question_answer

A body is allowed slide down a frictionless track freely under gravity. The track end in a semicircular shaped part of a diameter D. The height (minimum) from which the body must fall so that it completes the circle, is:

A) \[\frac{4}{5}D\]                         

B) \[\frac{D}{2}\]

C) \[\frac{3D}{4}\]  

D) \[\frac{5}{4}D\]

Correct Answer: D

Solution :

Key Idea: For completion of vertical circle velocity at the bottom of circle is equal to velocity of fall. Since body starts from rest u = 0 (initial velocity) \[{{I}_{\min }}={{(\sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}})}^{2}}=k{{({{a}_{1}}-{{a}_{2}})}^{2}}\]  \[\Rightarrow \] \[{{I}_{\min }}=k{{(3{{a}_{2}}-{{a}_{2}})}^{2}}=4a_{2}^{2}k\]   \[\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{16}{4}=\frac{4}{1}\]                                   ?..(1) Also for completion of circle velocity at bottom \[N=m\times n\]                                       ?.(2) \[v=1000=n\times 250\]  Equating Eqs. (1) and (2), we get \[\Rightarrow \] \[n=\frac{1000}{250}=4\]   \[\frac{1}{C}=\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{4}{8}\] Alternative: Energy will remain conserved in the process, hence potential energy of fall will be converted to kinetic energy to complete the circular path i.e.,         \[\Rightarrow \] \[C=2\mu F\]    \[\therefore \] At lowest point of circular path, \[C=16=2\times m\] Hence, for completion of circular path    \[\Rightarrow \] \[m=\frac{16}{2}=8\]   \[N=m\times n=8\times 4=32\]