How to Calculate Displacement , Average velocity , Angular & Linear Speed in a Uniform Circular Motion ?

How to Calculate Displacement  ?

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 :

Circular Motion

Circular Motion

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} $

How to Calculate Average Velocity ?

$ \displaystyle |\vec{V_{av}}| = \frac{|\vec{\Delta r} |}{\Delta t} $

$ \displaystyle |\vec{V_{av}}| = \frac{2 r sin\frac{\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\frac{\theta}{2}} $

How to Calculate Relation between Angular & 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 $

How to Calculate 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.

Circular Motion

Circular Motion

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 $

How to Calculate 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)

Solved Example : 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)

Also Read :

→ Non-uniform Circular Motion
→ Vertical circular motion with variable speed

Next Page →