Irodov 1.88: Spring Against Centrifugal Pull – Amit & Deepali Sinha Foundation

Problem Irodov 1.88

A smooth L-shaped rod lies in a horizontal plane, and a sleeve $A$ of mass $m$ slides on it, held by a weightless spring of stiffness $\varkappa$ anchored at the point $B$ where the rod’s two arms meet; the sleeve slides on the arm that runs out from $B$ . The whole assembly rotates at a constant angular velocity $\omega$ about a vertical axis through the point $O$ at the far end of the other arm. Find the elongation of the spring once the sleeve has settled into its steady position, and determine whether the result depends on the sense of rotation.

The shaft turns at ω about the axis through O. The L-shaped rod runs from O to the corner B, then outward, carrying the sleeve A and its spring anchored at B. The dashed ellipse is A's circle. — The shaft turns at $\omega$ about the axis through $O$ . The L-shaped rod runs from $O$ to the corner $B$ , then outward, carrying the sleeve $A$ and its spring anchored at $B$ . The dashed ellipse is $A$ ’s circle.

Given

$m$
$\varkappa$
$\omega$

Find

The elongation of the spring, and whether it depends on the sense of rotation

Solution

Step 1Working in the frame that rotates with the rod

Free body of the sleeve A in the rotating frame, seen from above with B to the left. The centrifugal force F_cf lies along OA, the spring force F_sp toward B, the normal force N across the arm. — Free body of the sleeve $A$ in the rotating frame, seen from above with $B$ to the left. The centrifugal force $\vect F_{cf}$ lies along $OA$ , the spring force $\vect F_{sp}$ toward $B$ , the normal force $\vect N$ across the arm.

The sleeve does not slide along the rod once the system has settled into steady rotation, so in the frame rotating with the rod at $\omega$ it sits in static equilibrium.

Three horizontal forces act on the sleeve: the centrifugal force, directed away from the axis along $OA$ ; the spring force, along the arm toward $B$ ; and the normal force from the smooth rod, perpendicular to the arm (the rod is smooth, so it exerts nothing along itself).

F_{cf} = m\omega^2 r \quad \text{(centrifugal, along $OA$ outward)}, \qquad F_{sp} = \varkappa x \quad \text{(spring, along the arm toward $B$)},

where $r=OA$ is the sleeve’s distance from the axis and $x$ is the spring’s elongation. Because $B$ is not on the axis, $\vect F_{cf}$ is not along the arm, and only its component along the arm can be balanced by the spring; the rest is taken by $\vect N$ .

Step 2The centrifugal pull along the arm

Let $l_0$ be the spring’s unstretched length, so the sleeve sits a distance $BA = l_0 + x$ from $B$ along the arm. The position vector of the sleeve from the axis is $\vect{OA} = \vect{OB} + \vect{BA}$ , and the two arms of the L are perpendicular, so $\vect{OB}$ has no component along the arm $BA$ . The component of $\vect{OA}$ along the arm is therefore just $BA$ , whatever the length of $OB$ , and the centrifugal force $m\omega^2\vect{OA}$ projects onto the arm as

F_{cf,\parallel} = m\omega^2\,(l_0 + x).

The perpendicular part, $m\omega^2\,OB$ , is exactly what the normal force $\vect N$ supplies, and it plays no further role.

Step 3Balancing the two forces

Equilibrium along the arm requires the spring force to supply exactly the centrifugal force’s component along it,

\varkappa x = m\omega^2(l_0+x)

(1)

Collecting the $x$ terms,

x(\varkappa - m\omega^2) = m\omega^2 l_0,

so the elongation is

x = \frac{m\omega^2 l_0}{\varkappa - m\omega^2}.

Step 4The elongation as a fraction of the natural length

Dividing by $l_0$ gives a result that does not need $l_0$ itself, only the ratio $\varkappa/m\omega^2$ : writing $\varepsilon = x/l_0$ for the elongation measured in units of the unstretched length,

\varepsilon = \frac{m\omega^2}{\varkappa - m\omega^2}.

Dividing numerator and denominator by $m\omega^2$ ,

\varepsilon = \dfrac{1}{\varkappa/m\omega^2 - 1}

(2)

Step 5Independence from the sense of rotation

Reversing the direction of rotation sends $\omega \to -\omega$ , but every occurrence of $\omega$ above appears only as $\omega^2$ : the centrifugal force $m\omega^2 r$ points outward regardless of which way the rod turns.

So the elongation $x$ , and hence $\varepsilon=x/l_0$ , is unchanged by the rotation direction, since the balance equation (1) that fixes $x$ depends on $\omega$ only through $\omega^2$ .

Check

As $\varkappa \to m\omega^2$ from above, $\varepsilon \to \infty$ : solving (1) for $x$ divides by $\varkappa - m\omega^2$ , and that factor vanishes exactly there, so no finite steady elongation exists once $\omega^2$ reaches $\varkappa/m$ . In the opposite limit $\omega\to0$ , $\varepsilon\to0$ : a stationary sleeve needs no elongation to stay put.

Remark

The fractional elongation $\varepsilon$ depends only on the single dimensionless ratio $\varkappa/m\omega^2$ : any two sleeve-spring systems sharing that ratio stretch by the same fraction of their natural length, however different their individual $m$ , $\varkappa$ , $\omega$ or $l_0$ happen to be. It is the same kind of grouping that turns a resonance condition into one dimensionless number instead of four separate ones.