B2.6 Solve horizontal and vertical uniform-circular-motion problems

Ontario Grade 12 Physics

Dynamics

SPH4U expectation B2.6

Uniform circular motion means motion along a circular path at constant speed. The object’s velocity still changes because its direction changes. Velocity is a vector: it has magnitude and direction. Speed is a scalar: it has magnitude only. In this lesson, the system is the object moving in the circle, and the reference frame is the ground or another stationary observer. At each point, choose the positive radial direction to point toward the circle’s centre. This makes inward acceleration and inward net force positive. You will use Grade 11 motion and force ideas to solve circular-motion problems algebraically.

What you will learn

Identify the system, reference frame, and inward direction in a circular-motion problem.
Use the relationship between speed, radius, and centripetal acceleration.
Apply Newton’s second law to horizontal and vertical circular motion.
Check the units, direction, and reasonableness of a calculated result.

1. Review the motion and choose directions

In straight-line motion, average speed is distance divided by time. For an object completing one circle, the distance for one turn is the circumference,

2\pi r

, where

r

is the radius in metres. If the object takes a time

T

in seconds to complete one turn, its speed is the circumference divided by that time. The period,

T

, is the time for one complete revolution.

The velocity at any point is tangent to the circle. The centre-directed acceleration is perpendicular to the instantaneous velocity. It points radially inward, even when the object’s speed stays constant. The name centripetal means centre-seeking; centripetal acceleration is not a new kind of force. It describes the direction and size of the object’s acceleration.

Before solving, state the system, reference frame, and positive direction. For the radial direction, choose inward as positive. A free-body diagram should show only forces acting on the chosen object. Do not add a separate ‘centripetal force’ arrow: the inward net force is made from the real forces in the diagram.

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

Speed is a scalar; velocity and acceleration are vectors.
For circular motion, velocity is tangent and acceleration points to the centre.
Inward is positive in the radial direction throughout this lesson.

2. The course relationships and horizontal circles

The radial acceleration depends on the object’s speed and the circle’s radius. Faster motion requires greater inward acceleration. For the same speed, a smaller circle also requires greater inward acceleration. The SI unit of acceleration is metres per second squared, written

\mathrm{m/s^2}

.

Newton’s second law says that net force equals mass times acceleration. Apply it along the inward radial direction. The inward components of the real forces must add to the required net inward force. Force is measured in newtons, where

1\,\mathrm{N}=1\,\mathrm{kg\,m/s^2}

.

In a horizontal circle at constant height, the vertical forces balance when there is no vertical acceleration. The horizontal inward force can then provide the required centripetal acceleration. For a vehicle turning on a level road, friction between the tires and road can supply that inward force. For an object moving in a horizontal circle on a string, the horizontal component of string tension may supply it. Identify the actual forces in the stated situation rather than assuming one force always provides the inward net force.

a_c=

\frac{v^2}{r}

, \sum F_{in}=ma_c=

\frac{mv^2}{r}

Use the inward direction when summing radial force components.
Centripetal force means the net inward force, not an additional force.
Keep speed in metres per second, radius in metres, mass in kilograms, and force in newtons.

3. Vertical circles and changing force components

In a vertical circle, inward still points toward the centre, but its direction changes around the path. The forces also have different radial components at different locations. At the top, inward is downward. At the bottom, inward is upward. The inward net force must equal

mv^2/r

at both locations.

For an object on a string, tension points along the string toward the centre. At the top of a vertical circle, both tension and weight point inward, so both contribute to the inward net force. At the bottom, tension points inward but weight points away from the centre. The inward net force is tension minus weight.

The same radial-force method applies to an object in contact with a track. Replace tension with the appropriate contact force from the track. Use only the forces that act on the object, and include their signs according to the chosen inward-positive direction. These relationships apply at a particular point; they do not say that the force is the same everywhere on a vertical path.

\sum F_{in}=

\frac{mv^2}{r}

At the top, inward is downward; at the bottom, inward is upward.
Weight has magnitude $mg$ and points downward.
Write a separate radial force equation for each location.

4. A reliable solving routine

First identify the object, the reference frame, and the location on the circle. Record the known values and the unknown. Draw a simple circle, mark its centre, and show the velocity tangent to the path. Add a free-body diagram with the real forces and mark inward as positive.

Next find the centripetal acceleration if the speed and radius are known. Then resolve forces along the inward direction and use Newton’s second law. If speed is unknown, rearrange the same relationships algebraically. Convert any given quantities to SI units before substituting.

Finally, report a magnitude with a direction, such as ‘

12\,\mathrm{N}

inward,’ when the question asks for a force. Check that the units match the requested quantity. A calculated acceleration should have units of

\mathrm{m/s^2}

; a force should have units of

\mathrm{N}

. Ask whether the direction and size make sense: more speed should require more inward acceleration, and opposing forces must reduce the net inward force.

A diagram and a stated sign convention prevent direction errors.
Substitute SI values with units shown.
Check units, direction, significant figures, and physical reasonableness.

Worked example

A vehicle on a level turn

A

1200\,\mathrm{kg}

car travels at

15.0\,\mathrm{m/s}

around a level circular curve of radius

50.0\,\mathrm{m}

. Find the inward net force required. Treat the car as the system and use the road as the reference frame.

Set the direction
Choose the inward radial direction as positive. The unknown is the net horizontal force toward the centre. On a level curve, vertical forces balance; the required radial net force is horizontal.
Find the acceleration
Use the centripetal-acceleration relationship with speed in metres per second and radius in metres.
$a_c=\frac{(15.0\,\mathrm{m/s})^2}{50.0\,\mathrm{m}}=4.50\,\mathrm{m/s^2}$
Find the net force
Multiply the car’s mass by its inward acceleration. The positive result indicates that the net force points inward. F_{in}=(1200 $\mathrm{kg}$ )(4.50 $\mathrm{m/s^2}$ )=5.40\times10^3 $\mathrm{N}$

Answer: The required net force is

5.40\times10^3\,\mathrm{N}

inward.

Check: The units reduce to kilograms times metres per second squared, or newtons. The force is inward, as required for turning. A car moving faster or turning on a smaller radius would require a larger inward force.

Worked example

Tension at the top of a vertical circle

A

0.500\,\mathrm{kg}

object on a light string moves at

6.00\,\mathrm{m/s}

at the top of a vertical circle of radius

2.00\,\mathrm{m}

. Find the string tension at that instant. Use the object as the system, the ground as the reference frame, and downward as inward.

Identify the inward forces
At the top, both tension and weight point downward, toward the centre. Their sum supplies the inward net force. Use $g=9.81\,\mathrm{m/s^2}$ .
$T+mg=\frac{mv^2}{r}$
Rearrange for tension
Subtract the weight from the required inward net force. This leaves the tension.
$T=\frac{mv^2}{r}-mg$
Substitute and calculate
Substitute the given values in SI units. Keep the weight term in newtons.
$T=\frac{(0.500\,\mathrm{kg})(6.00\,\mathrm{m/s})^2}{2.00\,\mathrm{m}}-(0.500\,\mathrm{kg})(9.81\,\mathrm{m/s^2})=4.10\,\mathrm{N}$

Answer: The string tension is

4.10\,\mathrm{N}

, directed downward toward the centre.

Check: Each term has units of newtons. The centripetal-force requirement is greater than the weight, so the remaining inward force is positive tension. The result is physically consistent with the chosen direction.

Worked example

Speed at the bottom of a vertical circle

A

0.800\,\mathrm{kg}

object moves at the bottom of a vertical circle of radius

1.50\,\mathrm{m}

. The string tension there is

18.0\,\mathrm{N}

. Find the object’s speed. Use the object as the system, the ground as the reference frame, and upward as inward.

Write the radial force equation
At the bottom, tension points upward toward the centre and weight points downward, away from the centre. Therefore, the inward net force is tension minus weight.
$T-mg=\frac{mv^2}{r}$
Solve for speed
Rearrange the equation to isolate the square of the speed, then take the positive square root because speed is a non-negative scalar.
$v=\sqrt{\frac{r(T-mg)}{m}}$
Substitute and calculate
Use $g=9.81\,\mathrm{m/s^2}$ . The tension exceeds the weight, so the net force is inward as assumed.
$v=\sqrt{\frac{(1.50\,\mathrm{m})[(18.0\,\mathrm{N})-(0.800\,\mathrm{kg})(9.81\,\mathrm{m/s^2})]}{0.800\,\mathrm{kg}}}=4.37\,\mathrm{m/s}$

Answer: The object’s speed is

4.37\,\mathrm{m/s}

.

Check: Inside the square root, the units reduce to

\mathrm{m^2/s^2}

, so the result is in metres per second. The tension is greater than the weight, giving a positive inward net force at the bottom.

Common mistakes and how to avoid them

Treating centripetal force as an extra force to add to tension, weight, or friction.

Correction: Centripetal force is the name for the net inward force. Add the real inward and outward force components to find it.

Drawing velocity toward the centre.

Correction: Velocity is tangent to the circular path. Acceleration and net force point inward.

Using the same weight sign at the top and bottom.

Correction: Set inward as positive at each location. Weight is inward at the top and outward at the bottom.

Assuming tension always equals the centripetal force.

Correction: Use the full radial force equation. Weight or other forces may also contribute to the net inward force.

Lesson summary

Uniform circular motion has constant speed, but velocity changes direction.
Centripetal acceleration points toward the centre and has magnitude $v^2/r$ .
Use Newton’s second law in the inward radial direction: the real forces’ inward net force equals $mv^2/r$ .
For vertical circles, identify inward direction and force components separately at each position.

Check your understanding

Question 1

An object moves at constant speed in a circle. Which statement describes its acceleration?

It is zero because the speed is constant.
It points tangent to the path.
It points toward the centre.
It points away from the centre.

Show answer and explanation

It points toward the centre.

The velocity direction changes during circular motion. The resulting acceleration points toward the centre, even when speed is constant.

Question 2

At the bottom of a vertical circle, a string’s tension is

20\,\mathrm{N}

and the object’s weight is

8.0\,\mathrm{N}

. What is the inward net force?

$28\,\mathrm{N}$
$12\,\mathrm{N}$
$8.0\,\mathrm{N}$
$20\,\mathrm{N}$

Show answer and explanation

12\,\mathrm{N}

At the bottom, tension is inward and weight is outward. The inward net force is

20\,\mathrm{N}-8.0\,\mathrm{N}=12\,\mathrm{N}

.

Question 3

If an object’s speed doubles while its circular-path radius stays the same, how does its centripetal acceleration change?

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.

Centripetal acceleration depends on the square of speed. Doubling speed multiplies the acceleration by four.

Key terms

Uniform circular motion: Motion around a circular path at constant speed.
Period: The time taken to complete one revolution, measured in seconds.
Centripetal acceleration: The acceleration directed toward the centre of a circular path.
Radial direction: A direction along a line from the object toward or away from the circle’s centre.
Net force: The vector sum of all forces acting on an 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 B2.6. 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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B2.6 · Solve horizontal and vertical uniform-circular-motion problems

Ontario Grade 12 Physics

SPH4U expectation B2.6

What you will learn

1. Review the motion and choose directions

2. The course relationships and horizontal circles

3. Vertical circles and changing force components

4. A reliable solving routine

A vehicle on a level turn

Tension at the top of a vertical circle

Speed at the bottom of a vertical circle

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