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JIPMER 2015
Exam
JIPMER
Year
2015
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2015 · Physics · Work, Power and Energy

A particle slides on surface of a fixed smooth sphere starting from...

A particle slides on surface of a fixed smooth sphere starting from topmost point. The angle rotated by the radius through the particle, when it leaves contact with the sphere, isA particle slides on surface of a fixed smooth sphere starting from topmost point. The angle rotated by the radius

Options

  1. Aθ=cos⁡−1(13)\theta=\cos ^{-1}\left(\frac{1}{3}\right)
  2. Bθ=cos⁡−1(23)\theta=\cos ^{-1}\left(\frac{2}{3}\right)
  3. Cθ=tan⁡−1(13)\theta=\tan ^{-1}\left(\frac{1}{3}\right)
  4. Dθ=sin⁡−1(43)\theta=\sin ^{-1}\left(\frac{4}{3}\right)

Correct answer

B. θ=cos⁡−1(23)\theta=\cos ^{-1}\left(\frac{2}{3}\right)

Step-by-step solution

See the diagram A particle slides on surface of a fixed smooth sphere starting from topmost point. The angle rotated by the radius Let the velocity be vv when the body leaves the surface. From free body diagram, mv2R=mgcos⁡θ\frac{m v^2}{R}=m g \cos \theta. ⇒v2=Rgcos⁡θ\Rightarrow \quad v^2=R g \cos \theta ...(i) Again from work energy principle, change in KE=\mathrm{KE}= work done ⇒12mv2−0=mg[R−Rcos⁡θ]v2=2gR(1−cos⁡θ)\Rightarrow \frac{1}{2} m v^2-0=m g[R-R \cos \theta] v^2=2 g R(1-\cos \theta) ...(ii) From Eqs. (i) and (ii)Rgcos⁡θ=2gR(1−cos⁡θ)3gRcos⁡θ=2gR⇒cos⁡θ=23⇒θ=cos⁡−1(23)\begin{aligned} R g \cos \theta & =2 g R(1-\cos \theta) \\ 3 g R \cos \theta & =2 g R \\ \Rightarrow \quad \cos \theta & =\frac{2}{3} \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)\end{aligned}
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