Motion in a Plane • Topic 3 of 3

Uniform Circular Motion

A stone tied to a string and whirled overhead, the tip of a fan blade, a satellite circling the Earth — all move along a circular path. When an object moves in a circle at constant speed, we call it uniform circular motion. The word uniform refers to the speed, not the velocity: even though the speed never changes, the direction of motion changes at every instant, so the velocity changes — which means the object is always accelerating.

Instead of measuring distance along the arc, it is easier to measure the angle swept at the centre. This is the angular displacement $\theta$, measured in radians (one full circle is $2\pi$ radians). The rate at which this angle is swept is the angular velocity $\omega=\dfrac{\theta}{t}$, with SI unit radian per second ($\text{rad/s}$). The ordinary (linear) speed $v$ along the circle of radius $r$ is linked to angular velocity by the simple and important relation $v=\omega r$.

Circular motion repeats itself, so we describe it with two more quantities. The time period $T$ is the time for one complete revolution, and the frequency $f$ is the number of revolutions per second, with $f=\dfrac{1}{T}$ (unit hertz, $\text{Hz}$). In one period the object sweeps $2\pi$ radians, so $\omega=\dfrac{2\pi}{T}=2\pi f$. These let us convert easily between revolutions, angles and time.

The most important idea is the acceleration. Because the velocity continuously changes direction, there is an acceleration directed at every instant towards the centre of the circle. This is the centripetal acceleration $a_c=\dfrac{v^2}{r}=\omega^2 r$. The word centripetal means centre-seeking. The force that produces it — tension in the string, gravity for a satellite, friction for a car turning — is the centripetal force $F_c=\dfrac{mv^2}{r}=m\omega^2 r$, always directed towards the centre. Without it, the object would fly off along a tangent in a straight line, exactly as Newton's first law predicts.

  • Uniform circular motion: constant speed, continuously changing direction (so the velocity changes).
  • Angular velocity: $\omega=\dfrac{\theta}{t}$, unit rad/s; linked to speed by $v=\omega r$.
  • Period and frequency: $f=\dfrac{1}{T}$ and $\omega=\dfrac{2\pi}{T}=2\pi f$.
  • Centripetal acceleration: $a_c=\dfrac{v^2}{r}=\omega^2 r$, directed towards the centre.
  • Centripetal force: $F_c=\dfrac{mv^2}{r}=m\omega^2 r$, also towards the centre.
Uniform circular motion: velocity along tangent, acceleration towards centrecentrerv (tangent)a_ca_c = v^2 / r = omega^2 r (towards centre)
1
Worked Example
A particle moves in a circle of radius $0.5\,\text{m}$ with a constant speed of $4\,\text{m/s}$. Find its centripetal acceleration.
Solution
  1. $a_c=\dfrac{v^2}{r}$ with $v=4\,\text{m/s}$, $r=0.5\,\text{m}$.
  2. $a_c=\dfrac{4^2}{0.5}=\dfrac{16}{0.5}$.
  3. $=32\,\text{m/s}^2$, directed towards the centre.

Answer: $a_c=32\,\text{m/s}^2$.

2
Worked Example
A body completes one revolution of a circle in $4\,\text{s}$. Find its angular velocity. (Take $\pi=3.14$.)
Solution
  1. $\omega=\dfrac{2\pi}{T}$ with $T=4\,\text{s}$.
  2. $\omega=\dfrac{2\times 3.14}{4}=\dfrac{6.28}{4}$.
  3. $=1.57\,\text{rad/s}$.

Answer: $\omega=1.57\,\text{rad/s}$.

3
Worked Example
An object moves in a circle of radius $2\,\text{m}$ with angular velocity $3\,\text{rad/s}$. Find its linear speed.
Solution
  1. Use $v=\omega r$ with $\omega=3\,\text{rad/s}$, $r=2\,\text{m}$.
  2. $v=3\times 2$.
  3. $=6\,\text{m/s}$.

Answer: Linear speed $=6\,\text{m/s}$.

4
Worked Example
A wheel rotates at $120\,\text{revolutions per minute (rpm)}$. Find its frequency in $\text{Hz}$ and its time period.
Solution
  1. Frequency $f=\dfrac{120}{60}=2\,\text{Hz}$ (revolutions per second).
  2. Time period $T=\dfrac{1}{f}=\dfrac{1}{2}$.
  3. $=0.5\,\text{s}$.

Answer: $f=2\,\text{Hz}$, $T=0.5\,\text{s}$.

5
Worked Example
A stone of mass $0.2\,\text{kg}$ is whirled in a circle of radius $1\,\text{m}$ at $5\,\text{m/s}$. Find the centripetal force.
Solution
  1. $F_c=\dfrac{mv^2}{r}$ with $m=0.2\,\text{kg}$, $v=5\,\text{m/s}$, $r=1\,\text{m}$.
  2. $F_c=\dfrac{0.2\times 5^2}{1}=\dfrac{0.2\times 25}{1}$.
  3. $=5\,\text{N}$, directed towards the centre.

Answer: $F_c=5\,\text{N}$.

6
Worked Example
A particle in a circle of radius $0.25\,\text{m}$ has angular velocity $4\,\text{rad/s}$. Find its centripetal acceleration using $a_c=\omega^2 r$.
Solution
  1. $a_c=\omega^2 r$ with $\omega=4\,\text{rad/s}$, $r=0.25\,\text{m}$.
  2. $a_c=4^2\times 0.25=16\times 0.25$.
  3. $=4\,\text{m/s}^2$.

Answer: $a_c=4\,\text{m/s}^2$.

Key Points

  • In uniform circular motion the speed is constant but the direction (and hence velocity) changes continuously.
  • Angular velocity $\omega=\dfrac{\theta}{t}$ (rad/s) links to linear speed by $v=\omega r$.
  • Period and frequency: $f=\dfrac{1}{T}$ and $\omega=\dfrac{2\pi}{T}=2\pi f$.
  • Centripetal acceleration $a_c=\dfrac{v^2}{r}=\omega^2 r$ always points towards the centre.
  • The centripetal force $F_c=\dfrac{mv^2}{r}=m\omega^2 r$ keeps the body on its circular path.
Tap an option to check your answer0 / 4
Q1.In uniform circular motion, which quantity stays constant?
Explanation: The speed is constant, but the direction (and so the velocity and the direction of acceleration) keeps changing.
Q2.The centripetal acceleration of a body in a circle of radius r at speed v is:
Explanation: Centripetal acceleration a_c = v^2 / r = omega^2 r, directed towards the centre.
Q3.The relation between linear speed v and angular velocity omega is:
Explanation: Linear speed equals angular velocity times the radius: v = omega r.
Q4.The centripetal acceleration in uniform circular motion is always directed:
Explanation: Centripetal means centre-seeking; the acceleration points towards the centre of the circle at all times.