B1.1 Analyse a device that applies linear or circular motion

Ontario Grade 12 Physics

Dynamics

Ontario Grade 12 Physics — study-guide label B1.1

A device that applies motion transfers energy and forces to an object so that the object moves in a planned way. A conveyor belt moves objects along a straight path. A rotating platform carries objects around a centre. Analysing such a device means identifying what moves, describing its motion, and connecting that motion to the device’s action. This lesson focuses on linear and circular motion. It uses familiar Grade 11 ideas about displacement, velocity, acceleration, and net force.

What you will learn

Describe how a device produces linear or circular motion.
Choose a system, reference frame, and positive direction before analysing motion.
Use motion and force relationships to connect a device’s design with the motion it produces.
Check calculations using units, direction, and physical reasonableness.

1. Start with the system and the motion

A system is the object or group of objects chosen for analysis. For a conveyor, the system might be one package. For a rotating platform, it might be one point on the platform or an object carried by it. State the system before using an equation, because forces and motion refer to that chosen system.

A reference frame is the viewpoint used to describe position and motion. In these examples, use the floor or surrounding room as a frame at rest. Choose a positive direction. For straight motion, it could be to the right. For circular motion, describe direction around the centre, such as clockwise, and identify the centre of the circular path.

A scalar has magnitude only, such as speed or time. A vector has magnitude and direction, such as displacement, velocity, acceleration, and force. Speed tells how fast something moves; velocity also tells its direction. Acceleration is the change in velocity over time, so a change in direction can mean acceleration even if speed stays constant.

\vec{a}=\frac{\Delta\vec{v}}{\Delta t}

Name the moving system and the frame before describing its motion.
State a positive direction and use it consistently.
Distinguish speed from velocity, and scalar quantities from vectors.

2. Describe linear motion and the forces that produce it

Linear motion follows a straight path. In a simple model, an object’s average velocity is its displacement divided by the time taken. Displacement is the change in position and includes direction. If velocity changes, the object accelerates. A positive acceleration points along the chosen positive direction; a negative acceleration points opposite to it.

A free-body diagram shows the forces acting on the chosen system as arrows. The arrow direction shows force direction, and the arrow length can represent relative size. The net force is the vector total of all forces on the system. If the net force is zero, velocity does not change. If the net force is not zero, the system accelerates in the direction of the net force.

Newton’s second law connects net force, mass, and acceleration. Mass is measured in kilograms, acceleration in metres per second squared, and force in newtons. One newton is one kilogram metre per second squared. In a device analysis, use this relationship to explain how the device’s force can change an object’s motion. Do not assume that a larger applied force always means a larger net force: other forces may oppose it.

\vec{F}_{\text{net}}=m\vec{a}

Linear displacement and velocity include direction.
A free-body diagram includes forces on the selected system, not forces the system applies to other objects.
Net force and acceleration point in the same direction.

3. Describe circular motion in a device

Circular motion follows a circular path around a centre. For an object moving at steady speed in a circle, its velocity still changes because its direction changes. Its acceleration points toward the centre. This inward acceleration is called centripetal acceleration; centripetal means centre-seeking.

For an object moving at speed

v

on a circle of radius

r

, the inward acceleration depends on the square of the speed and on the radius. A smaller radius or a higher speed requires greater inward acceleration. The net force must also point inward, since net force and acceleration have the same direction. The inward net force is sometimes called centripetal force. It is not an extra kind of force; it is the name for the net force directed toward the centre.

A device may supply this inward net force through contact, tension, or another force. Identify the actual force from the device’s design, then draw it on a free-body diagram. For example, a rotating platform can exert a contact force on an object it carries. The direction of that force must be consistent with the object’s circular path.

When analysing a device, connect its purpose to evidence in the motion description: the path, speed or velocity, and any change in velocity. Then identify the relevant forces and use the appropriate relationship. Equations describe a model; they do not replace checking that the model matches the device and the stated conditions.

a_{\text{c}}=\frac{v^2}{r}

In steady-speed circular motion, velocity changes direction even when speed is constant.
Centripetal acceleration and net force point toward the centre.
The device supplies an actual force; “centripetal force” describes the inward net force.

4. Analyse a device with a clear sequence

Use a consistent sequence. First, state the system, reference frame, and positive direction. Next, describe the path and identify known and unknown quantities. Draw a motion sketch or free-body diagram when it helps show direction. Select the relationship that matches the motion: average velocity for displacement over time, Newton’s second law for net force and acceleration, or the circular-motion relationship for inward acceleration.

Substitute values with SI units. Keep signs and directions visible, especially for straight-line motion. For circular motion, use the speed as a magnitude and state the inward direction separately. Round the final value to a sensible number of significant figures based on the given information.

Finally, check the result. The units should match the requested quantity. The sign or stated direction should agree with the chosen convention. Ask whether the size is plausible: for example, a faster circular motion should require more inward acceleration, while a larger radius at the same speed should require less.

\vec{v}_{\text{avg}}=\frac{\Delta\vec{x}}{\Delta t}

Choose the relationship that fits the path and the known quantities.
Carry units through substitution and report a direction for vector results.
Check units, direction, and whether the result changes sensibly when conditions change.

Worked example

A package on a powered conveyor

A package is the system. It moves right from rest on a straight conveyor and reaches

1.8\ \mathrm{m/s}

after

3.0\ \mathrm{s}

. In the room frame, right is positive. Find its average acceleration.

Set the model
The package follows a straight path. Its initial velocity is zero, and its final velocity is positive because it moves right. Average acceleration is the change in velocity divided by the time interval.
$a_{\text{avg}}=\frac{v_f-v_i}{\Delta t}$
Substitute with units
Use the stated velocities and elapsed time. The units reduce to metres per second squared.
$a_{\text{avg}}=\frac{1.8\ \mathrm{m/s}-0\ \mathrm{m/s}}{3.0\ \mathrm{s}}=0.60\ \mathrm{m/s^2}$

Answer: The package’s average acceleration is

0.60\ \mathrm{m/s^2}

to the right.

Check: The positive sign matches the chosen rightward direction. The units are acceleration units, and a modest speed increase over three seconds gives a plausible value.

Worked example

Force applied to a moving cart

A cart of mass

12\ \mathrm{kg}

is the system. On a level track, the device produces a net force of

30\ \mathrm{N}

to the right. Find the cart’s acceleration in the room frame, where right is positive.

Choose the relationship
The given force is the net force on the cart. Newton’s second law gives acceleration from net force divided by mass.
$a=\frac{F_{\text{net}}}{m}$
Substitute and identify direction
The force is positive in the selected direction. Divide by the cart’s mass and retain the direction in the result.
$a=\frac{+30\ \mathrm{N}}{12\ \mathrm{kg}}=+2.5\ \mathrm{m/s^2}$

Answer: The cart accelerates at

2.5\ \mathrm{m/s^2}

to the right.

Check: A newton per kilogram is equivalent to a metre per second squared. The positive result agrees with the rightward net force.

Worked example

An object carried by a rotating platform

A small object is the system. A platform carries it at steady speed

4.0\ \mathrm{m/s}

in a circle of radius

2.0\ \mathrm{m}

. Find its centripetal acceleration and state its direction.

Set the circular-motion model
The object follows a circular path in the room frame. Its speed is constant, but its velocity direction changes. The inward acceleration is found from speed and radius.
$a_{\text{c}}=\frac{v^2}{r}$
Substitute with SI units
Square the speed, then divide by the radius. The result has units of metres per second squared.
$a_{\text{c}}=\frac{(4.0\ \mathrm{m/s})^2}{2.0\ \mathrm{m}}=8.0\ \mathrm{m/s^2}$

Answer: The centripetal acceleration is

8.0\ \mathrm{m/s^2}

toward the centre of the circular path.

Check: The units reduce to acceleration units. The direction is inward, as required for circular motion; doubling the speed at the same radius would make the acceleration four times as large.

Common mistakes and how to avoid them

Treating speed and velocity as interchangeable.

Correction: Speed is a scalar. Velocity is a vector, so it includes direction. In circular motion, velocity changes direction even at steady speed.

Drawing the centripetal force as an extra force in addition to the device’s actual forces.

Correction: Draw the actual forces. Their inward vector total is the centripetal net force.

Using the object’s motion direction as the direction of circular acceleration.

Correction: For circular motion, centripetal acceleration points toward the centre, not along the path.

Using an applied force in Newton’s second law when the question requires net force.

Correction: Account for all forces on the system. Use their vector total as the net force.

Lesson summary

Begin by naming the system, reference frame, and positive direction.
Use displacement, velocity, acceleration, and net force to analyse straight-line motion.
For steady-speed circular motion, acceleration and net force point toward the centre.
Use SI units, show direction, and check the result for unit and physical reasonableness.

Check your understanding

Question 1

A device carries an object at steady speed around a circle. Which statement is correct?

The object has no acceleration because its speed is constant.
The object’s velocity changes direction, and its acceleration points toward the centre.
The object’s acceleration points along the direction of travel.
The object’s net force points away from the centre.

Show answer and explanation

The object’s velocity changes direction, and its acceleration points toward the centre.

Velocity includes direction. As the object moves around the circle, its velocity direction changes, so it accelerates inward.

Question 2

A cart has a net force of

18\ \mathrm{N}

to the left and a mass of

6.0\ \mathrm{kg}

. What is its acceleration?

$3.0\ \mathrm{m/s^2}$ to the left
$3.0\ \mathrm{m/s^2}$ to the right
$108\ \mathrm{m/s^2}$ to the left
$0\ \mathrm{m/s^2}$

Show answer and explanation

3.0\ \mathrm{m/s^2}

to the left

Newton’s second law gives acceleration as net force divided by mass. The magnitude is

18\div6.0=3.0\ \mathrm{m/s^2}

, in the direction of the net force.

Question 3

An object moves at the same speed on a circular path. If the path radius increases, what happens to its centripetal acceleration?

It increases in direct proportion to the radius.
It decreases because acceleration is the speed divided by radius.
It decreases because the squared speed is divided by a larger radius.
It stays the same because the speed is unchanged.

Show answer and explanation

It decreases because the squared speed is divided by a larger radius.

At fixed speed, centripetal acceleration is the squared speed divided by radius. A larger radius therefore gives a smaller acceleration.

Key terms

System: The object or group of objects selected for analysis.
Reference frame: The viewpoint used to describe an object’s position and motion.
Net force: The vector total of all forces acting on a system.
Centripetal acceleration: The inward acceleration of an object moving along a circular path.
Free-body diagram: A drawing that shows the forces acting on one selected system.

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About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B1.1. 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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B1.1 · Analyse a device that applies linear or circular motion

Ontario Grade 12 Physics

Ontario Grade 12 Physics — study-guide label B1.1

What you will learn

1. Start with the system and the motion

2. Describe linear motion and the forces that produce it

3. Describe circular motion in a device

4. Analyse a device with a clear sequence

A package on a powered conveyor

Force applied to a moving cart

An object carried by a rotating platform

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