question_answer

A 2 kg stone at the end of a string 1m long is whirled in a vertical circle. At some point its speed is 4 m/s. The tension of the string is 51.6 N at this instant. Therefore, the stone is

A)  At the top of the circle

B)  At the bottom of the circle  

C)  Half-way down

D)  None of these

Correct Answer: B

Solution :

 At the top,             \[T+mg=\frac{m{{v}^{2}}}{r}\] At the bottom,             \[T'-mg=\frac{mv{{'}^{2}}}{r}\] \[\therefore \]      \[T'-2\times 9.8=\frac{2\times {{(4)}^{2}}}{1}=32\] \[\Rightarrow \]   \[T'=32+19.6=\mathbf{51}\mathbf{.6}\,\,\mathbf{N}\] Thus, stone is at the bottom of the circle