The motion of a particle which is constrained to move in a circular path is quite interesting. It has direct bearing on the motion of such things as artificial and natural satellites, nuclear particles in accelerators, bodies whirling at the ends of the strings and flywheels spinning on the shafts.
Do you know? |
 Direction of motion changes continuously in circular motion. |
We all know that a ball whirled in a horizontal circle at the end of a string would not continue in a circular path if the string is snapped. Careful observation shows at once that if the string snaps, when the ball is at the point A, in Fig. 5.5 (b), the ball will follow the straight line path AB.
The fact is that unless a string or some other mechanism pulls the ball towards the centre of the circle with a force, as shown in Fig. 5.5 (a), ball will not continue along the circular path.
The force needed to bend the normally straight path of the particle into a circular path is called the centripetal force.
If the particle moves from A to B with uniform speed v as shown in Fig. 5.6 (a), the velocity of the particle changes its direction but not its magnitude. The change in velocity is shown in Fig. 5.6 (b). Hence, the acceleration of the particle is
Where ∆t is the time taken by the particle to travel from A to B. suppose the velocities at A and B are v1 and v2 respectively. Since the speed of the particle is v, so the time taken to travel a distance s, as shown in Fig. 5.6 (a) is
Let us now draw a triangle PQR such that PQ is parallel and equal to v1 and PR is parallel and equal to v2, as shown in Fig. 5.6 (b). we know that the radius of a circle is perpendicular to its tangent, so OA is perpendicular to v1 and OB is perpendicular to v2 (Fig. 5.6 a). Therefore, angle AOB equals the angle QPR between v1 and v2. Further, as v1 = v2 =v and OA = OB, both triangles are isosceles. From geometry, we know “two isosceles triangles are similar, if the angles between their equal arms are equal”. Hence, the triangle OAB of Fig. 5.6 (a) is similar to the triangle PQR of Fig. 5.6 (b). Hence, we can write
If the point B is close to the point A on the circle, as will be the case when ∆t ⇾0, the arc AB is of nearly the same length as the line AB. To that approximation, we can write AB = s, and after substituting and rearranging terms, we have, 
Putting this value for ∆v in the Eq. 5.12, we get
Where a is the instantaneous acceleration. As this acceleration is caused by the centripetal force, it is called the centripetal acceleration denoted by ac. This acceleration is directed along the radius towards the centre of the circle. In Fig. 5.6 (a) and (b), since PQ is perpendicular to OA and PR is perpendicular to OB, so QR is perpendicular to AB. It may be noted that QR is parallel to the perpendicular bisector of AB. As the acceleration of the object moving in the circle is parallel to ∆v when AB ⇾ 0, so centripetal acceleration is directed along radius towards the centre of the circle. It can, therefore, be concluded that:
Tid – bits |
 Banked tracks are needed for turns that are taken so quickly that friction alone cannot provide energy for centripetal force. |
The instantaneous acceleration of an object traveling with uniform speed in a circle is directed towards the centre of the circle and is called centripetal acceleration.
The centripetal force has the same direction as the centripetal acceleration and its value is given by
In angular measure, this equation becomes
Do You Know? |
 Curved flight at high speed requires a large centripetal force that makes the stunt dangerous even if the air planes are not so close. |
Example 5.2: A 1000 kg car is turning round a corner at 10 ms-1 as it travels along an arc of a circle. If the radius of the circular path is 10 m, how large a force a force must be exerted by the pavement on the tyres to hold the car in the circular path?
Solution: The force required is the centripetal force.
So
This force must be supplied by the frictional force of the pavement on the wheels.
Example 5.3: A ball tied to the end of a string, is swung in a vertical circle of radius r under the action of gravity as shown in Fig. 5.7. What will be the tension in the string when the ball is at the point A of the path and its speed is v at this point?
Solution: For the ball to travel in a circle, the force acting on the ball must provide the required centripetal force. In this case, at point A, two forces act on the ball, the pull of the string and the weight w of the ball. These forces act along the radius at A, and so their vector sum must furnish the required centripetal force. We, therefore have
