Question

A $1-kg $ stone at the end of $1 \,m$ long string is whirled in a vertical circle at a constant speed of $4\, ms ^{-1}$. The tension in the string is $6 \,N$ when the stone is

• A. At the top of the circle
• B. At the bottom of the circle
• C. Half way down
• D. None of above

Answer 1

Answer: (A)

$ \frac{m v^{2}}{R} =\frac{1 \times 4^{2}}{1}=16 \,N $
$m g=1 \times 10=10 \,N $
$ \Rightarrow T =m g \cos \theta+\frac{m v^{2}}{R}$
$ 6 =10 \cos \theta+16$
$ \operatorname{cosec} \theta=-1 $
$\theta =180^{\circ}$