Q. A $2\ kg$ stone at the end of a string $1\ m$ long is whirled in a vertical circle at a constant speed. The speed of the stone is $4\ m/sec$. The tension in the string will be $52\ N$, when the stone is
A) At the top of the circle
B) At the bottom of the circle
C) Halfway down
D) None of the above
Explanation
(b) $mg = 20N$ and $\frac{mv^{2}}{r} = \frac{2 \times (4)^{2}}{1} = 32N$
It is clear that $52\ N$ tension will be at the bottom of the circle. Because we know that $T_{Bottom} = mg + \frac{mv^{2}}{r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.