Question

A spring is attached to the center of a friction less horizontal turn table and at the other end a body of mass 2kg is attached. The length of the spring is 35cm. Now when the turn table is rotated with an angular speed of 10 rad s^–1, the length of the spring becomes 40 cm then the force constant of the spring is …………. N/m.

(A) 1.2 × 10³

(B) 1.6 × 10³

(C) 2.2 × 10³

(D) 2.6 × 10³

Answer 1

The correct option (B) 1.6 × 10³

Explanation:

radius of the rotational motion = r = 0.4m 

when the turn table rotates, the restoring force developed in the spring = centrifugal force.

∴ F_retoring = mω²r

given: m = 2kg

ω = 10rad/s

∴ F_restoring = 2 × 10² × 0.40

= 80 N

Now increase in the length of spring = 40 – 35 = 5 cm.

As F = K ∙ x

∴ Force constant = K = (F/x) = {80/(0.05)} = 1.6 × 10³N/m