Kvant Physics Problem 24
A body of mass $M$ is attached to an ideal spring of stiffness $k$, whose upper end is fixed.
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Problem
A support on which a body suspended from a spring rests is lowered with acceleration $a$. Initially, the spring is unstretched. After what time will the body lift off the support? To what maximum length will the spring be stretched? The mass of the body is $M$, the spring constant is $k$.
N. I. Goldfarb
Setup and Assumptions
A body of mass $M$ is attached to an ideal spring of stiffness $k$, whose upper end is fixed. The body initially rests on a rigid support that touches it from below. The spring is initially unstretched, so its natural length coincides with its initial length.
The support is then lowered vertically with constant downward acceleration $a$, starting from rest. Motion is considered in an inertial reference frame fixed to the ground. Gravity acts downward with acceleration $g$. The spring is massless and obeys Hooke’s law. Contact between the body and the support is perfectly smooth, and separation occurs when the normal reaction force becomes zero.
The unknowns are the time $t_\ast$ at which the body loses contact with the support and the maximum extension of the spring after this instant.
Physical Principles
Newton’s second law for the body while it remains in contact with the support is written in the vertical direction as $M \ddot{x} = \sum F$.
Hooke’s law for the spring force is $F_s = kx$, where $x$ is the extension measured from the unstretched length.
The condition for loss of contact is $N = 0$, where $N$ is the normal reaction force from the support.
After separation, the motion is governed solely by gravity and the spring force, so Newton’s second law becomes $M \ddot{y} = Mg - ky$ when downward displacement $y$ is measured from the initial unstretched position.
Mechanical energy in the post-separation phase is conserved in the form $E = \frac{1}{2}M v^2 + \frac{1}{2}k y^2 - M g y$.
Derivation
While the body remains in contact with the support, it follows the support’s motion. Taking downward displacement $x(t)$, the support accelerates as $x(t) = \frac{1}{2} a t^2$, and the body shares this motion.
Forces on the body during contact include gravity $Mg$ downward, spring force $kx$ upward, and normal force $N$ upward. Newton’s second law in the downward direction gives
$$M a = Mg - kx - N.$$
Separation occurs when $N=0$, giving the condition
$$M a = Mg - kx_\ast.$$
Solving for the extension at detachment yields
$$kx_\ast = M(g - a).$$
Since the support motion gives $x(t) = \frac{1}{2} a t^2$, the detachment time follows from
$$\frac{1}{2} a t_\ast^2 = \frac{M(g - a)}{k}.$$
This produces
$$t_\ast = \sqrt{\frac{2M(g - a)}{ka}}.$$
This expression is physically meaningful only when $g > a$, since otherwise the right-hand side becomes negative and contact is never lost.
At the moment of separation, the position and velocity are
$$y_1 = \frac{1}{2} a t_\ast^2, \quad v_1 = a t_\ast.$$
Substituting the expression for $t_\ast^2$ gives
$$y_1 = \frac{M(g - a)}{k}, \quad v_1 = \sqrt{\frac{2aM(g - a)}{k}}.$$
After separation, the motion satisfies
$$M \ddot{y} = Mg - ky.$$
Energy conservation gives
$$E = \frac{1}{2}M v_1^2 + \frac{1}{2}k y_1^2 - M g y_1.$$
Substituting $y_1$ and $v_1$ yields
$$E = \frac{M^2 a (g - a)}{k} + \frac{M^2 (g - a)^2}{2k} - \frac{M^2 g (g - a)}{k}.$$
The maximum extension corresponds to $v=0$, so $y_{\max}$ satisfies
$$\frac{1}{2}k y_{\max}^2 - M g y_{\max} = E.$$
Solving the quadratic equation gives
$$y_{\max} = \frac{Mg + \sqrt{M^2 g^2 + 2kE}}{k}.$$
Substituting $E$ and simplifying leads to cancellation of terms under the square root, producing
$$y_{\max} = \frac{M}{k}\left(g + \sqrt{g^2 - a^2}\right).$$
Result
The time of loss of contact is
$$t_\ast = \sqrt{\frac{2M(g - a)}{ka}}.$$
The maximum extension of the spring is
$$y_{\max} = \frac{M}{k}\left(g + \sqrt{g^2 - a^2}\right).$$
Numerically, these expressions correspond to quantities measured in seconds for $t_\ast$ and meters for $y_{\max}$ when $M$ is in kilograms, $k$ in newtons per meter, and $a$, $g$ in meters per second squared.
Sanity Checks
Dimensional consistency holds since $t_\ast^2$ scales as $\frac{M g}{k a}$, which reduces to $\frac{\mathrm{kg}\cdot \mathrm{m/s^2}}{\mathrm{N/m}\cdot \mathrm{m/s^2}} = \mathrm{s^2}$, and $y_{\max}$ scales as $\frac{Mg}{k}$, which has units of length.
In the limit $a \to 0$, detachment time diverges, consistent with the support being static and continuously supporting the body. In the same limit, the maximum extension reduces to $y_{\max} \to \frac{2Mg}{k}$, corresponding to oscillation about the equilibrium position reached from initial compression-free release.
The sign in the detachment condition is the most sensitive point; reversing the force balance would incorrectly predict detachment for all accelerations or eliminate it entirely. The transition condition depends critically on comparing inertial acceleration with gravitational acceleration.