Kinematics Question 537
Question: 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/sec2) [AIIMS 1982]
Options:
A) Top of the circle
B) Bottom of the circle
C) Half way down
D)None of the above
Show Answer
Answer:
Correct Answer: A
Solution:
$ 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 $
