Motion in a Plane · Lesson 12 of 13

Angular Speed, Period and Circular-Motion Problems

“Connect angular and linear measures and solve circular-motion and arc-distance applications.”

Learning Objectives

• Convert radians, revolutions, angular speed, period and frequency. • Derive v = ωR and a_c = ω²R. • Apply circular-motion relations with consistent units. • Compare arc distance, displacement and average rates for a circular route.

Describe a turn by an angle

A radian measures the central angle as arc length divided by radius: θ = s/R. One complete turn has arc length 2πR, so it contains 2π radians, or 360°. Radians are dimensionless ratios, but the label rad identifies an angular quantity. The relation s = Rθ requires θ in radians. Using a degree value directly in that relation would introduce a factor-of-180/π error.

Angular speedLaTeX

ω = \frac{d θ}{d t}, v = \frac{d s}{d t} = R \frac{d θ}{d t} = R ω

Here θ is an angular coordinate increasing in the direction of motion, so ω represents positive angular speed. For constant speed it equals angular distance divided by elapsed time.

For a signed angle measured counterclockwise positive, dθ/dt can be positive or negative, and angular speed is its magnitude. Angular velocity is an axial vector perpendicular to the plane, assigned by the right-hand rule: curl your right-hand fingers in the rotation direction and your thumb gives the axis direction. For a circle in the x–y plane, counterclockwise rotation has angular velocity along +z. The present calculations mainly use the non-negative angular speed.

One revolution sets the period

Period and frequencyLaTeX

f = \frac{1}{T}, ω = \frac{2 π}{T} = 2 π f, v = \frac{2 π R}{T} = 2 π R f

T is the time per revolution in seconds. Frequency f is revolutions per second, measured in hertz (Hz). Angular speed is measured in rad/s.

Acceleration in angular formLaTeX

a_{c} = \frac{( R ω ) ^{2}}{R} = R ω^{2} = \frac{4 π ^{2} R}{T ^{2}} = 4 π^{2} R f^{2}

These are equivalent forms of the same uniform-circular acceleration. Choose the one matching the given data.

A small object and a larger one can share angular speed while having different linear speeds: at the same ω, v is proportional to radius. At fixed v, acceleration v²/R decreases as radius increases. At fixed ω, acceleration ω²R increases with radius. Always identify which quantity is being held fixed before describing a radius dependence.

An insect moving in a circular groove

Problem
An insect completes seven revolutions in 100 s in a groove of radius 12 cm. Find angular speed, linear speed and acceleration magnitude.

1.Frequency is 7/100 = 0.07 Hz; period is 100/7 = 14.29 s.
2.Angular speed is 2π × 0.07 = 0.4398 rad/s.
3.Radius is 0.12 m. Linear speed is 0.4398 × 0.12 = 0.05278 m/s, or 5.278 cm/s.
4.Acceleration magnitude is 0.4398² × 0.12 = 0.02321 m/s², or 2.321 cm/s². Its direction continually points toward the centre, so the acceleration vector is not constant.

A stone on an 80 cm string

Problem
A stone travels uniformly in a horizontal circle of radius 80 cm, making 14 revolutions in 25 s. Find acceleration.

1.Frequency is 14/25 = 0.56 Hz, so angular speed is 2π × 0.56 = 3.5186 rad/s.
2.Radius is 0.80 m. Acceleration is ω²R = 3.5186² × 0.80 = 9.904 m/s².
3.The acceleration is directed horizontally inward at every instant. Stating only 9.904 m/s² supplies magnitude but omits direction.
4.The linear speed is ωR = 2.815 m/s; inserting it into v²/R gives the same acceleration.

Distance around a circle is not its chord

For endpoints separated by a central angle θ between 0 and 2π, an arc of that angular length has distance Rθ. The endpoint displacement magnitude is 2R|sin(θ/2)|. The arc depends on which part of the circle was followed, while displacement depends only on endpoints. After a complete turn the arc length is 2πR and displacement is zero. For a semicircle, distance is πR and displacement magnitude is 2R.

Arc and chordLaTeX

s = R θ, ∣Δ r ∣ = 2 R sin \frac{θ}{2}

θ is in radians. For multiple turns, total angular distance includes the full turns but the endpoints determine displacement.

A cycling route through a park

Problem
A cyclist travels from the centre O to edge P of a 1 km radius park, follows the circumference counterclockwise through a quarter-turn to Q, then returns along QO. Total time is 10 min. Find displacement and average rates.

1.The two radial legs total 2 km. The quarter-circle arc is (π/2) × 1 km. Total distance is 2 + π/2 = 3.5708 km.
2.The cyclist returns to O, so net displacement is the zero vector and average velocity is zero.
3.Ten minutes is 1/6 h. Average speed is (2 + π/2)/(1/6) = 21.4248 km/h, or 5.9513 m/s.
4.The quarter-turn route must be specified; a three-quarter-turn route between the same edge points would have a larger distance but the same zero net displacement for the complete trip.

A useful unit checklist

Convert centimetres to metres when using m/s², convert kilometres per hour to metres per second by dividing by 3.6, and convert revolutions per minute to hertz by dividing by 60. Multiply frequency by 2π to obtain angular speed. Do not insert a revolution count directly where a radian angle is required. A factor of 2π is often the clue when two answers differ unexpectedly.

Quick check

A particle completes one revolution every 2 s. What is its angular speed?

Key Takeaways

• One revolution equals 2π radians. • Frequency is 1/T and angular speed is 2π/T. • Linear speed is ωR and inward acceleration is ω²R. • Keep fixed-v and fixed-ω comparisons distinct. • Arc length determines distance; endpoint geometry determines displacement.

Previous · Lesson 11

Uniform Circular Motion and Centripetal Acceleration

Next · Lesson 13

Chapter Summary and Practice

Describe a turn by an angle

One revolution sets the period

Distance around a circle is not its chord

A useful unit checklist

Quiz

Key Takeaways