JEE MainPhysicsLaws of Motion
A small block of mass m is attached to one end of a light spring of spring constant k and natural length L . The other end of the spring is fixed to a point on a smooth horizontal table. The block is made to rotate in a horizontal circle about the fixed point such that the total length of the spring becomes nL (where n > 1 ). The angular velocity of the block is
Options
- Ak(n-1) m
- Bk m
- Cmn k(n-1)
- Dk(n-1) mn
Correct answer
D. k(n-1) mn
Step-by-step solution
Let the angular velocity of the block be . The natural length of the spring is L and its stretched length is nL . Elongation in the spring, x = nL - L = (n-1)L . The spring force provides the necessary centripetal force for circular motion. The radius of the circular path is the total stretched length of the spring, r = nL . Equating the spring force to the centripetal force: kx = m ^2 r k(n-1)L = m ^2(nL) Solving for : ^2 = k(n-1)L mnL = k(n-1) mn = k(n-1) mn Answer: k(n-1) mn