We will derive the **Relationship between linear displacement** **and** **angular displacement**.

**Relationship formula (to be derived)**

The relationship is actually this:

Linear Displacement = Angular Displacement x Radius of the Circular path

**Derivation [step-by-step]**

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 in 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,Sineθ = PQ /OQ = s/r ________________ (1)

[ When P and Q are very close then PQ can be considered as astraight line = s.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**

Reference post: linear and circular motion – 3 selected quantities & relationships