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derive relation between angular acceleration and linear acceleration  
Asked by singhparansh | 18 Dec, 2021, 05:15: PM
answered-by-expert Expert Answer
Let us consider a circular lemma as shown in fifure that rotates about an axis passing through centre of mass.
Let z-axis coincide with the axis of rotation so that lemma is in x-y plane.
 
At a point P , we see that tangential velocity vector begin mathsize 14px style v with rightwards arrow on top end style is mutually perpendicular to angular velocity vector begin mathsize 14px style omega with rightwards arrow on top end style
and radial position vector begin mathsize 14px style r with rightwards arrow on top end style .
If the lemma makes one complete rotation in T seconds , then linear displacement made by point P in tangential direction is ( 2π r )
 
Hence velocity v in tangential direction , v = ( 2π r ) / T
 
But angular velocity ω = 2π / T 
 
Hence from above two expressions, we get , v = ω r
 
Since magnitude of velocity v in tangential direction is produt of magnitude of angular velocity and radial distance
and also direction of velocity v is mutually perpendicular to vector begin mathsize 14px style r with rightwards arrow on top end style
and vector begin mathsize 14px style omega with rightwards arrow on top end style , then we get velocity vector begin mathsize 14px style v with rightwards arrow on top end style in terms of vector product as
begin mathsize 14px style v with rightwards arrow on top end style = begin mathsize 14px style omega with rightwards arrow on top end style × begin mathsize 14px style r with rightwards arrow on top end style
Acceleration begin mathsize 14px style a with rightwards arrow on top end style = ( d begin mathsize 14px style v with rightwards arrow on top end style / dt ) in tangential direction is determined by differentiating abvove expression .
begin mathsize 14px style fraction numerator d v with rightwards arrow on top over denominator d t end fraction space equals space fraction numerator d omega with rightwards arrow on top over denominator d t end fraction cross times r with bar on top space plus space space fraction numerator d r with rightwards arrow on top over denominator d t end fraction cross times space omega with rightwards arrow on top end style
since  ( dbegin mathsize 14px style r with rightwards arrow on top end style / dt )= 0 , we get begin mathsize 14px style fraction numerator d v with rightwards arrow on top over denominator d t end fraction space equals space fraction numerator d omega with rightwards arrow on top over denominator d t end fraction cross times r with bar on top space end style
where begin mathsize 14px style alpha with rightwards arrow on top end style = ( dbegin mathsize 14px style omega with rightwards arrow on top end style  / dt ) is angular acceleration.
Hence by magnitude ,  a = α r  is the relation between angular acceleration and acceleration in tangential direction .
Answered by Thiyagarajan K | 19 Dec, 2021, 09:47: AM

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