A body of mass 1 kg is attached to the lower end of a spring of natural length 0.6 m and spring constant 100 N m ⁻¹ . The upper end of the spring is fixed to a rigid ceiling. The body is made to move in a horizontal circle such that the spring acts as a conical pendulum, maintaining a constant angle of 60^ with the vertical. Taking g = 10 m s ⁻² , the constant angular speed of the body is:

Step-by-step solution

Let the extension in the spring be x . The spring force is F_s = kx . In the vertical direction, the component of the spring force balances the weight of the body: kx = mg 100 x (60^ ) = 1 10 100 x 1 2 = 10 50x = 10 x = 0.2 m The radius R of the horizontal circular path is the horizontal component of the stretched spring's length: R = (l + x) R = (0.6 + 0.2) (60^ ) = 0.8 3 2 = 0.4 3 m In the horizontal direction, the component of the spring force provides the necessary centripetal force: kx = mR ^2 100 0.2 (60^ ) =