Question

A 2kg stone at end of a string 1m long is whirled in a vertical circle at a constant speed.The spped of a stone is 4m/s.The tension is in the string will be 52N,when the stone is ​

Answer 1

the stone is at the top of the circle

Answer 2

Answer:

At the top of the circular path

Explanation:

Given,

• Mass of the stone = m = 2 kg
• Speed of the stone = v = 4 m/s.
• Tension in the string = 52 N
• Length of the sting = R = 1 m

Now the stone is moving in the circular path, therefore a force is acting on the    stone is centripetal force,

Centripetal force acting on the stone due to the circular motion,

$\therefore F_c\ =\ \dfrac{mv^2}{R}\\\Rightarrow F_c\ =\ \dfrac{2\times 4^2}{1}\\\Rightarrow F_c\ =\ 32\ N\\\therefore T\ >\ F_c\,\,\,\,\,(52\ N\ >\ 32\ N)$

Hence the tension force given is greater than the centripetal force acting on the stone in only one condition if the stone is at the top of the vertical circular path.