Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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 :

Answers (1)

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

 

Posted by

Safeer PP

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

NTA JEE Main admit card 2026 out for January 28, 29
anam-khan

JEE Main 2026 January 24 shift 2 rated moderate in difficulty; chemistry, maths 'time-consuming'
anam-khan

JEE Main 2026 Session 1 logs 96.26% attendance till Jan 24, over 10 lakh appear: NTA

Latest Question