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, \[{{T}_{B}}=6\,mg\] \[\therefore \] \[6\,mg-mg=\frac{m{{v}^{2}}}{r}\] \[\Rightarrow \] \[5\,mg=\frac{m{{v}^{2}}}{r}\] \[\therefore \] \[v=\sqrt{5gr}\]