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A particle rotates in U.C.M with ten...

A particle rotates in U.C.M with tengential velocity 'v' along a ghorizontal circle of diameter 'D' . Total angular displacement of the particle in time 't' is

A

`vt`

B

`((v)/(D))-t`

C

`(vt)/(2D)`

D

`(2vt)/(D )`

Text Solution

Verified by Experts

The correct Answer is:
D
In U.C.M . `v= r omega `
` therefore omega =(v )/(r ) = (v )/((D )/(2))=(2v )/(D)`
`:' omega =(theta )/(t ) theta = omega t = ((2v)/(D ))t`
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