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  • A particle is released on a vertical smooth semicircular track from point X so that OX makes angle Theta from the vertical (see figure). Th

A particle is released on a vertical smooth semicircular track from point X so that OX makes angle \Theta from the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y where OY makes angle \Phi with the horizontal. Then :

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figure?

  let velocity at point Y  is \theta

 From energy conservatiuon 

1/2 mv^2 = mg (R \cos \theta -R \sin \phi ) \: \: or \: \: mv^ 2 = 2mg (R\cos \theta -R \sin \phi ).......(1)At Y 

mg \sin \phi -N = \frac{mv^2}{R }

mg \sin \phi = \frac{mv^2 }{R } = 2mg ( \cos \theta- \sin \phi )\\ 3 \sin \phi = 2 \cos \theta \\ \sin \phi = 2/3 \cos \theta

 

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Safeer PP

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