-->

Laws Of Motion

Question
CBSEENPH11016969

Derive the relation between linear speed and angular speed.

Solution
Let a particle be moving in a circular orbit of radius r.
Let at any instant 't' the particle be at A and in time ∆t the particle undergoes angular displacement ∆θ and reach point B. 

AB with overparenthesis on top space equals space straight r space triangle straight theta space space space space space space space space space space space space space space space space space space... space left parenthesis 1 right parenthesis

Dividing space both space sides space by space increment straight t comma space we space have

fraction numerator AB with overparenthesis on top over denominator increment straight t end fraction space equals space straight r space fraction numerator space triangle straight theta over denominator increment straight t end fraction space space space space space space space space space... space left parenthesis 2 right parenthesis

Taking space the space limit space increment straight t space rightwards arrow 0 space on space both space sides comma space

limit as increment straight t space rightwards arrow space 0 of space fraction numerator AB with overparenthesis on top over denominator increment straight t end fraction space equals stack space lim with increment straight t space rightwards arrow space 0 below space straight r space fraction numerator space triangle straight theta over denominator increment straight t end fraction

space space space space space space space space space space space space space space space space space space space space space space equals space straight r stack space lim with increment straight t space rightwards arrow space 0 below space fraction numerator space triangle straight theta over denominator increment straight t end fraction space space space space space space space space space... space left parenthesis 3 right parenthesis

If space increment straight t space rightwards arrow space 0 comma space arc space AB space becomes space equal space to space segment space
AB space which space is space equal space to space the space linear space displacement.

Therefore comma space

stack lim space with increment straight t space rightwards arrow 0 below space fraction numerator AB with overparenthesis on top over denominator increment straight t end fraction space equals space stack lim space with increment straight t space rightwards arrow 0 below space fraction numerator AB over denominator increment straight t end fraction space equals space straight v space

and

straight r space stack lim space with increment straight t space rightwards arrow 0 below space fraction numerator space triangle straight theta over denominator increment straight t end fraction space equals space straight r space dθ over dt space equals space rω space

Therefore comma space equation space left parenthesis 3 right parenthesis space becomes

straight v space equals space rω