JEE MainPhysicsLaws of Motion
A small bob of mass m is suspended from a rigid ceiling by a light spring of natural length L and spring constant k . The bob is made to move in a horizontal circle with a constant angular speed , acting as a conical pendulum. The cosine of the angle that the spring makes with the vertical is
Options
- Amg(k - m ^2) k m ^2 L
- Bg ^2 L
- Cmg(k + m ^2) k m ^2 L
- Dmg kL
Correct answer
A. mg(k - m ^2) k m ^2 L
Step-by-step solution
Let the stretched length of the spring during the motion be L' . The radius of the horizontal circular path is r = L' . Let T be the tension in the spring. The horizontal component of tension provides the centripetal force: T = m ^2 r T = m ^2 (L' ) T = m ^2 L' From Hooke's Law, the tension is also given by the elongation of the spring: T = k(L' - L) Equating the two expressions for tension: m ^2 L' = k(L' - L) L'(k - m ^2) = kL L' = kL k - m ^2 The vertical component of tension balances the weight of the bob: T =