A spring mass system (mass m, spring constant k and natural length l) rests in equilibrium on horizo

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

When the disc rotates, the mass must move outward to stretch the spring, because an unstretched spring provides zero force. The mass settles at a new equilibrium distance from the center, which is the original length $l$ plus the stretch $x$.

Q: What exactly does 'relative change in length' mean?

Relative change is a dimensionless ratio that compares the change in a quantity to its original value. Here, the change in length is the extension $x$, and the original length is $l$. So, the relative change is $x/l$.

Q: Why do we ignore $m\omega^2$ in the denominator but not in the numerator?

We are not ignoring it completely; we are comparing it to $k$. The denominator is $k - m\omega^2$. Since $k$ is huge compared to $m\omega^2$, subtracting $m\omega^2$ from $k$ barely changes the value of $k$ (like subtracting 1 from 10000). However, in the numerator, $m\omega^2$ is the only term, so it dictates the entire magnitude of the numerator and cannot be ignored.

Q: Does gravity affect this setup?

No, because the disc is horizontal. Gravity acts downwards (perpendicular to the disc), and the normal force from the disc perfectly balances it. The motion and spring stretching happen entirely in the horizontal plane.

Q: What if we solved this in the rotating frame of reference?

You would get the exact same result! In the rotating frame, the mass is at rest. The outward centrifugal pseudo-force $m\omega^2(l+x)$ is perfectly balanced by the inward spring force $kx$. Equating them gives the same starting equation.

Question

A spring mass system (mass $m$, spring constant $k$ and natural length $l$) rests in equilibrium on horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity $\omega$, $(k >> m\omega^2)$, the relative change in the length of the spring is best given by the option

Options:

(A) $\frac{m\omega^2}{3k}$

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

(C) $\sqrt{\frac{2}{3}}\left(\frac{m\omega^2}{k}\right)$

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

Sigma Percentile · Accepted Answer

### The Setup: A Dance of Inertia and Restoring Force Imagine a classic physics scenario: a horizontal disc, smooth and frictionless, with a spring anchored exactly at its center. At the other end of this spring lies a block of mass $m$. The spring has a natural, unstretched length of $l$ and a stiffness defined by its spring constant $k$. When the disc is perfectly still, the mass sits quietly at a distance $l$ from the center. But physics gets interesting when things start to move. Suppose the disc begins to spin around its central axis with a constant angular velocity $\omega$. What happens to the mass? ### The Physics of Rotation As the disc spins, the mass $m$ is forced to travel in a circular path. By Newton's First Law, the mass wants to travel in a straight line, effectively trying to fly off the disc tangentially. This tendency is often intuitively felt as an outward "centrifugal" push. However, the mass is tethered by the spring. To keep the mass moving in a circle, a force must constantly pull it towards the center. This is the **centripetal force**. The only thing capable of providing this inward pull is the spring. But a spring only pulls when it is stretched! Therefore, the mass slides outward, stretching the spring by some amount, let's call it $x$. The new radius of the circular path is no longer just $l$, but $r = l + x$. ### The Master Equation The required centripetal force to keep the mass in this new circular orbit is given by: $$F_c = m\omega^2r = m\omega^2(l + x)$$ The restoring force provided by the stretched spring is governed by Hooke's Law: $$F_s = kx$$ For the mass to remain in a stable circular orbit (equilibrium in the rotating frame), these two forces must perfectly balance each other: $$kx = m\omega^2(l + x)$$ ### Algebraic Manipulation The question asks for the **relative change in length**, which is mathematically defined as the change in length divided by the original length, or $\frac{x}{l}$. Let's rearrange our master equation to isolate this ratio. First, expand the right side: $$kx = m\omega^2l + m\omega^2x$$ Next, bring all terms containing $x$ to one side: $$kx - m\omega^2x = m\omega^2l$$ Factor out the $x$: $$x(k - m\omega^2) = m\omega^2l$$ Now, divide both sides by $l$ and by the bracketed term to isolate $\frac{x}{l}$: $$\frac{x}{l} = \frac{m\omega^2}{k - m\omega^2}$$ ### The Crucial Approximation We have an exact expression, but it doesn't quite look like the options. This is where we use the critical piece of information given in the problem: $k >> m\omega^2$. This inequality tells us that the spring is very stiff compared to the rotational forces at play. Let's divide the numerator and the denominator of our fraction by $k$ to see how this helps: $$\frac{x}{l} = \frac{\frac{m\omega^2}{k}}{1 - \frac{m\omega^2}{k}}$$ Since $k$ is much greater than $m\omega^2$, the fraction $\frac{m\omega^2}{k}$ is a very tiny number, much less than $1$ ($\frac{m\omega^2}{k} << 1$). Therefore, in the denominator, subtracting this tiny number from $1$ leaves us with something that is practically just $1$: $$1 - \frac{m\omega^2}{k} \approx 1$$ Applying this approximation, our complex fraction simplifies beautifully: $$\frac{x}{l} \approx \frac{m\omega^2}{k}$$ This elegant result shows that for a stiff spring, the relative stretch is directly proportional to the rotational kinetic energy factor ($m\omega^2$) and inversely proportional to the spring's stiffness ($k$).

Select Answer:

Visualized Solution

Visualizing the Setup

Dynamics of Rotation

Force Balance Equation

Algebraic Manipulation

Finding Relative Change

Applying the Approximation

Final Result

The Way Forward

The Sigma Insight: Dynamics of Circular Motion

The Setup

The Physics of Rotation

The Master Equation

Algebraic Manipulation

The Crucial Approximation

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Select Answer:

Visualized Solution

Visualizing the Setup

Dynamics of Rotation

Force Balance Equation

Algebraic Manipulation

Finding Relative Change

Applying the Approximation

Final Result

The Way Forward

The Sigma Insight: Dynamics of Circular Motion

The Setup

The Physics of Rotation

The Master Equation

Algebraic Manipulation

The Crucial Approximation

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