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# CBSE Class 10 Answered

derive relation between angular acceleration and linear acceleration
Asked by singhparansh | 18 Dec, 2021, 05:15: PM
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 is mutually perpendicular to angular velocity vector
and radial position vector .
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
and vector , then we get velocity vector in terms of vector product as
= ×
Acceleration  = ( d / dt ) in tangential direction is determined by differentiating abvove expression .
since  ( d / dt )= 0 , we get
where = ( d  / 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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