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 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}} $
Relation between Angular Speed & 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)
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)
Q: A particle moving uniformly in a circle is in rotational equilibrium . is it true or false ?
Sol: It is true . A particle is said to be in rotational equilibrium when its angular acceleration is zero . Since the particle moves uniformly , its angular velocity is constant in magnitude and direction and so its angular acceleration is zero . So it is true that a particle in uniform circular motion is in rotational equilibrium .
Q: Which of the following statements is false for a particle moving in a circle with a constant angular speed ?
(a) The velocity vector is tangent to the circle
(b) The acceleration vector is tangent to the circle
(c) The acceleration vector points to the centre of the circle
(d) The velocity and acceleration vectors are perpendicular to each other
Ans: (b)
Also Read :
| → Non-uniform Circular Motion → Vertical circular motion with variable speed |