B2.7 · Investigate centripetal force, acceleration, radius, period, and speed

Ontario Grade 12 Physics

Dynamics

Centripetal force, acceleration, radius, period, and speed

What you will learn

• Describe circular motion using a clear system, reference frame, and direction convention.
• Relate centripetal acceleration, speed, radius, and period.
• Explain how the net inward force produces centripetal acceleration.
• Plan an investigation that tests relationships among the quantities in circular motion.

1. Start with the system and motion

• Velocity is tangent to the circular path; centripetal acceleration points inward.
• Centripetal force means the net inward force, not an extra force to add to a force diagram.

2. Connect radius, period, speed, and acceleration

• Use metres for radius, seconds for period, metres per second for speed, and metres per second squared for acceleration.
• The period is the time for one revolution, not the time for a fraction of a revolution.

3. Relate acceleration to net inward force

• Centripetal force is the net inward force required for the circular motion.
• Keep measured evidence distinct from calculated predictions and proposed procedures.

4. Use the relationships to interpret changes

• At fixed radius, acceleration is proportional to speed squared.
• At fixed speed, acceleration decreases when radius increases.

Circular-motion quantities

Worked example

Find speed and acceleration from a period

1. Define the motion
   The system is the marker, viewed from a room-fixed frame. Choose inward as positive at the marker's current position. The unknowns are speed and the magnitude of centripetal acceleration.
2. Find speed
   One revolution covers the circumference. Divide that distance by the period, keeping metres and seconds in the substitution.
   $$v=\frac{2\pi r}{T}=\frac{2\pi(0.80\,\mathrm{m})}{2.0\,\mathrm{s}}=2.5\,\mathrm{m/s}$$
3. Find inward acceleration
   Use the speed and radius. The acceleration points toward the centre, so its direction is inward.
   $$a_c=\frac{v^2}{r}=\frac{(2.5\,\mathrm{m/s})^2}{0.80\,\mathrm{m}}=7.8\,\mathrm{m/s^2}$$

Worked example

Find the required net inward force

1. Define the system
   The system is the moving object in a room-fixed frame. Inward is positive. The known values are mass, speed, and radius; the unknown is the net inward force.
2. Calculate acceleration
   First determine the inward acceleration from speed and radius. Its direction is toward the circle's centre.
   $$a_c=\frac{v^2}{r}=\frac{(3.0\,\mathrm{m/s})^2}{1.5\,\mathrm{m}}=6.0\,\mathrm{m/s^2}$$
3. Apply Newton's second law
   The net inward force equals mass times inward acceleration. Keep the direction with the result.
   $$F_c=ma_c=(0.50\,\mathrm{kg})(6.0\,\mathrm{m/s^2})=3.0\,\mathrm{N}$$

Worked example

Compare two periods at the same radius

1. Set the comparison
   The system is the same object in the same room-fixed frame. Inward is positive at each position. Radius is held constant, and the two periods are given.
2. Calculate the initial values
   Use the circumference divided by period for speed, then use speed squared divided by radius for acceleration.
   $$v_1=\frac{2\pi(0.60\,\mathrm{m})}{1.5\,\mathrm{s}}=2.5\,\mathrm{m/s},\quad a_1=\frac{(2.5\,\mathrm{m/s})^2}{0.60\,\mathrm{m}}=10\,\mathrm{m/s^2}$$
3. Calculate the later values
   Repeat with the shorter period. The shorter time for the same circumference should produce a larger speed and acceleration.
   $$v_2=\frac{2\pi(0.60\,\mathrm{m})}{1.0\,\mathrm{s}}=3.8\,\mathrm{m/s},\quad a_2=\frac{(3.8\,\mathrm{m/s})^2}{0.60\,\mathrm{m}}=24\,\mathrm{m/s^2}$$

Common mistakes and how to avoid them

Lesson summary

• Circular motion involves changing velocity direction, even when speed is steady.
• Centripetal acceleration and net centripetal force point toward the circle's centre.
• Speed is circumference divided by period; acceleration depends on speed squared and radius.
• Use stated conditions, SI units, and measured evidence carefully when investigating relationships.

Check your understanding

Key terms

Centripetal

Directed toward the centre of a circular path.

Period

The time for one complete revolution.

Radius

The distance from the centre of a circle to its path.

Scalar

A quantity with size but no direction.

Vector

A quantity with both size and direction.

Continue through SPH4U

View the complete SPH4U Ontario Grade 12 Physics curriculum and lessons

• B1.1 · Analyse a device that applies linear or circular motion
• B1.2 · Assess impacts of linear- and circular-motion technologies
• B2.1 · Use terminology for frames, components, friction, and circular motion
• B2.2 · Solve projectile and relative-motion problems with two-dimensional vectors
• B2.3 · Solve two-dimensional force and friction problems
• B2.4 · Predict and investigate forces acting on systems of objects

About this lesson and its review

Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 12 Physics (SPH4U), expectation B2.7. 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.

Official curriculum reference

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