A particle of mass m is fixed to one end of a light spring having force constant k and unstretched l

Q: Why is the radius $l+x$ and not just $l$?

When the system rotates, the mass tends to move outward due to inertia, stretching the spring. The mass only settles into a stable circular orbit once the spring has stretched enough to provide the necessary centripetal force. Thus, the actual radius of rotation includes this stretch $x$.

Q: Why don't we consider gravity in this problem?

The problem explicitly states that the system is in a 'gravity free space'. This simplifies the scenario, meaning we don't have to worry about the weight of the mass or the pendulum-like swinging motion.

Q: What if the spring was compressed instead of stretched?

In a rotating system like this, the required centripetal force is always directed inward, meaning the spring must pull the mass inward. A spring only pulls when it is stretched. Therefore, it cannot be compressed in this setup.

Q: Can the denominator $k - m\omega^2$ be negative?

If $m\omega^2 > k$, the required centripetal force increases faster with distance than the spring force can provide. In reality, the spring would stretch until it breaks or loses its elastic properties. A stable circular orbit is impossible in this regime.

Q: Is the angular velocity $\omega$ constant?

Yes, the problem implies a steady state where the system 'rotates in a circle' with a given angular speed $\omega$. We assume no energy is lost and the rotation is uniform.

Question

A particle of mass $m$ is fixed to one end of a light spring having force constant $k$ and unstretched length $l$. The other end is fixed. The system is given an angular speed $\omega$ about the fixed end of the spring such that it rotates in a circle in gravity free space. Then, the stretch in the spring is

Options:

(A) $\frac{ml\omega^2}{k+m\omega}$

(B) $\frac{ml\omega^2}{k+m\omega^2}$

(C) $\frac{ml\omega^2}{k-m\omega^2}$

(D) $\frac{ml\omega^2}{k-m\omega}$

Sigma Percentile · Accepted Answer

### The Setup: A Spinning Spring

Imagine you are in the vast emptiness of space, far away from any gravitational pull. You have a spring of natural length $l$ and stiffness $k$, with one end pinned down and a mass $m$ attached to the other. Now, you give the mass a push, setting it into a continuous spin with an angular velocity $\omega$.

As the mass spins, it doesn't just stay at distance $l$. Inertia—the tendency of the mass to keep moving in a straight line—causes it to pull outward. The spring stretches by some amount, let's call it $x$. So, the mass is now tracing a circle of a new, larger radius: $R = l + x$.

### The Master Equation: Balancing Forces

For any object to maintain a circular path, it absolutely must have a force pulling it towards the center. This is the famous **centripetal force**, given by the formula $F_c = mR\omega^2$. If this force suddenly vanished, the mass would fly off tangentially into the void.

In our setup, what is acting as the invisible string pulling the mass inward? It's the spring! As the spring stretches by $x$, it fights back with a restoring force according to Hooke's Law: $F_s = kx$.

Since the spring is the sole provider of the centripetal force, we can equate the two:

$$kx = m(l + x)\omega^2$$

### The Final Calculation: Isolating the Stretch

Now, it's just a matter of simple algebra to find our unknown, $x$. Let's expand the right side of our master equation:

$$kx = ml\omega^2 + mx\omega^2$$

We want all the $x$ terms on one side. Let's subtract $mx\omega^2$ from both sides:

$$kx - mx\omega^2 = ml\omega^2$$

Factoring out $x$, we get:

$$x(k - m\omega^2) = ml\omega^2$$

Finally, dividing by the term in the parentheses gives us the exact stretch in the spring:

$$x = \frac{ml\omega^2}{k - m\omega^2}$$

This elegant expression tells us exactly how much the spring will yield under the rotational inertia. Notice the denominator: $k - m\omega^2$. If the rotation is too fast, and $m\omega^2$ approaches $k$, the stretch approaches infinity! The spring would simply fail to hold the mass in a circle.

Select Answer:

Visualized Solution

Visualizing the Rotating System

Centripetal Force Requirement

Equating the Forces

Solving for Extension $x$

Final Expression for Stretch

Physical Significance

The Sigma Insight: Dynamics of Circular Motion

The Setup

The Master Equation

The Final Calculation

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