Uniform Circular Motion - Definition, Examples, Diagram, Characteristic, Examples, FAQs

What Is Uniform Circular Motion?

Circular motion is the motion of a body travelling in a path around a fixed point in the shape of a circle. Uniform circular motion implies that if the body travels equal distances along the circumference of the circle in equal intervals of time, then the motion is said to be uniform circular motion. Naturally, this means that it is a type of circular motion in which the speed is maintained while the direction of the velocity is altered.

Uniform circular motion examples:

(i) Motion of the moon around the Earth
(ii) Motion of a satellite around its planet

Note: Circular motion is also known as accelerated motion

Uniform Circular Motion Formula

Uniform circular motion Formula Class 9 or other classes are given here.

In uniform circular motion, a particle moves with constant speed.

• Angular displacement $\Delta \theta=\frac{\operatorname{Arc}\left(P P^{\prime}\right)}{r}$
• Angular velocity $\omega=\frac{\Delta \theta}{\Delta t}=\frac{2 \pi}{T}=2 \pi v$
• Linear speed $v=r_\omega$

• Centripetal acceleration is due to a change in direction of velocity and is always directed towards the centre.

  $
  a={v^2/r}=r \omega^2=4 \pi^2 v^2 r=v \omega
  $

Terms Related to Uniform Circular Motion

The term related to circular motion is given below:

Radius vector

The vector joining the centre of the circular path to the position on the circular path is called the radius vector.

Angular position

The angle made by the radius vector with a reference line (arbitrarily chosen diameter) is called angular position. The direction of angular position can be clockwise or anticlockwise depending upon the choice of frame of reference. The angular position of the particle at position "P" is denoted by the y $\text { angle } \theta$ in the diagram above.

Angular displacement

The change in angular position is called angular displacement. It is the angle through which the radius vector rotates during the given circular motion.

The angular displacement between positions 'P' and 'Q' is denoted $\text { by } \Delta \theta$ in the diagram above with the S.I unit of angular position and angular displacement is Radian.

Angular velocity

Denoted by $\omega$ (omega)
$\omega$-Rate of change of angular displacement.
Average angular velocity-
$
\omega_{\text {avg }}=\frac{\Delta \theta}{\Delta t}
$

Instantaneous angular velocity

$
\omega=\frac{d \theta}{d t}
$
S.I. units- Radian per second (rad per sec )
$\omega$ is a vector quantity
The direction of $\omega$ is given by the Right-hand rule.

According to the right-hand rule, if you hold the axis with your right hand and rotate the fingers in the direction of motion of the rotating body, then the thumb will point in the direction of the angular velocity.

Angular Acceleration

The rate of change of angular velocity with time is said to be Angular Acceleration.

$
\alpha=\frac{\Delta \omega}{\Delta t}
$

• SI units- $\operatorname{rad} .(\mathrm{sec})^{-2}$
• Angular Acceleration is a vector quantity.

The direction of Angular Acceleration

a) If angular velocity is increasing then the direction of Angular Acceleration is in the direction of angular velocity.

b) If angular velocity is decreasing then the direction of Angular Acceleration is in the direction which is opposite to the direction of angular velocity.

Time is taken to complete one rotation

Formula-

$
T=\frac{2 \pi}{\omega}
$

Where $\omega=$ angular velocity
If $\mathrm{N}=\mathrm{no}$. of revolutions and total time then
$
T=\frac{t}{N} \text { or } \quad\left(\omega=\frac{2 \pi N}{t}\right)
$

• S.I unit - seconds (s)

Frequency

The total number of rotations in one second.
Formula-
$
\nu=\frac{1}{T}
$

• S.I. unit $=$ Hertz

We can write the relation between angular frequency and frequency as
$
w=2 \pi \nu
$

Centripetal Acceleration and Tangential acceleration

Centripetal acceleration: When a body is moving in a uniform circular motion, a force is responsible for changing the direction of its velocity. This force acts towards the centre of the circle and is called centripetal forceThe acceleration produced by this force is centripetal acceleration.

Formula-

$
a_c=\frac{V^2}{r}
$

Where $a_c=$ Centripetal acceleration,
$\mathrm{V}=$ linear velocity
$r=$ radius

Tangential acceleration: During circular motion, if the speed is not constant, then along with centripetal acceleration, there is also a tangential acceleration, which is equal to the rate of change of magnitude of linear velocity.

$a_t=\frac{\mathrm{d} v}{\mathrm{~d} t}$

Relation Between Angular Acceleration and Tangential Acceleration

$\overrightarrow{a_t}=\vec{\alpha} \times \vec{r}$

Where:

• $\overrightarrow{a_t}=$ Tangential acceleration (linear acceleration along the tangent to the circular path)
• $\vec{\alpha}=$ Angular acceleration vector (rate of change of angular velocity)
• $\vec{r}^*=$ Position vector (radius vector from axis of rotation to the point of interest)

Total acceleration

The vector sum of Centripetal acceleration and tangential acceleration is called Total acceleration.

Formula-

$a_n=\sqrt{a_c^2+a_t^2}$

The angle between Net acceleration and tangential acceleration $(\theta)$

From the above diagram-
$
\tan \theta=\frac{a_c}{a_t}
$

Also Check-

Uniform Circular Motion Questions and Answers

Example 1: If a body moving in a circular path maintains a constant speed of 10 ms-1, then which of the following correctly describes the relation between acceleration and radius?

1)

2)

3)

4)

Solution:

$a=\frac{v^2}{r}$

Figure Shows Centripetal acceleration

$\begin{aligned}
& a=\frac{v^2}{r} \because|\vec{v}|=\text { constant } \\
& a \propto \frac{1}{r} \text { or } a r=\text { constant }
\end{aligned}$

Hence, the graph between a and r will be a hyperbola.

Example 2: A Point P moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of $P$ is such that it sweeps out a length that is in metres and $t$ is in seconds. The radius of the path is $\mathbf{2 0 ~ m}$, The acceleration (in $\mathrm{m} / \mathrm{s}^2$ ) of $P$ When $t=2 s$ is nearly

1) 14

2) 13

3) 12

4) 7.2

Solution:

$
\begin{aligned}
& \text { As } S=t^3+3 \\
& V=\frac{d s}{d t}=3 t^2+0 \\
& \Rightarrow v=3 t^2
\end{aligned}
$
tangential acceleration
$
\begin{aligned}
& =a_t=\frac{d v}{d t}=\frac{d\left(3 t^2\right)}{d t} \\
& a_t=6 t
\end{aligned}
$

At $t=2 \mathrm{sec}$
$
\begin{aligned}
& v=3(2)^2=12 \mathrm{~ms}^{-1} \\
& a_t=6 \times 2=12 \mathrm{~ms}^{-2}
\end{aligned}
$
$
\begin{aligned}
& \therefore \text { centripetal acceleration }=\vec{a}_c=\frac{v^2}{r}=\frac{(12)^2}{20}=\frac{144}{20} \\
& a_c=7.2 \mathrm{~ms}^{-2}
\end{aligned}
$

$\therefore$ Net acceleration
$
\begin{aligned}
& a=\sqrt{a_c^2+a_t^2}=\sqrt{7.2^2+12^2} \\
& a_c \simeq 14 \mathrm{~ms}^{-2}
\end{aligned}
$

Example 3: A particle is moving with speed varying as v = 2t, then the angle which the resultant acceleration makes with the radial direction (R=1m) at t = 2 is

$\begin{aligned} & \text { 1) } \tan ^{-1}\left(\frac{1}{2}\right) \\ & \text { 2) } \tan ^{-1}\left(\frac{1}{6}\right) \\ & \text { 3) } \tan ^{-1}\left(\frac{1}{8}\right) \\ & \text { 4) } \tan ^{-1}\left(\frac{1}{4}\right)\end{aligned}$

Solution:

The angle between Total acceleration and centripetal acceleration is given by
$
\tan \phi=\frac{a_t}{a_c}=\frac{r^2 \alpha}{V^2}
$
where
$\alpha=$ angular acceleration
$V=$ velocity
$r=$ radius of circle
So From the below figure

$\begin{aligned}
& \tan \theta=\frac{a_t}{a_r} \\
& a_t=\frac{\mathrm{d} v}{\mathrm{~d} t}=2 \mathrm{~m} / \mathrm{s}^2 \\
& a_r=\frac{v^2}{R} \\
& a_r=\frac{4 t^2}{1}=4 \times 2^2=16 \\
& \therefore \tan \theta=\frac{2}{16}=\frac{1}{8} \\
& \therefore \theta=\tan ^{-1}\left(\frac{1}{8}\right)
\end{aligned}$

Example 4: A particle is moving with a constant speed of 8 m/s in a circular path of radius 1 m. What will be the displacement of the particle in 1 sec?

1) 2 sin 80

2) 2 sin 40

3) 4 sin 80

4) 4 sin 40

Solution:

Displacement in Circular Motion -

$
\Delta r=2 r \sin \frac{\theta}{2}
$
$\Delta r=$ displacement
$\theta=$ Angle between two vectors
- wherein
$
\text { If }\left|\overrightarrow{r_1}\right|=\left|\overrightarrow{r_2}\right|=r
$

Let the angular displacement of the particle from $\mathrm{A}$ to $\mathrm{B}$ will be
$
2 R \sin \frac{\Theta}{2}
$

Length of circular arc $A B=8 * 1=8 \mathrm{~m}$
$
\begin{aligned}
& \text { Angle } \Theta=\frac{\text { arc length }}{\text { Radius }}=\frac{\Theta}{1}=8 \mathrm{rad} \\
& d=2 R \sin \frac{\Theta}{2}=2 * 1 * \sin \frac{8}{2} \\
& 2 \sin 4^0
\end{aligned}
$

Hence, the correct answer is option (2).

Uniform circular motion involves a particle moving with constant speed along a circular path, whose velocity continuously changes due to the continuous change in direction. A simple example is the merry-go-round in which objects move in circles at an even speed. This concept also applies to natural phenomena like planets orbiting the sun. This understanding of uniform circular motion allows us to comprehend how forces and motion work in circular paths, which is essential for our daily lives and for grasping bigger movements in cosmology.

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