Table of contents 
Circular Motion 
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Circular motion is described as a movement of an object while rotating along a circular path. Circular motion can be either uniform or nonuniform. During uniform circular motion the angular rate of rotation and speed will be constant while during nonuniform motion the rate of rotation keeps changing.
Some of the most common examples of circular motion include manmade satellite that revolves around the earth, a rotating ceiling fan, a moving car’s wheel, the blades in a windmill, and gears in gas turbines.
A particle is said to execute circular motion when it moves along the circumference of a circular path. An important aspect of circular motion is that the direction of motion is changing continuously unlike in the case of linear motion. Hence circular motion can also be described in terms of angular variables.
This motion refers to the circular motion if the magnitude of the velocity of the particle in circular motion remains constant. The nonuniform circular motion refers to the circular motion when the magnitude of the velocity of the object is not constant. Another special kind of circular motion is when an object rotates around itself also known as spinning motion.
If a particle is moving in a circle, it must have some acceleration acting towards the centre which is making it move around the centre. Since this acceleration is perpendicular to the velocity of a particle at every instant, it is only changing the direction of velocity and not magnitude and that’s why the motion is uniform circular motion. We call this acceleration centripetal acceleration (or radial acceleration), and the force acting towards the centre is called centripetal force.
In the case of uniform circular motion, the acceleration is:
ar = v^{2}r = ω^{2}r
If the mass of the particle is m, we can say from the second law of motion that:
F = ma
mv^{2}r= mω^{2}r
This is not a special force, actually force like tension or friction may be a cause of origination of centripetal force. When the vehicles turn on the roads, it is the frictional force between tyres and ground that provides the required centripetal force for turning.
Note: So if a particle is moving in a uniform circular motion:
 Its speed is constant
 Velocity is changing at every instant
 There is no tangential acceleration
 Radial (centripetal) acceleration = ω^{2}r
 v = ωr
In case of nonuniform circular motion, there is some tangential acceleration due to which the speed of the particle increases or decreases. The resultant acceleration is the vector sum of radial acceleration and tangential acceleration.
Following are the examples of uniform circular motion:
Q.1. During the course of a turn, an automobile doubles its speed. How much must additional frictional force the tires provide if the car safely makes around the curve? Since Fc varies with v2, an increase in velocity by a factor of two must be accompanied by an increase in centripetal force by a factor of four.
A satellite is said to be in geosynchronous orbit if it rotates around the earth once every day. For the earth, all satellites in geosynchronous orbit must rotate at a distance of 4.23×107 meters from the earth’s center. What is the magnitude of the acceleration felt by a geosynchronous satellite?
Solution: The acceleration felt by any object in uniform circular motion is given by
a = v^{2 }÷ r
We are given the radius but must find the velocity of the satellite. We know that in one day, or 86400 seconds, the satellite travels around the earth once. Thus:
Q.2. A cyclist moves around a circular path of a radius of 50 m with a speed of 10 m/s.
(a) Find the acceleration of the cyclist?
(b) Considering the combined mass of the cyclist and cycle to be 120 kg, what is the net force applied on them?
Solution: (a) A movement with constant speed around a circular path yields a radially inward acceleration called centripetal acceleration whose magnitude is found as below
(b) The net force that acts on the object to keep it moving along a circular track is called centripetal force whose magnitude is found using Newton's second law as below
F_{net} = ma_{c} = 120 × 2 = 240N
Q.3. A race car travelling around a circular path of a radius of 400 m with a speed of 50 m/s. Find the centripetal acceleration of the car.
Solution: In a circular motion with constant speed the centripetal acceleration is a_{c} = v^{2}/r. Therefore, substituting the numerical values into this equation we have
This acceleration is toward the center of the circle.
Q.4. A racer is moving with a constant tangential speed of 50 m/s, takes one lap around a circular track in 40 seconds. Calculate the magnitude of the acceleration of the car.
Solution: Speed v = 50 m/s
T = 40 seconds.
We know that,
Acceleration,
a = v^{2}/r
T = 2 πr/v
Therefore, r = Tv/2π
When both the formulas are combined, we get
a = v^{2}/(Tv/2π) = v/(T/2π)
= 50/[ 40/6.28]
a = 7.86 m/s^{2}
Q.5. An object moving in a circular motion has a centripetal acceleration of 20 m/s^{2}. If the radius of the motion is 0.5 m, calculate the frequency of the motion.
Solution: Given:
Acceleration = 20 m/s^{2}
Radius = 0.5 m
We know that
a = v^{2}/r
20 = v^{2}/0.5
v = 3.16 m/s
v = 2 πr/t
3.16 = (2)(3.14)(0.5)/t
3.16 = 3.14 / t
T = 1.006
Frequency, f = 1/t = 1/1.006
Therefore, f = 0.99 Hz.
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