For an object to move along a curved circular path, the direction of its velocity must change. It is because at each point on the circular path tangent will give the direction. A change in velocity will cause the acceleration which will not be in the same direction as the velocity. Therefore for an object to move along a circular path, there must be an acceleration that will always be perpendicular to the velocity. The circular motion may be uniform as well as non –uniform. In this topic, we will discuss the uniform circular motion formula with examples. Let us learn the concept!

**Uniform Circular Motion **

**Concept of Uniform Circular Motion:**

Uniform circular motion means that the magnitude of the velocity will always be constant. But the direction of the velocity will change at a constant rate from every point. It means that the path of the object will form a circle. And the object will complete the repeated trips around the path in the same amount of time every time.

The term circular is applicable to describe the motion in a curved path. The motion of any object along some circular path, covering equal distance along the circumference in the same time interval is known as the uniform circular motion. In any such motion, the speed remains constant, with constantly changing direction.

The tangential speed at every point on the circumference will be constant in a uniform circular motion. This tangential velocity vector is tangent at every point over the circumference.

Also, the acceleration vector is therefore always directed toward the center of the circle formed by the motion of the object. This acceleration is either known as the “radial acceleration”, because it is at a certain radius from a central point, or also as “centripetal acceleration”, which means that it is “center seeking”.

**The formula for Uniform Circular Motion:**

If the radius of the circular path is R, and the magnitude of the velocity of the object is V. Then the radial acceleration of the object will be:

**\(a_{rad} = \frac {v^2} {R}\)**

Again, this radial acceleration will always be perpendicular to the direction of the velocity. Its SI unit is \(m^2s^{-2}\).

The radial acceleration can also be expressed with the help of the period of the motion i.e. T. This period T is the amount of time taken to complete a revolution. Its unit is seconds.

If the magnitude of the velocity of an object traveling in uniform circular motion is v, then the velocity will be equal to the circumference C of the circle divided by the period. Thus,

**\(V = \frac{C}{T}\)**

The circumference of the circle is equal to pi Π multiplied by the radius R.

So, **C = 2Π R**

At any point in the motion, therefore the velocity will be,

**\(V = \frac{2\pi R} {T}\)**

Using this value in the equation for radial acceleration, we will get,

**\(a_{rad} = \frac{4{\pi}^2 R}{T^2}\)**

Where,

\(a_{rad}\) | Radial acceleration |

R | The radius of the circular path |

T | Time Period |

V | Velocity |

C | Circumference |

** ****Solved Examples**

Q.1: A player is moving with a constant tangential speed of 50 m per second. He takes one lap around a circular track in 40 seconds. Calculate the magnitude of the acceleration of the player.

Solution:

Given parameters:

The magnitude of Velocity, V = 50 m per second

Time period, T = 40 seconds.

We know that,

\(V = \frac{2\pi R} {T}\)

\(Therefore, R = \frac{T\times V}{2 \pi}\)

Putting values,

\(R = \frac{40\times 50}{2 \times 3.14}\)

R = 31.84

Now,

\(a_{rad} = \frac{4{\pi}^2 R}{T^2} put R = \frac{T\times V}{2 \pi} \)we get,

= \(\frac{4{\pi}^2 R}{T^2} \)

= \(\frac{4{\pi}^2\times \frac{T\times V}{2 \pi} }{T^2} \)

= \(\frac{2\times \pi\times V}{T}\)

= \(\frac{2\times 3.14\times 50}{40}\)

= 7.86

Acceleration will be \(7.86 ms^{-2}\)

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