Question

A stone of mass 2 kg is attached to the string of length 1 meter .The string can with stand maximum tension 200 N.what is the maximum speed that stone can have during the whirling motion?

Answer 1

I already answered your same question-
Now again-

At bottom most point the tension will be maximum !!

when we draw Free body diagram we get-

$Tmax = \frac{m {v}^{2} }{r} + mg \\ 200 = \frac{2 {v}^{2} }{1} + 20 \: taking \: g = 10 m {sec}^{ - 2} \\ 180 = 2 {v}^{2} \\ 90 = {v}^{2} \\ v = \sqrt{90} \\ v = 3 \sqrt{10} m {sec}^{ - 1}$

Answer 2

we can solve such problems by considering two points,

1. changing velocity,constant tension ,
2. constant velocity,changing tension,(ucm).

since here it is asked about the velocity, it's defiantly a vertical circular motion and need to be solved
considering the second point,

we know tht in vcm, tension at highest point is 0 and at bottom point it's maximum

thus,at bottom point we can write,

centripital force=tension _mg
now substitute the value and get the answer.