Situation

A small ball of mass $m$ moves in uniform circular motion on the smooth inner surface of a conical container. It travels in a horizontal circle of radius $r$ at speed $v$ , and the slant side of the cone makes an angle $θ$ with the horizontal. Determine the speed and the period of the motion.

Step 1 of 6 — try it yourself before you open the next one.

Now you try (p.66)

On a smooth rail where a slope meets a vertical semicircle of radius

r

at the lowest point

A

, a small ball of mass

m

is released from rest at point

P

on the slope, at a height

h

above

A

, and travels around to the highest point

B

. Determine the speed

v

of the ball as it passes

B

, in terms of

g

h

and

r

.
Answer. The rail is smooth, so mechanical energy is conserved. Taking

A

as the zero of height,

B

is at height

2 r

m g h = \frac{1}{2} m v^{2} + m g (2 r)

v^{2} = 2 g h - 4 g r \Rightarrow v = 2 g (h - 2 r)

The mass cancels. Note the condition hidden in the algebra: the ball only completes the loop if

v^{2} \geq g r

at the top, which needs

2 g (h - 2 r) \geq g r

, that is

h \geq 2.5 r

. Releasing it from exactly the height of

B

is not enough — a common and expensive mistake in roller-coaster design.

One end of a thread of length

l

[m] is fixed to the ceiling, and a small ball of mass

m

[kg] attached to the other end undergoes uniform circular motion in a horizontal plane. Let the angle between the thread and the vertical direction be

θ

, the acceleration due to gravity be

g

[m/s²]. (i) Determine the magnitude of the tension in the thread in N. (ii) Determine the period of the circular motion in s.

In a New Context

The roller-coaster loop. A roller-coaster loop is motion in a vertical plane, like the water in the bucket. But the car's speed is determined by energy conservation, rather than being set by the rider. A circular loop has the same radius throughout. At the top, the car needs a minimum speed $v^{2} = g r$ . By the bottom it is moving much faster, so the track pushes on the riders with a normal force $N = m g + m v^{2} / r$ — about six times their weight. Real loops are clothoid (teardrop) curves instead: the small radius at the top lowers the speed needed there, and the large radius at the bottom lowers the force — which one fixed radius cannot do.

Explain why a car that just maintains contact with the track at the top of a circular loop produces a large normal force on the riders at the bottom.

Answer, with the arithmetic. At the top, just maintaining contact means $v_{top}^{2} = g r$ . Falling to the bottom drops the car through a height $2 r$ , and energy conservation gives $v_{bottom}^{2} = v_{top}^{2} + 4 g r = g r + 4 g r = 5 g r$ . At the bottom the centre is above the car, so $N - m g = m v^{2} / r$ and $N = m g + m (5 g r) / r = 6 m g$ . So the rider is pressed into the seat at six times their own weight — and the number does not depend on the radius, which is why simply making a circular loop bigger does not help. The clothoid solves it by using two different radii: small where the speed must be low, large where the force must be small.

🧪 Nature of Science — Models — roller coasters

Early roller coasters at Coney Island were rapidly improved after unsafe designs. In 1895, the Flip-Flap Railway introduced a small circular loop that was exciting but uncomfortable and dangerous. In 1901, the Loop-the-Loop used a smoother oval loop to reduce forces on passengers. Later, the Drop-the-Dip added higher speeds and introduced lap bars for passenger safety.

An answer to "your turn". The assumption removed is that the radius is a single constant. The circular model has one $r$ that appears in both conditions at once — the minimum speed at the top ( $v^{2} = g r$ , which wants $r$ small) and the force at the bottom ( $N = m g + m v^{2} / r$ , which wants $r$ large). One number cannot satisfy both, so a circular loop is always a compromise, and the compromise is what hurt the passengers on the Flip-Flap. Letting the radius vary around the loop turns one constrained number into a curve that can be small at the top and large at the bottom. The physics did not change at all; the model was simply allowed one more degree of freedom.

Your turn: Which assumption of the circular-loop model is removed in a variable-radius loop?

✅ Lesson question, answered

Why can water remain inside a bucket at the top of a vertical circle?

At the top of the circle, the centre of the circular path lies below the water, so the centripetal acceleration — and therefore the resultant radial force on the water — must be directed vertically downward, toward the centre. The water's weight acts downward, and the bucket also exerts a downward normal force while contact is maintained.

At the minimum speed required to maintain contact, the water is on the point of losing contact with the bucket, so the normal force is zero. The water's weight alone then provides the required resultant radial force, giving $v_{m i n} = g r$ .

If $v \geq g r$ , the water follows the circular path and remains inside the bucket; if $v < g r$ , it loses contact and falls out. Note what this means: the water does not stay in despite gravity — it stays in because of gravity, which at the top is pointing exactly the way the circular motion requires.

Key terms

conical pendulum: — A bob on a string tracing a horizontal circle, the string making a constant angle with the vertical.
vertical circular motion: — Motion in a vertical circle: the speed changes around the loop, so it is not uniform.
minimum speed at the top: — v_min = √(gr), the speed at which the tension or normal force just falls to zero.
normal force: — The contact force perpendicular to a surface; it can push but never pull.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

A bucket of water is swung in a vertical circle. At the top of the circle, the resultant force on the water is directed:

An object of mass m moves in a vertical circle of radius r. The minimum speed at the top for the string to stay taut is:

For a conical pendulum, the vertical forces balance but the horizontal ones do not. This is because:

A 0.20 kg ball on a string of length 0.50 m moves in a vertical circle at a constant 3.0 m/s. Comparing the tension at the top and at the bottom:

A car just maintains contact at the top of a circular loop of radius r. The normal force on the riders at the bottom is about:

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

(Exercise a) One end of a thread of length

l

[m] is fixed to the ceiling, and a small ball of mass

m

[kg] attached to the other end undergoes uniform circular motion in a horizontal plane. Let the angle between the thread and the vertical be

θ

and the acceleration due to gravity be

g

. (i) Determine the magnitude of the centripetal force in N. (ii) Determine the speed of the small ball

v

in terms of

l

g

and

θ

1.2

(Exercise b) One end of a thread of length

l

[m] is fixed to the ceiling, and a small ball of mass

m

[kg] attached to the other end undergoes uniform circular motion on a smooth horizontal plane while receiving a normal force from the surface. Let the angle between the thread and the vertical be

θ

, the angular velocity be

ω

[rad/s], and the gravitational acceleration be

g

. (i) Determine the magnitude of the tension in the thread in N. (ii) Determine the magnitude of the normal force the small ball receives from the horizontal plane in N.

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

(Exercise c) A small ball of mass

m

undergoes uniform circular motion at a height

h

inside a conical container with a smooth inner surface. Let the angle between the axis of the cone and the slant side be

θ

and the gravitational acceleration be

g

. (i) Determine the magnitude of the normal force

N

in terms of

m

θ

and

g

. (ii) Determine the speed

v

in terms of

h

and

g

. (iii) Determine the period

T

using

h

θ

and

g

2.2

A ball of mass 0.30 kg on a string of length 0.40 m moves in a vertical circle. At the top of the circle its speed is 5.0 m/s. Determine the tension in the string at the top, and then at the bottom of the circle. Take

g = 9.8

m/s².

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

A roller-coaster car starts from rest at a height

h

and runs down a smooth track into a vertical circular loop of radius

r

. Derive the minimum value of

h

for which the car just completes the loop, and explain why releasing it from the height of the top of the loop is not enough.

3.2

Compare horizontal circular motion (a conical pendulum) with vertical circular motion. Explain why one is uniform and the other is not, what follows for the tension in each case, and what this means for how each problem must be solved.

📝 Exam-style question

A small ball of mass 0.50 kg is tied to a string and moved in a vertical circle of radius 0.80 m. At the lowest point of the circle its speed is 4.0 m/s. Determine the tension in the string at the lowest point.

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