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Derive the relation between Linear displacement and Angular displacement

Last updated on February 8th, 2024 at 10:26 am

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.

Circular motion showing angular and linear displacement
angular and linear displacements

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)  
[ When P and Q are very close then PQ can be considered as a straight 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.

See also  How uniform acceleration is different from nonuniform acceleration?

Linear Displacement = Angular Displacement x Radius of the Circular path

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

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