B3.3 Derive course-level uniform-circular-motion relationships

Ontario Grade 12 Physics

Dynamics

How speed, period, radius, and inward acceleration are connected

In SPH3U, you used displacement, velocity, acceleration, and Newton’s laws to describe motion. This lesson applies those ideas to an object moving around a circle at constant speed. The speed stays the same, but the velocity changes because its direction changes. We will use geometry and algebra to derive relationships for speed and acceleration. These are course-level derivations; no calculus is needed.

What you will learn

Distinguish the scalar speed of an object from its changing velocity vector in uniform circular motion.
Derive relationships among speed, radius, period, and frequency.
Use velocity-vector changes to derive the inward acceleration relationship without calculus.
Connect inward acceleration to the net force that produces it.

1. Set up the motion and review the key ideas

Take the moving object as the physical system. Use a reference frame fixed to the centre of the circle, with the circle lying in a fixed plane. A reference frame is the viewpoint and coordinate system used to describe motion. At each point, define the positive radial direction as pointing outward from the centre. The inward direction is therefore negative radial. The direction of motion, called the tangential direction, changes continuously around the circle.

Uniform circular motion means motion along a circular path at constant speed. Speed is a scalar: it has magnitude but no direction. Velocity is a vector: it has magnitude and direction. So an object in uniform circular motion has constant speed but changing velocity. Acceleration describes a change in velocity, including a change in direction.

Let

r

be the radius in metres,

T

the period in seconds, and

f

the frequency in hertz. The period is the time for one complete revolution. Frequency is the number of revolutions per second. Let

v

be the constant speed in metres per second. A full revolution covers one circumference,

2\pi r

, where

\pi

is the circle constant.

f=\frac{1}{T}

The system is the moving object; the frame is fixed at the circle’s centre.
Speed is constant, but velocity changes direction.
Use SI units: radius in metres, time in seconds, speed in metres per second, and frequency in hertz.

2. Derive the speed and period relationships

During one revolution, the object travels a distance equal to the circle’s circumference. Average speed is distance divided by elapsed time. Since the speed is constant, average speed over a revolution equals the object’s speed at every point. Dividing the circumference by the period gives the first relationship.

Frequency counts revolutions per second, while period gives seconds per revolution. They are reciprocals. Replacing the period with the reciprocal of frequency gives an equivalent expression for speed. These expressions show that, for the same period, a larger radius means a larger speed.

v=\frac{2\pi r}{T}=2\pi r f

One revolution has path length $2\pi r$ .
The period and frequency are reciprocals.
The speed relationship can be written using either period or frequency.

3. Derive inward acceleration from velocity changes

Acceleration is the change in velocity divided by the time taken. To see its direction and size for circular motion, compare the velocity at two nearby points. Each velocity vector is tangent to the circle, and each has magnitude

v

. The velocity vectors point in slightly different directions.

For a small turn through angle

\Delta\theta

, the two velocity vectors and their difference form a triangle. The velocity-vector triangle has the same shape as the triangle made by the two radii. In the small-angle limit used in the course-level derivation, this gives

\Delta v/v=\Delta s/r

. Here,

\Delta v

is the magnitude of the velocity change and

\Delta s

is the short arc length between the points.

Divide both sides by the time interval

\Delta t

. The ratio

\Delta s/\Delta t

is the speed along the path, so it equals

v

. The ratio

\Delta v/\Delta t

gives the acceleration magnitude as the interval becomes small. Substitution gives

a=v^2/r

. The change in velocity points toward the centre, so the acceleration does too. With outward chosen as positive radial, its radial component is negative.

Using the speed relationships from the previous section gives equivalent forms in terms of period or frequency. These equations describe the acceleration required to keep changing the direction of motion. The acceleration is larger at higher speed and smaller at larger radius when speed is held fixed.

a_c=\frac{v^2}{r}=\frac{4\pi^2r}{T^2}=4\pi^2rf^2

The velocity vectors are tangent to the circle; their change points inward.
Inward acceleration is perpendicular to the instantaneous velocity.
The sign of the radial component depends on the chosen direction; the acceleration magnitude is positive.

4. Connect acceleration to net force

Newton’s second law relates the net force on a system to its mass and acceleration. In circular motion, the net force must point toward the centre because the acceleration points inward. Its magnitude is mass times the inward acceleration.

Centripetal means centre-directed. It names the direction of the net force or acceleration; it is not an extra kind of force. The inward net force may be supplied by one force or by the combined effect of several forces. The equation does not identify which physical force supplies it. That depends on the situation.

Keep the vector direction clear. If inward is positive for a force calculation, the radial acceleration is positive. If outward is positive, the radial components of acceleration and net force are negative. The magnitude relationship remains the same.

F_{\text{net},c}=ma_c=\frac{mv^2}{r}

The net force points inward and causes the inward acceleration.
Centripetal describes a direction, not a separate force.
Choose a sign convention and use it consistently.

Worked example

Speed from period and radius

A point on a rotating platform follows a circle of radius

0.45\ \mathrm{m}

. It completes one revolution in

1.8\ \mathrm{s}

. Find its speed.

Define the system and values
The system is the point on the platform, viewed from a frame fixed at the circle’s centre. Let outward be positive radial. The radius is $0.45\ \mathrm{m}$ and the period is $1.8\ \mathrm{s}$ . We want the scalar speed.
Choose the relationship
One revolution covers the circumference, so dividing that distance by the period gives the speed.
$v=\frac{2\pi r}{T}$
Substitute and calculate
Use the given values with SI units. Keep guard digits during the calculation, then report two significant figures because both measured values have two.
$v=\frac{2\pi(0.45\ \mathrm{m})}{1.8\ \mathrm{s}}=1.5708\ldots\ \mathrm{m/s}\approx1.6\ \mathrm{m/s}$

Answer: The point’s speed is

1.6\ \mathrm{m/s}

. Speed has no direction.

Check: Metres divided by seconds give the correct speed unit. A circumference of about

2.8\ \mathrm{m}

covered in

1.8\ \mathrm{s}

gives a speed near

1.6\ \mathrm{m/s}

, so the result is reasonable.

Worked example

Inward acceleration from speed and radius

A cart moves at

3.6\ \mathrm{m/s}

around a circular track of radius

2.4\ \mathrm{m}

. Find its acceleration magnitude and direction.

Define the system and values
The system is the cart, described from a frame fixed at the track’s centre. Let outward be positive radial. The speed is $3.6\ \mathrm{m/s}$ and the radius is $2.4\ \mathrm{m}$ . We seek the acceleration magnitude and direction.
Use the derived relationship
The velocity direction changes as the cart follows the circle. The resulting acceleration points toward the centre and has magnitude speed squared divided by radius.
$a_c=\frac{v^2}{r}$
Substitute and assign direction
Substitute the values with units. Since outward is positive, inward acceleration has a negative radial component.
$a_c=\frac{(3.6\ \mathrm{m/s})^2}{2.4\ \mathrm{m}}=5.4\ \mathrm{m/s^2}\quad\text{inward}$

Answer: The acceleration magnitude is

5.4\ \mathrm{m/s^2}

, directed toward the centre. Its radial component in the chosen sign convention is

-5.4\ \mathrm{m/s^2}

.

Check: The units reduce to metres per second squared. The inward direction matches the required change in velocity. The result is plausible: increasing speed raises the acceleration strongly because speed is squared.

Worked example

Net force from period

A

0.20\ \mathrm{kg}

object moves uniformly in a circle of radius

0.75\ \mathrm{m}

with period

1.5\ \mathrm{s}

. Find the magnitude of the net inward force.

Define the system and values
The system is the object in a frame fixed at the circle’s centre. Choose inward as positive radial for this calculation. The mass is $0.20\ \mathrm{kg}$ , the radius is $0.75\ \mathrm{m}$ , and the period is $1.5\ \mathrm{s}$ . We want the net force magnitude.
Select the period form
The acceleration relationship in terms of period can be combined with Newton’s second law. This avoids rounding an intermediate speed.
$F_{\text{net},c}=m\frac{4\pi^2r}{T^2}$
Substitute and calculate
Insert the values with SI units. The given values have two significant figures, so report the force to two significant figures.
$F_{\text{net},c}=(0.20\ \mathrm{kg})\frac{4\pi^2(0.75\ \mathrm{m})}{(1.5\ \mathrm{s})^2}=2.6319\ldots\ \mathrm{N}\approx2.6\ \mathrm{N}$

Answer: The net force has magnitude

2.6\ \mathrm{N}

and points toward the centre.

Check: The units reduce to kilograms times metres per second squared, or newtons. The force is inward, consistent with the inward acceleration. A positive magnitude of a few newtons is reasonable for the stated mass and motion.

Common mistakes and how to avoid them

Saying that velocity is constant because speed is constant.

Correction: Velocity includes direction. Its direction changes continuously during circular motion, so velocity changes even when speed does not.

Treating centripetal force as an additional force that must be added to a force diagram.

Correction: Centripetal force is the name for the inward net force. Identify the actual forces in the situation, then find their inward resultant.

Using the diameter instead of the radius in the acceleration relationship.

Correction: Use the radius, measured from the centre to the path. Check the stated geometry before substituting.

Giving the acceleration direction as tangent to the circle.

Correction: The velocity is tangent. The acceleration from the changing velocity direction points toward the centre.

Lesson summary

Uniform circular motion has constant speed but changing velocity.
The period and frequency are reciprocals, and one revolution covers a circumference.
The inward acceleration magnitude is speed squared divided by radius.
The net inward force equals mass times inward acceleration; centripetal describes its direction.

Check your understanding

Question 1

An object moves at constant speed around a circle. Which statement is correct?

Its velocity is constant because its speed is constant.
Its velocity changes because its direction changes.
Its acceleration is zero because its speed is constant.
Its acceleration points along its velocity.

Show answer and explanation

Its velocity changes because its direction changes.

Velocity is a vector. Its direction changes around the circle, so velocity changes and there is acceleration toward the centre.

Question 2

A moving object’s speed doubles while its circular path radius stays the same. What happens to the inward acceleration magnitude?

It doubles.
It becomes half as large.
It becomes four times as large.
It stays the same.

Show answer and explanation

It becomes four times as large.

The relationship uses the square of speed. Doubling speed multiplies the acceleration by four when radius is unchanged.

Question 3

For an object moving in a circle, which direction does the net force point?

Toward the centre of the circle.
Away from the centre of the circle.
Tangent to the circle in the direction of motion.
It must point upward.

Show answer and explanation

Toward the centre of the circle.

The net force points in the direction of the acceleration. The acceleration required for uniform circular motion is inward.

Key terms

Uniform circular motion: Motion along a circular path at constant speed.
Period: The time required for one complete revolution.
Frequency: The number of complete revolutions per second.
Centripetal: Directed toward the centre of a circular path.
Radial direction: A direction along a line from the circle’s centre to the moving object.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B3.3. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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B3.3 · Derive course-level uniform-circular-motion relationships

Ontario Grade 12 Physics

How speed, period, radius, and inward acceleration are connected

What you will learn

1. Set up the motion and review the key ideas

2. Derive the speed and period relationships

3. Derive inward acceleration from velocity changes

4. Connect acceleration to net force

Speed from period and radius

Inward acceleration from speed and radius

Net force from period

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through SPH4U

About this lesson and its review