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.

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

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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In
one complete rotation, angular distance traveled is 2$\pi$ and time is
time period (T) then,

Angular Speed = $\frac{2\pi }{T}$

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.

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

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,

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.

From this Relation, we can find Angular speed.

$\omega$ 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).

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 is given by,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.

$\omega$ = $\frac{V}{r}$

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.

The problems on angular speed are given below :

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

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

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

r = 0.60m and V = 10m/s

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

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.

ω = 88 rad./s

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.

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