A particle rotates in U.C.M. with tangential velocity ‘v’ along a

1.	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(A) vt (B) \((\frac{v}{D})-t\)(C) \(\frac{vt}{2D}\) (D) \(\frac{2vt}{D}\)
Answer» (D) \(\frac{2vt}{D}\) θ = \(\frac{s}t\) s = vt and r = \(\frac{D}2\) θ = \(\frac{2vt}{D}\)

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(A) vt (B) \((\frac{v}{D})-t\)(C) \(\frac{vt}{2D}\) (D) \(\frac{2vt}{D}\)

Answer»

(D) \(\frac{2vt}{D}\)

θ = \(\frac{s}t\)

s = vt and r = \(\frac{D}2\)

θ = \(\frac{2vt}{D}\)

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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(A) vt (B) \((\frac{v}{D})-t\)(C) \(\frac{vt}{2D}\) (D) \(\frac{2vt}{D}\)

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