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A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is

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Question

A particle rotates in U.C.M. with tangential velocity V along a horizontal circle of diameter ‘D' . Total angular displacement of the particle in time 't' is..........

Options

  • vt

  • (v/D)-t

  • vt/2D

  • 2vt/D

MCQ
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Solution

`(2("vt"))/D`

Velocity= v ; Diameter= D →Radius= `D/2`

`omega=theta/t andv=r*omega`

`therefore v=D/2*theta/t`

`therefore theta=(2vt)/D`

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2015-2016 (March)

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