Kinematics Question 539
Question: A 2 kgstone at the end of a string 1 m long it’s 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 [AIIMS 1982]
Options:
A) At the top of the circle
B) At the bottom of the circle
C) Halfway down
D) None of the above
Show Answer
Answer:
Correct Answer: B
Solution:
$ mg=20N $ and $ \frac{mv^{2}}{r}=\frac{2\times {{(4)}^{2}}}{1}=32N $ .
It’s clear that 52 N tension will be at the bottom of the circle..
Because we know that $ {T _{\text{Bottom}}}=mg+\frac{mv^{2}}{r} $
BETA
