# Kinematics of Circular Motion

##### Uniform circular motion :

Displacement $\displaystyle (\vec{\Delta r} )$
When particle moves in a circular path describing an angle θ during time t as shown in the Figure from the position 1 to the position 2, we see that the magnitude of the position vector $\displaystyle (\vec{r} )$ (that is equal to the radius of the circle) remains constant $\displaystyle \Rightarrow |\vec{r_1}| = |\vec{r_2}| = r$ and the direction of the position vector changes from time to time.
The change of position vector or the displacement $\displaystyle \vec{S}$ of the particle from position 1 to the position 2 is given by referring the Figure :  As $\displaystyle \vec{\Delta r} = \vec{r_2} - \vec{r_1}$ $\displaystyle \Delta r = |\vec{\Delta r}| = |\vec{r_2} - \vec{r_1}|$ $\displaystyle \Delta r = \sqrt{r_1^2 + r_2^2 - 2r_1 r_2 cos\theta}$

Putting r1 = r2 = r we obtain $\displaystyle \Delta r = \sqrt{r^2 + r^2 - 2r . r cos\theta}$ $\displaystyle = \sqrt{2 r^2 (1 - cos\theta )}$ $\displaystyle \Delta r = \sqrt{2 r^2 ( 2 sin^2\frac{\theta}{2} )}$ $\displaystyle \Delta r = S = 2 r sin\frac{\theta}{2}$

##### Average Velocity $\displaystyle |\vec{V_{av}}| = \frac{|\vec{\Delta v} |}{\Delta t} = \frac{2 V sin\theta /2}{t}$

The distanced covered by the particle during the time t is given as
d = length of the arc AB
⇒ d = r θ
Therefore, the ratio of distance to the magnitude of displacement for any angular distance is given as $\displaystyle \frac{d}{S} = \frac{r \theta}{2 r sin\frac{\theta}{2}}$ $\displaystyle \frac{d}{S} = \frac{\theta}{2 sin\theta/2}$

##### Angular and Linear Speed

If the particle moves with constant linear speed V the angular speed of the particle also remains constant $\displaystyle Angular \; Speed = \frac{Angular \; distance \; covered}{Time \; elapsed}$ $\displaystyle \omega = \frac{\theta}{t}$

Since r θ = length of the Arc AB

Linear Speed $\displaystyle V = \frac{length \; of \; arc \; AB}{time \; t}$ $\displaystyle V = \frac{r \theta}{t}$

Eliminating t from the previous equation we obtain $\displaystyle \omega = \frac{V}{r}$ $\displaystyle V = r \omega$

##### Change in Velocity

We want to know the magnitude and direction of the change in the velocity of the particle as it moves from A (position 1) to B (position 2) during time t as shown in Figure.  The change in velocity vector is given as

As $\displaystyle \vec{\Delta V} = \vec{V_2} - \vec{V_1}$

As $\displaystyle |\vec{\Delta V}| = |\vec{V_2} - \vec{V_1} |$ $\displaystyle \Delta V = \sqrt{V_1^2 + V_2^2 - 2 V_1 V_2 cos\theta}$

Putting V1 = V2 = V we obtain $\displaystyle \Delta V = \sqrt{V^2 + V^2 - 2 V . V cos\theta}$ $\displaystyle = \sqrt{2 V^2 (1 - cos\theta )}$ $\displaystyle \Delta V = \sqrt{2 V^2 ( 2 sin^2\frac{\theta}{2} )}$ $\displaystyle \Delta V = 2 V sin\frac{\theta}{2}$

gives the magnitude of $\displaystyle \vec{\Delta V}$ and the direction of $\displaystyle \vec{\Delta V}$ is shown in Figure that can be given as $\displaystyle \phi = \frac{180-\theta}{2} = 90 -\theta/2$

### Centripetal Acceleration

If θ tends to zero, φ → 90°, that means, the direction of change of velocity vector becomes perpendicular to the instantaneous tangential velocity vector $\displaystyle \vec{V}$ .

In the other words, we can say that, at any instant the change of velocity vector $\displaystyle \vec{\Delta V}$ of the particle executing uniform circular motion is always directed radially.

That means the acceleration vector $\displaystyle \vec{a} = \frac{\vec{\Delta V}}{\Delta t}$ of the particle is radially inwards. Therefore the particle is accelerating towards the centre.

Hence this type of acceleration of the particle in a circular path is known as Centripetal Acceleration. The magnitude of centripetal acceleration ar is given as $\displaystyle a_r = \frac{|\vec{\Delta V}|}{\Delta t} = \frac{2 V sin (\theta/2)}{t}$

when θ is very small during very small time interval $\displaystyle sin\theta/2 = \theta /2$ $\displaystyle a_r = \frac{2 V (\theta/2)}{t} = \frac{V \theta}{t}$

Putting $\displaystyle \frac{\theta}{t} = \omega = \frac{V}{r}$

We obtain , $\displaystyle a_r = \frac{V^2}{r}$ $\displaystyle a_r = \omega ^2 r$ (Since V = r ω)

Illustration : Find the magnitude of average acceleration of the tip of the second hand during 10 seconds.

Solution: Average acceleration has the magnitude $\displaystyle a = \frac{\Delta V}{\Delta t} = \frac{2 V sin (\theta/2)}{\Delta t}$

Putting V = π/300  m/sec (obtained earlier), Δt = 15 seconds and θ = 60°, we obtain $\displaystyle a = \frac{2 (\pi/300) sin 30}{15}$ $\displaystyle a = \frac{\pi}{4500} m/s^2$

Exercise : An astronaut is rotating in a rotor of radius 4 m . If he can withstand upto acceleration of 10 g, then what is the maximum number of permissible revolutions ? (g = 10 m/s2)

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