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Angular Speed

When we see the merry go round or a giant wheel moving in the constant circular path, where motion is controlled at the center, we get curious to know the concept of motion in circular path.
Also, there are motions like motion of car on the curved path, motion of Moon around Earth and motion of Earth around Sun, we don’t see those motions controlled at the center.

How these motions are being controlled? There comes the term centripetal force. This centripetal force develops due to the object being continually accelerated, even though moving with constant speed. The centripetal force acts towards center of circular motion and keeps the object in circular path, also this type of motion follows the Newton’s first laws of motion.
According to the first law of motion the unbalanced force keeps the body moving in straight line path, hence for body in circular path the presence of unbalanced (centripetal) force is must.

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Angular Speed Formula

Angular Speed ($\omega$) is the scalar measure of rotation rate.

In one complete rotation, angular distance traveled is 2$\pi$ and time is time period (T) then,
Angular Speed = $\frac{2\pi }{T}$

Hence, angular speed $\omega$ = 2$\pi$ f
where $\frac{1}{T}$ = f (frequency)

Thus, the rotation rate is also expressed as angular frequency.
Average speed = $\frac{Distance}{Time}$ = $\frac{Circumference}{Time}$

Finding Angular Speed

Relation Between Angular Speed and Linear Speed

Let the body be moving in circular path of radius r and angular displacement be θ than we have, angle, θ = $\frac{arc}{radius}$

We know that linear speed,V =$\frac{S}{t}$,
where S is linear displacement of arc, and
θ = $\frac{S}{r}$

Therefore, linear speed V =$\frac{(\theta .r)}{t}$
= r . ($\frac{\theta}{t}$)

V = r $\omega$

Hence, Angular speed,

$\omega$ = $\frac{V}{r}$
where V = linear speed

This is the relation between angular speed, linear speed and radius of circular path.
From this Relation, we can find Angular speed.

Units of the above terms used:

$\omega$ is the angular frequency or angular speed and this measured in radians per second.

T is nothing but the time period and measured in seconds.

f is the linear frequency (measured in hertz).

v is considered to be the tangent velocity and is measured in meters/second.

r is the radius of circular path (measured in meters).

Angular Speed of Earth

The Earth takes 365.25 days to complete the revolution around the Sun, now convert days to seconds,

T = 365.25 x 24 x 60 x 60 = 31557600 seconds
Angular speed = $\frac{2\pi }{T}$

Hence,
$\omega$ = 2 x $\frac{3.14}{365.25 \times 24 \times 60 \times 60}$

Therefore,
$\omega$ = 1.99 x $10^{-7}$ radians /seconds.

Angular Speed Units

Angular Speed is given by,
$\omega$ = $\frac{V}{r}$
where,
V = linear speed
r = radius of circular path

We have linear speed, V = r . ($\frac{\theta}{t}$)

$\therefore$ $\omega$ = ($\frac{\theta}{t}$)
$\theta$ is expressed in radians.

So the basic unit for Angular Speed is Radians per second or rad/s or rps.

Angular Speed Problems

The problems on angular speed are given below :

Solved Examples

Question 1: 10 kg mass is moving in circular path of radius 5m, completes one round in 4 seconds. Calculate the linear speed and angular speed of mass.
Solution:

m = 10kg, r = 5m and t = 4 seconds, V = ?

Linear speed, V = $\frac{2\pi r}{t}$  = 2 x 3.14 x $\frac{5}{4}$

V = 7.85 m/s

Now, ω = $\frac{V}{r}$ = $\frac{7.85}{5}$ = 1.57 rad. /s

Question 2: Calculate the angular speed of an hour hand.
Solution:

Angular distance = 2 π & T = 12 hours = 12 x 60 x 60 s = 43200 s (this is the time for one complete circular motion)

Now, ω =  $\frac{2\pi }{t}$  = 2 x $\frac{3.14}{43200}$  = 1.45 x $10^{-4}$ rad. /s

Question 3: What is the angular speed of earth of rotation of earth on its axis?
Solution:

Angular distance = 2 π = 2 x 3.14 = 6.28 rad.

Time period (T) = 24 hours = 24 x 60 x 60 = 86400 s

Now, ω = $\frac{2\pi }{t}$ = $\frac{6.8}{86400}$ = 7.27 x $10^{-5}$ rad. /s

Question 4: A wheel of 0.60m in radius is moving with a speed of 10m/s. Find the angular speed.
Solution:

r = 0.60m and V = 10m/s

Now, ω = $\frac{V}{r}$ = $\frac{10}{0.60}$ = 16.67 rad./s

Question 5: Moon is continuously revolving round the earth without falling towards it. Justify, your reason.
Solution:

The weight of moon provides the necessary centripetal force that keeps it revolving in circular path. Since the weight of moon gets used in providing it the necessary centripetal force, so it revolves round the earth with out falling at the centre.

Question 6: The angular speed of wheel is 88 rad./s. Calculate the number of revolutions made by it in one second.
Solution:

1 rad./s = $\left ( \frac{1}{2\pi } \right )$ rev. /s
88 rad./s = $\frac{88}{(2 \times \frac{22}{7})}$ rev./s = 14 rev/s.