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AP EAMCET2005PhysicsMotion in Two Dimensions

At a given instant of time the position vector of a particle moving in a circle with a velocity 3 i -4 j +5 k is i +9 j -3 k . Its angular velocity at that time is

Options

  1. A(13 i +29 j -31 k ) 146
  2. B(13 i -29 j -31 k ) 146
  3. C(13 i +29 j -31 k ) 146
  4. D(13 i +29 j +31 k ) 146

Correct answer

B. (13 i -29 j -31 k ) 146

Step-by-step solution

Angular momentum, aligned & L =m r v & but L =I & m r^2 w=m r v & = r v r^2 = r v | r |^2 & r = i +9 j -8 k , v =3 i -4 j +5 k & r v = | array ccc i & j & k 1 & 9 & -8 3 & -4 & 5 array | & =13 i -29 j -31 k & = 13 i -29 j -31 k [ 1^2+9^2+(-8)^2 ]^2 & = 13 i -29 j -31 k 146 & aligned

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