Irodov 1.140: When the Spring Pulls It Off – Amit & Deepali Sinha Foundation

Problem Irodov 1.140

A small bar $A$ rests on a frictionless horizontal table. It is held by a horizontal thread to a fixed bracket $P$ on the wall, and separately, by a second thread running over a light, frictionless pulley mounted at the table’s edge, to a hanging weight $B$ of the same mass $m$ as the bar. A light spring, of natural (non-deformed) length $l_0=50$ cm and stiffness $\varkappa=5mg/l_0$ , connects the bar to a fixed point $O$ . The spring is initially undeformed, so $O$ sits at height $l_0$ directly above the bar’s starting position. At some instant the thread $PA$ is burned, and the bar begins to slide. Find the bar’s speed at the instant it lifts off the table.

Bar A rests on a fixed table, held by a thread to the wall point P, a spring of natural length l_0 hanging from the ceiling point O, and a thread over an edge pulley to the hanging weight B. — Bar $A$ rests on a fixed table, held by a thread to the wall point $P$ , a spring of natural length $l_0$ hanging from the ceiling point $O$ , and a thread over an edge pulley to the hanging weight $B$ .

Given

$l_0 = 50$ cm
$\varkappa = 5mg/l_0$
$m_A = m_B = m$

Find

Find

v

, the bar’s speed at the instant it leaves the table

Solution

Step 1The bar and the weight move together

Once thread $PA$ is burned, the only motion the table allows bar $A$ is horizontal, so let $x$ be the distance $A$ has slid toward the pulley at some later instant. The pulley is weightless and the thread from $A$ to $B$ is inextensible, so whatever length disappears on $A$ ’s side reappears on $B$ ’s side: $B$ has descended the same distance $x$ , and the two bodies share one speed, $v_A = v_B = v$ .

Step 2Energy from rest to a general position

The spring’s far end $O$ is fixed directly above the bar’s starting point, at height $l_0$ , because the spring is given as non-deformed there. Once the bar has slid a distance $x$ , the segment from $O$ to the bar, $x$ horizontally and $l_0$ vertically, has length

L = \sqrt{x^2+l_0^2}. \tag{1}

The table is smooth, the pulley and thread ideal, and the spring conservative, so mechanical energy is conserved between the start, at rest with the spring undeformed, and this general position. The bar’s kinetic energy is $mv^2/2$ and so is $B$ ’s, since both share speed $v$ ; the spring stores $\varkappa(L-l_0)^2/2$ ; and $B$ has fallen height $x$ , releasing gravitational energy $mgx$ .

mv^2 + \dfrac{1}{2}\varkappa(L-l_0)^2 = mgx.

Solving for $v^2$ ,

v^2 = gx - \frac{\varkappa}{2m}(L-l_0)^2

(2)

Step 3The vertical balance that ends the contact

Forces on bar A while it is still on the table, a distance x from under O: the normal N, weight m g, tension T and spring force F_sp along the hypotenuse L at angle θ to the leg l_0. — Forces on bar $A$ while it is still on the table, a distance $x$ from under $O$ : the normal $\vect N$ , weight $m\vect g$ , tension $\vect T$ and spring force $\vect F_{\mathrm{sp}}$ along the hypotenuse $L$ at angle $\theta$ to the leg $l_0$ .

The spring is stretched from its natural length $l_0$ to the current length $L$ , so by Hooke’s law it pulls the bar toward $O$ with a force proportional to that stretch.

F_{\mathrm{sp}} = \varkappa(L-l_0)

The pull is directed along the spring, at the angle $\theta$ from the vertical, and the right triangle in the figure gives that angle’s cosine as the ratio of the vertical leg to the hypotenuse.

\cos\theta = \frac{l_0}{L}

The spring force’s vertical component is therefore

F_{\mathrm{sp}}\cos\theta = \varkappa(L-l_0)\,\frac{l_0}{L}

While the bar stays on the table its vertical acceleration is zero, so the forces normal to the table balance: the contact force $N$ and the spring’s vertical component together support the weight $mg$ . Horizontal forces, the tension $T$ toward the pulley and the spring’s horizontal component $F_{\mathrm{sp}}\sin\theta$ , do not enter this balance at all.

N + \varkappa(L-l_0)\frac{l_0}{L} = mg.

The bar breaks off exactly when $N$ reaches zero, so at that instant

mg = \varkappa(L-l_0)\frac{l_0}{L}.

Substituting $\varkappa = 5mg/l_0$ and canceling $mg$ ,

1 = \frac{5(L-l_0)}{L} \quad\Longrightarrow\quad L = 5(L-l_0) \quad\Longrightarrow\quad 4L = 5l_0.

L = \frac{5l_0}{4}

(3)

Then from (1), with $L = 5l_0/4$ ,

x^2 = L^2 - l_0^2 = \left(\frac{25}{16}-1\right)l_0^2 = \frac{9}{16}l_0^2 \quad\Longrightarrow\quad x = \frac{3}{4}l_0.

Step 4The speed at breakoff

Substitute $x = 3l_0/4$ and $L-l_0 = l_0/4$ from (3) into the energy relation (2), using $\varkappa/m = 5g/l_0$ from the given stiffness:

v^2 = g\cdot\frac{3l_0}{4} - \frac{5g}{2l_0}\left(\frac{l_0}{4}\right)^2 = \frac{3}{4}gl_0 - \frac{5}{32}gl_0 = \frac{24-5}{32}gl_0 = \frac{19}{32}gl_0.

v = \sqrt{\frac{19gl_0}{32}}

(4)

Evaluating with $l_0 = 0.50$ m and $g = 9.8$ m/s $^2$ :

v = \sqrt{\frac{19\times9.8\times0.50}{32}} = \sqrt{2.91} \approx 1.7\ \text{m/s}.

Check

$gl_0$ carries units of m $^2$ /s $^2$ , so its square root is a speed, as it must be. The geometry is consistent too: $x = 3l_0/4$ is less than $L = 5l_0/4$ , a genuine right triangle, and at $x=0$ the balance gives $N=mg$ , its full undisturbed value, falling to zero only at the computed $x$ , so the breakoff point is the first and only one reached as the bar slides out.

Remark

The mass $m$ cancels completely out of the final answer: it appears in the kinetic energy, in the weight $mgx$ , and, through $\varkappa = 5mg/l_0$ , in the spring term, and all three cancel together. The breakoff speed depends only on $g$ and $l_0$ , not on how heavy the bar or the weight are, which is exactly what naming the stiffness proportional to $m$ in the problem was built to produce.