Question

A particle of 10g is attached to one end of a rubber string of length 50cm. It is whirled into a horizontal circle about the other end of string with a uniform angular speed of 300 r.p.m. The rubber string has a force constant of 0.05 kgfcm^-1, g = 10ms^-2. The tension in string is

(1) 5 N

(2) 6.25 N

(3) 9 N

(4) 12.5 N

Answer 1

(2) 6.25 N

k = Force constant of rubber string = 0.05 kgf cm^-1 

= 0.05 x 10 / 10^-2 Nm^-1 = 50 Nm^-1

ω = 300r.p.m. = 5r.p.s. = 10 rad s^-1

Let x be increase in length of rubber string when particle attached to it moves in a horizontal circle. The radius of circle = r = (0.5+x)m. The tension in string = T = mr ω² = kx