question_answer

A stone tied at one end of a long string is rotated in a vertical circle about the other end. The minimum speed of the stone at the lowest point so, as to complete the circular path will be

A) \[\sqrt{gr}\]                      

B)        \[\sqrt{2gr}\]

C) \[\sqrt{3gr}\]                    

D)        \[\sqrt{5gr}\]

Correct Answer: D

Solution :

Key Idea: Law of conservation of energy holds valid. The speed of body at the  lowest point B, can be calculated from the law of conservation of energy, when   body moves from B to A, it rises through a distance 2r  and  acquires  a potential energy   of \[mg\times 2r.\] Thus, kinetic energy at B = kinetic energy at A + potential energy at A \[\frac{1}{2}mv_{B}^{2}=\frac{1}{2}mv_{A}^{2}+2mgr\] where, \[{{v}_{A}}\sqrt{gr},\]the critical speed, then \[\frac{1}{2}mv_{B}^{2}=\frac{1}{2}m(gr)+2mgr=\frac{5}{2}mgr\] \[\Rightarrow \]               \[{{v}_{B}}=\sqrt{5\,gr}\] Note: If \[{{\text{V}}_{\text{B}}}\]is less than \[\sqrt{5\,gr},\] then \[{{V}_{A}}\]will be less than \[\sqrt{gr}\]and the string will become slack at the highest point. Alternative: At the lowest point B, tension and centripetal force balance the weight of stone. \[{{T}_{B}}-mg=\frac{m{{v}^{2}}}{r}\] But at the lowest point, \[{{T}_{B}}=6\,mg\]                 \[\therefore \] \[6\,mg-mg=\frac{m{{v}^{2}}}{r}\]                 \[\Rightarrow \]               \[5\,mg=\frac{m{{v}^{2}}}{r}\]                 \[\therefore \]  \[v=\sqrt{5gr}\]