Uniform circular motion :
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 (that is equal to the radius of the circle) remains constant and the direction of the position vector changes from time to time.
The change of position vector or the displacement of the particle from position 1 to the position 2 is given by referring the Figure :
Putting r1 = r2 = r we obtain
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
Angular and Linear Speed
If the particle moves with constant linear speed V the angular speed of the particle also remains constant
Since r θ = length of the Arc AB
Eliminating t from the previous equation we obtain
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
Putting V1 = V2 = V we obtain
gives the magnitude of and the direction of is shown in Figure that can be given as
If θ tends to zero, φ → 90°, that means, the direction of change of velocity vector becomes perpendicular to the instantaneous tangential velocity vector .
In the other words, we can say that, at any instant the change of velocity vector of the particle executing uniform circular motion is always directed radially.
That means the acceleration vector 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
when θ is very small during very small time interval
We obtain ,
(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
Putting V = π/300 m/sec (obtained earlier), Δt = 15 seconds and θ = 60°, we obtain
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)