CIE AS/A Level Physics 12.1 Kinematics of uniform circular motion Study Notes- 2025-2027 Syllabus
CIE AS/A Level Physics 12.1 Kinematics of uniform circular motion Study Notes – New Syllabus
CIE AS/A Level Physics 12.1 Kinematics of uniform circular motion Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on AS/A Level Physics latest syllabus with Candidates should be able to:
- define the radian and express angular displacement in radians
- understand and use the concept of angular speed
- recall and use ω = 2π / T and v = rω
Definition of the Radian and Angular Displacement
The radian is the SI unit of angular displacement. One radian is defined as the angle subtended at the centre of a circle by an arc whose length is equal to the radius of the circle.
Mathematically:
\( \mathrm{\theta = \dfrac{s}{r}} \)
- \( \mathrm{\theta} \) = angular displacement (in radians)
- \( \mathrm{s} \) = arc length (in metres)
- \( \mathrm{r} \) = radius of the circle (in metres)
Conversion Between Degrees and Radians:
\( \mathrm{2\pi \ radians = 360^\circ} \Rightarrow 1 \ rad = \dfrac{180}{\pi}^\circ \)
Key Idea:
An angular displacement of 1 radian corresponds to an arc length equal to the radius of the circle. It is a natural measure of angle used in circular motion and oscillations.
Example
A wheel of radius \( \mathrm{0.5 \ m} \) rolls through an arc length of \( \mathrm{1.0 \ m.} \) Calculate the angular displacement in radians.
▶️ Answer / Explanation
Using \( \mathrm{\theta = \dfrac{s}{r}} \):
\( \mathrm{\theta = \dfrac{1.0}{0.5} = 2.0 \ radians.} \)
Hence, the wheel turns through 2 radians.
Example
A car moves around a circular track of radius \( \mathrm{50 \ m.} \) It travels an arc length of \( \mathrm{78.5 \ m.} \) through what angle, in radians and degrees, has it moved?
▶️ Answer / Explanation
Using the formula \( \mathrm{\theta = \dfrac{s}{r}} \):
\( \mathrm{\theta = \dfrac{78.5}{50} = 1.57 \ radians.} \)
Convert to degrees:
\( \mathrm{1.57 \times \dfrac{180}{\pi} = 90^\circ.} \)
Hence: The car moves through an angle of \( \mathrm{1.57 \ rad} \) (or \( \mathrm{90^\circ.} \))
Key Idea: An angular displacement of 1 radian corresponds to an arc length equal to the radius. Here, the car covers a quarter of the circle, or \( \mathrm{\tfrac{\pi}{2}} \) radians.
Concept of Angular Speed (ω)
The angular speed (or angular velocity in magnitude) of a rotating body is the rate of change of angular displacement with respect to time.
\( \mathrm{\omega = \dfrac{\Delta \theta}{\Delta t}} \)
- \( \mathrm{\omega} \) = angular speed (in radians per second, rad/s)
- \( \mathrm{\Delta \theta} \) = change in angular displacement (radians)
- \( \mathrm{\Delta t} \) = time interval (seconds)
For uniform circular motion:
\( \mathrm{\omega = \dfrac{\theta}{t}} \)
Key Idea:
Angular speed measures how rapidly an object is rotating — it represents how many radians it sweeps out per second.
Example
A fan blade rotates through \( \mathrm{10 \ radians} \) in \( \mathrm{2.0 \ s.} \) Calculate its angular speed.
▶️ Answer / Explanation
Using \( \mathrm{\omega = \dfrac{\theta}{t}} \):
\( \mathrm{\omega = \dfrac{10}{2.0} = 5.0 \ rad/s.} \)
Hence, the angular speed = 5 rad/s.
Example
A rotating platform makes \( \mathrm{15 \ revolutions} \) every second. Calculate the angular speed in radians per second.
▶️ Answer / Explanation
Each revolution corresponds to an angular displacement of \( \mathrm{2\pi \ radians.} \)
\( \mathrm{\omega = 2\pi f = 2\pi \times 15 = 94.2 \ rad/s.} \)
Hence: The angular speed of the platform is \( \mathrm{94.2 \ rad/s.} \)
Key Idea: Angular speed measures how quickly an object rotates. One revolution per second equals \( \mathrm{2\pi \ rad/s.} \)
Relations Between Angular Speed, Frequency, and Linear Speed
(a) Relation Between ω and Period T:
- One complete revolution corresponds to an angular displacement of \( \mathrm{2\pi \ radians.} \)
- If the time taken for one revolution is \( \mathrm{T} \), then:
\( \mathrm{\omega = \dfrac{2\pi}{T}} \)
- Alternatively, if the frequency of rotation is \( \mathrm{f = \dfrac{1}{T}}, \)
- Then \( \mathrm{\omega = 2\pi f.} \)
(b) Relation Between Linear Speed and Angular Speed:
For an object moving in a circle of radius \( \mathrm{r} \):
\( \mathrm{v = r\omega} \)
- \( \mathrm{v} \) = linear speed (m/s)
- \( \mathrm{r} \) = radius of circular path (m)
- \( \mathrm{\omega} \) = angular speed (rad/s)
The faster the angular rotation (larger \( \mathrm{\omega} \)) or the larger the radius, the greater the linear speed along the circular path.
Angular Quantities and Relations
| Quantity | Symbol | Formula | SI Unit | Description |
|---|---|---|---|---|
| Angular displacement | \( \mathrm{\theta} \) | \( \mathrm{\theta = \dfrac{s}{r}} \) | radian (rad) | Angle swept at the centre by an arc of length \( \mathrm{s.} \) |
| Angular speed | \( \mathrm{\omega} \) | \( \mathrm{\omega = \dfrac{\theta}{t}} \) | rad/s | Rate of change of angular displacement. |
| Relation with period | \( \mathrm{\omega} \) | \( \mathrm{\omega = \dfrac{2\pi}{T}} \) | rad/s | Angular speed for uniform circular motion. |
| Linear speed | \( \mathrm{v} \) | \( \mathrm{v = r\omega} \) | m/s | Speed of a point moving on the circular path. |
Example
A point on the rim of a rotating wheel moves at a linear speed of \( \mathrm{4.0 \ m/s.} \) If the wheel’s radius is \( \mathrm{0.8 \ m,} \) find the angular speed and the time for one revolution.
▶️ Answer / Explanation
Using \( \mathrm{v = r\omega} \):
\( \mathrm{\omega = \dfrac{v}{r} = \dfrac{4.0}{0.8} = 5.0 \ rad/s.} \)
Now, \( \mathrm{\omega = \dfrac{2\pi}{T}} \Rightarrow T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{5.0} = 1.26 \ s. \)
Hence: Angular speed = \( \mathrm{5.0 \ rad/s,} \) and period = \( \mathrm{1.26 \ s.} \)
Example
A stone is whirled in a horizontal circle of radius \( \mathrm{1.5 \ m.} \) making 2 revolutions per second. Find:
- (a) its angular speed,
- (b) its linear speed.
▶️ Answer / Explanation
(a) Angular speed:
Frequency \( \mathrm{f = 2 \ rev/s.} \)
\( \mathrm{\omega = 2\pi f = 2\pi \times 2 = 12.57 \ rad/s.} \)
(b) Linear speed:
\( \mathrm{v = r\omega = 1.5 \times 12.57 = 18.85 \ m/s.} \)
Hence: Angular speed = \( \mathrm{12.6 \ rad/s,} \) Linear speed = \( \mathrm{18.9 \ m/s.} \)
Key Idea: Angular and linear speeds are directly proportional — the larger the radius, the higher the tangential speed for the same rotation rate.