question_answer

A parade is kept at rest at the top of a sphere of diameter 42 m. When disturbed slightly, it slides down. At what height h from the bottom, the particle will leave the sphere?                   [BHU M-2003]

A)  \[14\,\,m\]                                       

B)  \[28\,\,m\]

C)  \[35\,\,m\]                                       

D)  \[7\,\,m\]

Correct Answer: C

Solution :

                 Key Idea: The sphere will fly off when the normal reaction at the surface of sphere is zero. Let \[P\] be the instantaneous position of the particle during its sliding from the top\[A\]. \[N\]be the normal reaction at the surface of sphere, thus supplies the necessary centripetal force thus.                                 \[mg\,\,\cos \,\theta -N=\frac{m{{v}^{2}}}{R}\] When\[N=0\], particle will fly off                                                 \[mh\,\,\cos \,\,\theta \frac{m{{v}^{2}}}{R}\]                                 \[\frac{1}{2}mg\,R\,\cos \,\,\theta =\frac{1}{2}m{{v}^{2}}\]                         ?(1) Fall in height in sliding from \[A\] to \[P\] is          \[AC=AO-CO=R-R\,\,\cos \,\,\theta \] Loss in potential energy is                                 \[mg\,\,R\left( 1-\cos \,\,\theta  \right)=\frac{1}{2}m{{v}^{2}}\]                ?(2) From Eqs. (1) and (2), we get                                 \[\frac{1}{2}mg\,\,R\,\,\cos \,\,\theta =mg\,\,R\left( 1-\cos \,\,\theta  \right)\]                                 \[\Rightarrow \]                               \[\cos \,\,\theta =\frac{2}{3}\] Vertical height of point \[P\] from bottom                 \[BC=BO+OC=R+R\,\,\cos \,\,\theta \] =\[=R+R\left( \frac{2}{3} \right)=\frac{5}{3}R\] Given,                   \[R=21\,\,m\]                                                 \[BC=\frac{5}{3}\times 21=35\,\,m\]