One end of a massless spring of spring constant k and natural length l₀ is fixed while the other end is connected to a small object of mass m lying on a frictionless table. The spring remains horizontal on the table. If the object is made to rotate at an angular velocity ω about an axis passing through fixed end, then the elongation of the spring will be :

Solution

Physics sir · 4 steps

The spring pulls inward, and the circle's radius is the stretched length l₀ + x, not l₀.

1. What supplies the force. The table is frictionless and the spring is horizontal, so the spring force kx is the only horizontal force on the object. It points towards the fixed end and supplies the whole centripetal force.
2. Radius of the circle. The object moves on a circle of radius l₀ + x, where x is the elongation. The needed centripetal force is mω²(l₀ + x).

   kx = mω²(l₀ + x)

3. Solve for x. Collect the x terms: kx - mω² x = mω² l₀, so x(k - mω²) = mω² l₀.
4. So. The elongation is x = (mω² l₀)/(k - mω²), which is option (c).

The idea to keep. Whenever a spring holds a rotating mass, the radius grows as the spring stretches. Write the centripetal force with radius l₀ + x, then solve for x.

Answer from the NTA final answer key. Solution by PrepWiser.