Question

14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path

Answer 1

Given, the mass is 14.5 kg, the initial length of each wire is 1.0 m, the angular velocity is 2  rev/s . The cross sectional area of the wire is 0.065  cm 2 .

The force acting on the wire when the mass is at its lowest point is given by the equation,

F=mg+mr ω 2

Substituting the given values in the above expression, we get:

F=14.5 kg×9.8+14.5 kg×1 m ( 2  rev/s ) 2 =200.1 N

Young’s modulus is given by the equation:

Y= FL AΔL ΔL= FL AY

Substituting the values in the above expression, we get:

ΔL= 200.1 N×1 m ( 0.65× 10 −4   m 2 )( 2× 10 11  Pa ) =1.539× 10 −4  m

Hence, the elongation of the wire when the mass is at the lowest point of it’s path is 1.539× 10 −4  m.