Irodov 1.142: The Cost of Spinning Up Slowly – Amit & Deepali Sinha Foundation

Problem Irodov 1.142

A sleeve of mass $m$ rides on a smooth (frictionless), massless horizontal rod $AB$ that can rotate about a fixed vertical axis through its end $A$ ; the rod itself carries no moment of inertia. A weightless spring of natural length $l_0$ and stiffness $\varkappa$ connects the sleeve to $A$ along the rod. Starting from rest, the rod is spun up slowly (quasi-statically) until it reaches angular velocity $\omega$ . Find the work that must be done on the system to reach that final state.

Top view: the ringed dot at A is the rotation axis. The rod AB turns at ω, carrying the sleeve m held at radius r by the spring force F_sp, the spring stretched beyond its natural length l_0. — Top view: the ringed dot at $A$ is the rotation axis. The rod $AB$ turns at $\omega$ , carrying the sleeve $m$ held at radius $r$ by the spring force $F_{sp}$ , the spring stretched beyond its natural length $l_0$ .

Given

$m$
$l_0$
$\varkappa$
$\omega$

Find

A

, the work needed to reach

\omega

Solution

Step 1Quasi-static radial balance fixes the working radius

Because the spin-up is slow, at every instant the sleeve is essentially in radial equilibrium at whatever angular velocity $\Omega$ the rod currently has: the spring’s pull supplies exactly the centripetal force the sleeve needs at radius $r(\Omega)$ . Writing $\varkappa$ times the extension against $m\Omega^2 r$ and evaluating at the final angular velocity $\omega$ ,

\varkappa\bigl(r-l_0\bigr) = m\omega^2 r. \tag{1}

Solving for $r$ and introducing $\eta \equiv m\omega^2/\varkappa$ ,

\begin{align*}\varkappa r - \varkappa l_0 &= m\omega^2 r\\ r(\varkappa-m\omega^2) &= \varkappa l_0\end{align*}

r = \dfrac{\varkappa l_0}{\varkappa-m\omega^2} = \dfrac{l_0}{1-\eta}

(2)

Step 2The final extension

The spring’s extension at the end of the process follows from the equilibrium relation (1), rewritten with $\eta$ in place of $m\omega^2/\varkappa$ .

\begin{align*}\varkappa(r-l_0) &= m\omega^2 r = \varkappa\eta r\\ r-l_0 &= \eta r = \frac{\eta l_0}{1-\eta}\end{align*}

Step 3Work equals the final mechanical energy

The rod is smooth and massless, so it carries no kinetic energy of its own and applies no friction, and the spring is weightless (ideal); the only places the input work can go are the sleeve’s kinetic energy and the spring’s elastic energy, and nothing is dissipated. Because the process is quasi-static, the sleeve’s speed at the end is purely tangential, $v=\omega r$ , with no leftover radial component.

\begin{align*}A &= \underbrace{\frac12 m\omega^2 r^2}_{\text{kinetic energy}} + \underbrace{\frac12\varkappa(r-l_0)^2}_{\text{spring energy}}\\ &= \frac12 m\omega^2\left(\frac{l_0}{1-\eta}\right)^2+\frac12\varkappa\left(\frac{\eta l_0}{1-\eta}\right)^2\end{align*}

Writing $m\omega^2=\varkappa\eta$ in the first term puts both terms over the same factor $\varkappa l_0^2/(1-\eta)^2$ ,

\begin{align*}A &= \frac{\varkappa l_0^2}{2(1-\eta)^2}\Bigl(\eta+\eta^2\Bigr)\end{align*}

A = \dfrac{\varkappa l_0^2\,\eta(1+\eta)}{2(1-\eta)^2}, \qquad \eta = \dfrac{m\omega^2}{\varkappa}

(3)

Check

As $\eta\to0$ (a very stiff spring or a slow final speed), the extension vanishes and $A\to \varkappa l_0^2\eta/2=m\omega^2 l_0^2/2$ , exactly the kinetic energy of a mass moving on a circle of the spring’s natural radius $l_0$ , which is what the sleeve should look like when it barely moves off $l_0$ . Dimensionally $[\varkappa][l_0^2]=(\text{N/m})(\text{m}^2)=\text{N}\!\cdot\!\text{m}$ , an energy, as required.

Remark

The work diverges as $\eta\to1$ , i.e. as $m\omega^2\to\varkappa$ : the spring’s restoring force per unit extension can no longer keep pace with the centripetal demand at large $r$ , so no finite radius satisfies (1) and the sleeve cannot be spun up quasi-statically past that speed at all. This is the same runaway that appears in any spring-supplied centripetal problem (a conical pendulum on a spring, a rotating governor) once the required force starts growing faster with radius than the restoring force does.