question_answer

A section of fixed smooth circular track of radius R in vertical plane is shown in the figure. A block is released from position A and leaves the track at B. The radius of curvature of its trajectory when it just leaves the track at B is

A) \[R\]                 

B)        \[R/4\]

C) \[R/2\]  

D)        None of these

Correct Answer: C

Solution :

[c] \[OC=R\,\cos 53{}^\circ =\frac{3R}{5};\] \[OE=R\cos 37{}^\circ =\frac{4R}{5}\] So,     \[CE=\frac{4R}{5}-\frac{3R}{5}=\frac{R}{5}\] From A to B using energy conservation \[\frac{1}{2}m{{v}^{2}}=mg\frac{R}{5}\Rightarrow v=\sqrt{\frac{2gR}{5}}\] Let radius of curvature at B be r, \[r=\frac{{{v}^{2}}}{{{a}_{n}}}\] \[\Rightarrow \,\,\,r=\frac{{{v}^{2}}}{g\cos 37{}^\circ }=\frac{{{\left( \sqrt{\frac{2gR}{5}} \right)}^{2}}}{g\cos 37{}^\circ }=\frac{\frac{2gR}{5}}{\frac{4g}{5}}=\frac{R}{2}\]