Kinematics Question 566
Question: The maximum and minimum tension in the string whirling in a circle of radius 2.5 m with constant velocity are in the ratio 5 : 3 then it’s velocity is [Pb. PET 2003]
Options:
A) $ \sqrt{98}m/s $
B)$ 7m/s $
C) $ \sqrt{490}m/s $
D) $ \sqrt{4.9} $
Show Answer
Answer:
Correct Answer: A
Solution:
In this problem it’s assumed that particle although moving in a vertical loop but it’s speed remain constant. Tension at lowest point $ {T _{\max }}=\frac{mv^{2}}{r}+mg $ Tension at highest point $ {T _{\min }}=\frac{mv^{2}}{r}-mg $
$ \frac{{T _{\max }}}{{T _{\min }}}=\frac{\frac{mv^{2}}{r}+mg}{\frac{mv^{2}}{r}-mg}=\frac{5}{3} $ by solving we get,$ v=\sqrt{4gr} $
$ =\sqrt{4\times 9.8\times 2.5} $
$ =\sqrt{98}m/s $
