Question

A $1\text{kg}$ stone at the end of a $1\text{m}$ long string is whirled in a vertical circle at a constant speed of $4\text{m/s}$. The tension in the string is $6\text{N}$ when the stone is (Take $g=10{\text{m/s}}^2$)

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

Answer 1

The correct option is A At the top of the circle.

Suppose the string is making an angle $\theta$ with the verical
On balancing the forces along the length of the string,
$T=mg\cos \theta +\frac{m{v}^2}{R}$ ----- (1)
where, $R=1\text{m},v=4\text{m/s},m=1\text{kg}$ (given)
Tension $T=6\text{N}$
Putting all these values in equation (1), we get:
$6=1\times 10\times \cos \theta +\frac{1\times 4^2}{1}$
$\Rightarrow 6=10\cos \theta +16$
$\Rightarrow \cos \theta =-1$
$\Rightarrow \theta ={180}^{\circ }$ {With the vertical}
So, the stone will be at the top of the circle.