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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). The normal reaction of the track on the particle vanishes at point Y where OY makes angle \Phi with the horizontal. Then :

  • Option 1)

    \sin \Phi =\cos \Theta

  • Option 2)

    \sin \Phi = \frac{1}{2}\cos \Theta

  • Option 3)

    \sin \Phi = \frac{2}{3}\cos \Theta

  • Option 4)

    \sin \Phi = \frac{3}{4}\cos \Theta

 

Answers (1)

As we have learned

If only conservative forces act on a system, total mechnical energy remains constant -

K+U=E\left ( constant \right )

\Delta K+\Delta U=0

\Delta K=-\Delta U

-

 

  let velocity at point Y  is \theta

 From energy conservatiuon 

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

 

 

 

 

 


Option 1)

\sin \Phi =\cos \Theta

Option 2)

\sin \Phi = \frac{1}{2}\cos \Theta

Option 3)

\sin \Phi = \frac{2}{3}\cos \Theta

Option 4)

\sin \Phi = \frac{3}{4}\cos \Theta

Posted by

Vakul

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