In this post we will focus on the ** definitions of circular motion and linear motion** first which will show their basic differences as well.

In addition to this we’ll derive couple of

*among the quantities of these 2 types of motion.*

**interesting relationships**

*We will derive the following:*

a)

**angular displacement and linear displacement relation**

b)

**angular velocity and linear velocity relation**

c)

**angular acceleration and linear acceleration relation**

## Circular Motion

Circular motion of an object is its movement along the circumference of a circle.

You can define it as the rotation of an object along a circular path.

It can be uniform, with constant angular rate of rotation, or non-uniform with a changing rate of rotation.

## Linear Motion

Linear motion or Rectilinear motion is one dimensional straight-line motion.

This type of motion describes the one dimensional movement of a particle or a body.

A body experiences rectilinear motion if any two particles of the body travel the same distance along two parallel straight lines.

## Quantities of Linear Motion

Lets list down the variables of Linear motion first.

The most basic variables are Distance traveled, Displacement, time etc.The next level variables of linear motion are Speed, Velocity, Acceleration, Momentum etc.We discussed already in a previous **post** about the difference between distance traveled and displacement and the difference between speed and velocity.

## Quantities of Circular Motion

The variables mentioned in the Linear motion section have their individual counterparts in case of Circular Motion.

We will discuss some of these in this post and draw a relation with their Linear Motion counterparts.

## Angular Displacement and Linear Displacement Relation – circular motion vs linear motion

Say a point object is moving in a circular path of radius r. Let’s say the center of this circular path is O.

At some point of time say the particle is at point P on the circumference of the circle. At this moment the radial vector of this motion is OP.

After a time gap of t the position of the object becomes Q (obviously on the circumference of the same circle) and the radial vector is OQ.

Now say the angle between OP and OQ is **θ** (**theta**). Hence the angular displacement of the particle in time duration t is **θ** .

In addition to this angular displacement the particle is having a linear displacement as well when traveling from point P to Q.

Here PQ is the linear displacement, and say it is designated as **s. **

Now consider the right angled triangle ΔOPQ.

Here using the Trigonometry we get, **Sine** **θ = PQ /OQ = s/r ________________ (1) [ Note: OQ is the radius r of the circle ]**

Now, as per Trigonometry, if the angle θ is very small then Sine θ is equal to θ.

So from equation (1), we get

**θ = s/r i.e., s= θ r ______________(2).**

Here we get the relation between linear displacement and angular displacement.

**Linear Displacement =**** Angular Displacement x Radius of the Circular path**

**Angular Velocity and Linear Velocity **relation – linear motion vs circular motion

During the circular motion an object experiences angular displacement as well as linear displacement, therefore it will have angular velocity as well as linear velocity at any position on the circle.

The direction of the linear velocity of the rotating object at any point is along the tangent to the circle drawn at the point where the object is positioned at that moment on the circumference of the circle.

And angular velocity is being made by changing the angular position with time. Angular velocity is identified with the sign **omega** (**ω**). *And **ω** = **θ /t** *_________(3).

At the same time, we can calculate the linear velocity as the following: linear velocity(v) = linear displacement per unit time = s/t,

**v = s /t = (θ r)/t= (θ /t). r=ω.r ****or, v=ω.r ___________________(4)Linear Velocity **

**= Angular Velocity x radius**

## A**ngular Acceleration and Linear Acceleration relation**

Say at point P the linear velocity of the object in rotation is V1. In addition to this at point P angular velocity of the rotating object is **ω**1.

Similarly at point Q these are V2 and **ω**2 respectively.

So from equation 4 you can say that V1= **ω**1.r and V2 = **ω**2.r

We already said that the time taken to travel from P to Q is t.

Therefore we get the equation of linear acceleration (f)= (V2-V1)/t = (**ω**2. r – **ω**1.r)/t= (**ω**2-**ω**1)r/t = α r

where Alpha (α) = (**ω**2-**ω**1)/t = Rate of change of Angular Velocity = Angular Acceleration.

So we see** f = α r**

Linear Acceleration

= Angular Acceleration **x** Radius

## Numerical problems – circular motion

**Assuming you know these:**** RPM means Revolutions per minute,**** RPS means revolutions per second and 180 degree = π radian**and

**π=22/7**

Q1 ) An electric fan was rotating with a speed of 210 rpm. Its velocity is increased to 630 rpm in 11 seconds with the help of a regulator.

Calculate the angular acceleration of the fan.

Solution:

The initial angular velocity = **ω**1 = 210 rpm = 210/60 rps = (2 π .210)/60 radian/second = 7 π rad/s

And the final angular velocity = **ω2** = 630 rpm = 630/60 rps = (2 π .630)/60 radian/second = 21 π rad/s

Time elapsed =t= 11 seconds.

**So Angular acceleration of the fan (α) **

= change in angular velocity/time elapsed

i.e. (**ω2** –**ω1)/t =(21π – 7π ) /11 rad/s^2 = 14π /11 rad/s^2 ****= (14/11) X(22/7) rad/s^2 = 4 rad/s^2 (answer)**

** related posts**:

Centripetal force

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