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1.A particle is moving in circular path, the angular position of the particle at point 'P' withe 2.The position of the particle at point 'Q' with respect to OX axis is3.The change in angular position of the particle moving from P to Q is = |
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Answer» The velocity in the TANGENTIAL direction about the center is v. We resolve this velocity along and perpendicular to chord OB. We get these components as vcosθ and vsinθ.The momentum of mvsinθ about the origin WOULD be zero. We CALCULATE the momentum of mvcosθ about the origin as mvsinθ×2acosθ. Here 2acosθ is the length of the chord OB. Thus we get the momentum as 2mvacos 2 θ or mva(1+cos2θ)(Note that cos2θ=2cos 2 θ−1) |
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