A mass 5kg is whirled in a horizontal circle @ 1 end of a string 50cm long,the other end being fixed. If the string when hanging vertically will just support a load of 200kg mass without breaking,find the maximum whirling speed in revolution per second and the maximum angular velocity.

Answer 1

Breaking tensile strength of string = Tmax = 200 kg*9.8 m/s^2 = 1960 N

Then whirling horizontally,

T sin A = M g
T cos A = M V^2/R
T^2 = M^2*[g^2 + (V^2/R)^2]
T = Tmax = M* sqrt [g^2 + (V^2/R)^2]
= 1960 N
Tmax/M = 392 m/s^2
= sqrt [g^2 + (V^2/R)^2]
Solve for V