A stone of mass m is whirled in a horizontal circle of radius R at a height H above the ground. The rotational speed of the stone is gradually increased until the string breaks. Upon breaking, the stone flies off horizontally and strikes the ground at a horizontal distance D from the point where the string broke. The maximum tension the string could withstand before breaking is:

Step-by-step solution

When the string breaks, the stone undergoes horizontal projectile motion. The time of flight t to fall a vertical height H is given by: H = 1 2 g t^2 t = 2H g The horizontal velocity v of the stone at the instant the string breaks is: v = D t = D g 2H Just before breaking, the tension T in the string provided the necessary centripetal force: T = m v^2 R Substituting the expression for v^2 : T = m R ( D^2 g 2H ) = m D^2 g 2 H R Answer: m D^2 g 2 H R