Question

One end of a massless spring of spring constant k and natural length \(l_0\) 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 :

• A. \(\frac{k-mω^2l_0}{mω^2}\)
• B. \(\frac{mω^2l_0}{k+mω^2}\)
• C. \(\frac{mω^2l_0}{k-mω^2}\)
• D. \(\frac{(k+mω^2l_0)}{mω^2}\)

Answer 1

The Correct Option is C

To find the elongation of the spring when an object of mass m is rotating with angular velocity ω about an axis passing through the fixed end of the spring, we need to consider the forces acting on the object due to the spring and the circular motion.

Concepts Used:

• Centripetal Force: For an object moving in a circle of radius r with an angular velocity ω, the required centripetal force is mω^2r.
• Hooke's Law: If the spring is extended by a length x from its natural length l_0, the restoring force it exerts is kx.

Since the object is in circular motion, the centripetal force is provided by the spring. The effective radius of rotation will be the new length of the spring, which is l = l_0 + x.

Equation Setup:

Equating the centripetal force to the spring force, we have:

mω^2(l_0 + x) = kx

Solving for x:

1. Expand and rearrange: mω^2l_0 + mω^2x = kx
2. Bring terms involving x together: mω^2l_0 = kx - mω^2x
3. Factor out x: mω^2l_0 = x(k - mω^2)
4. Solve for x: x = \frac{mω^2l_0}{k - mω^2}

Thus, the elongation of the spring is \(\frac{mω^2l_0}{k - mω^2}\).

Conclusion:

The correct answer is \(\frac{mω^2l_0}{k - mω^2}\). This answer matches with the given correct option, confirming our solution is correct.