Q. A$1\ kg$ stone at the end of $1\ m$ long string is whirled in a vertical circle at constant speed of $4\ m/sec$. The tension in the string is $6\ N$, when the stone is at $(g = 10\ m/\sec^{2}{\ )}$
A) Top of the circle
B) Bottom of the circle
C) Half way down
D) None of the above
Explanation
(a) $mg = 1 \times 10 = 10N,\frac{mv^{2}}{r} = \frac{1 \times (4)^{2}}{1} = 16$
Tension at the top of circle $= \frac{mv^{2}}{r} - mg = 6N$
Tension at the bottom of circle $= \frac{mv^{2}}{r} + mg = 26N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.