Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

myvalq · 2009-05-19 14:02:11

A mass (weight 100g) is hanged on a spring inside a cubic box (side a=1m). Stiffness of the spring is 10N*m-1. Equilibrium state of the spring is right in a centre of the box.
Now, we let the box fall to the ground from height h. Describe a behaviour of the mass inside the box. We can assume that the box is very heavy and the spring is an ideal one.

So, I think that during the fall, there appears a weightlessness => There is no gravity on the mass, but there is still the force of the elasticity, so the spring goes as high as it can. Is that right?

After it hits the ground, there will be a gravity plus the inertia force acting, but I can't find out the maximum amplitude, could someone give me some hint please?

Juliet16 · 2009-06-25 13:33:32

After box falls on the ground, its velocity is v=sqrt(2gh). Velocity of a mass is the same, its energy is E=mv^2/2+mga. Spring's maximal energy: E=kA^2/2=m(v^2/2+ga) .

Martin · 2009-07-02 04:06:14

myvalq wrote:
A mass (weight 100g) is hanged on a spring inside a cubic box (side a=1m). Stiffness of the spring is 10N*m-1. Equilibrium state of the spring is right in a centre of the box.
Now, we let the box fall to the ground from height h. Describe a behaviour of the mass inside the box. We can assume that the box is very heavy and the spring is an ideal one.

So, I think that during the fall, there appears a weightlessness => There is no gravity on the mass,

"Weight" is the force exerted on an object by gravity, so no mass is "weightless" unless it is not in the presence of a gravitational field. Unfortunately, the term "weightless" often is misunderstood and misused.

Gravity is always acting on the hanging mass, both before and after the box is released. So, even during free-fall, the hanging mass—any mass—has "weight" (though its apparent weight—what you would measure if you were to weigh it while in its reference frame—would indeed be zero).

However, in this example, the falling hanging mass is not in free-fall, because it’s attached to a spring which exerts a (Hooke’s Law) force on it. Hence, while falling, its "apparent weight" is not zero, and it’s not "weightless" (except for the moments that the Hooke’s Law force of the spring cycles through zero).

but there is still the force of the elasticity, so the spring goes as high as it can. Is that right?

"As high as it can?" I don’t understand your question.

As soon as the box is released, the spring begins to pull down on the box and up on the mass. But as it pulls, its length decreases (the upper end moves down and the lower end moves up) which, in turn, reduces the amount of elastic force it exerts on both the box and the mass. Simultaneously, gravity begins to pull down on them both (the box and the mass).

After it hits the ground, there will be a gravity plus the inertia force acting, but I can't find out the maximum amplitude,

I don’t quite understand what you’re saying.

could someone give me some hint please?

The best way to approach this is to realize that you are dealing with a two-body problem: A mass m connected via an ideal spring to a box M, both of which are subjected to the spring’s elastic force as well as the earth’s gravitational force.

Initially, the spring hangs down a distance equal to its no-load length l, plus an additional amount d due to the weight of the mass m. Specifically (from Hooke's Law), d = mg/k, where k is the spring constant (stiffness), and g is the acceleration due to gravity.

Let the top wall of the box lie along the x-axis of a reference frame whose y-axis points down towards the ground. Assume that the top of the spring, which is connected to the middle of the top wall of the box, is initially located at the origin of the reference frame, and denote its ordinate (which will change once the box is released) by ${y_{M}(t)}$ . Denote the ordinate of the bottom of the spring (which is connected to the mass) by ${y_{m}(t)}$ . Then, assuming that both the box and the mass can be modeled as point masses, Newton’s Second Law yields the following equations of motion for the box and the mass for t>0:

$(1)\,\,\,\,\,\,M\frac{d^2 y_{M}(t)}{dt^2}=Mg+kz(t)$

and

$(2)\,\,\,\,\,\,m\frac{d^2 y_{m}(t)}{dt^2}=mg-kz(t)$

respectively, where

$(3)\,\,\,\,\,\,z(t)=y_{m}(t)-y_{M}(t)-l$

is the amount that the spring is stretched beyond its no-load length (i.e., the amount of the spring’s "extension"), and serves to couple together the two equations of motion (1) and (2).

Note the force terms on the right hand sides of both equations of motion:

• They contain a force due to gravity, as well as a Hooke’s Law force due to the extension of the spring beyond its no-load length.

• The magnitudes of the gravitational forces acting on the box and the mass differ, while the magnitudes of the Hooke’s Law forces are the same.

• The directions of the gravitational forces acting on the box and the mass are the same, while the directions of the Hooke’s Law forces are opposite.

This tells you that while gravity is pulling down on both the box and the mass, attempting to accelerate them both at g, the spring is adding to gravity’s attempt to accelerate the box by providing a downward acceleration component of kz/M, while opposing gravity’s attempt to accelerate the mass by providing an upward acceleration component of kz/m.

We could proceed with just these two equations of motion and determine the general behavior of the box and the mass. However, without adding too much more complexity, we can refine our model, and get better results.

While the assumption that the mass can be treated as a point mass is a good approximation, treating the box as a point mass would introduce a glaring error in the results. A more accurate description of the box consists of treating it as four point masses, each one-quarter the size of the total mass M, each located at the mid point of each of the box’s four walls. Doing so leads to replacing (1) with four separate equations:

$(1a)\,\,\,\,\,\,\,\frac{M}{4}\frac{d^2 y_{M}(t)}{dt^2}=\frac{M}{4}g+\frac{kz(t)}{4}$

$(1b)\,\,\,\,\,\,\,\frac{M}{4}\frac{d^2 y_{M1}(t)}{dt^2}=\frac{M}{4}g+\frac{kz(t)}{4}$

$(1c)\,\,\,\,\,\,\,\frac{M}{4}\frac{d^2 y_{M2}(t)}{dt^2}=\frac{M}{4}g+\frac{kz(t)}{4}$

$(1d)\,\,\,\,\,\,\frac{M}{4}\frac{d^2 y_{M3}(t)}{dt^2}=\frac{M}{4}g+\frac{kz(t)}{4}$

where $y_{M}(t)$ denotes (as before) the ordinate of the (now M/4) point mass at the top wall of the box, and $y_{Mi}(t)$ for i=1,2,3 denotes the ordinate of each of the other three M/4 point masses corresponding to each of the box’s other three walls (clockwise from the top wall).

Now if we add (1) and (2), we get

$(4)\,\,\,\,\,\,{(m+M)\frac{d^2 r(t)}{dt^2}}={(m+M)}g$

where

$(5)\,\,\,\,\,\,r(t)=\frac{my_{m}(t) + \frac{M}{4}[y_{M}(t)+y_{M1}(t)+y_{M2}(t)+y_{M3}(t)]}{m+M}$

which happens to be the Center of Mass of the combined box/hanging mass system.

Note that while (1) and (2) each contain both an external force (gravity’s "agent" is external to the box/hanging mass system) and an internal force (the "agents" of the Hooke’s Law force are the box and the mass—the internal constituents of the box/hanging mass system), (4) contains only an external force. The solution to (4) is

$(6)\,\,\,\,\,\,r(t)=\frac{1}{2}gt^2 + r(t=0)$

and describes the motion of the (Center of Mass of the) combined box/hanging mass as it falls to the ground under the influence of the external force of gravity. Since

$(7)\,\,\,\,\,\,y_{m}(t=0)=l+d$

and

$(8a)\,\,\,\,\,\,y_{M}(t=0)=0$

$(8b)\,\,\,\,\,\,y_{M1}(t=0)=\frac{w}{2}$

$(8c)\,\,\,\,\,\,y_{M2}(t=0)=w$

$(8d)\,\,\,y_{M3}(t=0)=\frac{w}{2}$

where w is the length of each of the box’s walls, from (5) we see that

$(9)\,\,\,\,\,\,r(t=0)=\frac{m(l+d)+M\frac{w}{2}}{m+M}$

so that (6) becomes

$(10)\,\,\,r(t)=\frac{1}{2}gt^2+\frac{m(l+d)+M\frac{w}{2}}{m+M}$ .

Had we not refined the analysis to account for the box’s distributed mass, the "glaring error" I mentioned would have manifested itself by a lack of the Mw/2 term in these expressions for the Center of Mass. (Note that w/2 is the initial Center of Mass of the box, and l+d is the initial Center of Mass of the hanging mass.)

Now, multiply (1) by m, (2) by M, and subtract the results:

$(11)\,\,\,\,mM\frac{d^2 y_{m}}{dt^2}-m\frac{M}{4}[\frac{d^2 y_{M}}{dt^2}+\frac{d^2 y_{M1}}{dt^2}+\frac{d^2 y_{M2}}{dt^2}+\frac{d^2 y_{M3}}{dt^2}]=-k(m+M)z.$

Because of symmetry and the constraint that the box remains intact (i.e., the box’s walls remain rigidly connected, because they constitute the box, and are not separate, independent masses),

$(12a)\,\,\,\,\,y_{M1}(t)=y_{M3}(t)=y_{M}(t)+\frac{w}{2}$

and

$(12b)\,\,\,\,\,y_{M2}(t)=y_{M}(t)+w$ .

Using (12) along with (3), (11) becomes

$(13)\,\,\,\,\,\mu\frac{d^2 z(t)}{dt^2}=-kz$

where

$(14)\,\,\,\,\,\mu=\frac{mM}{m+M}$

is referred to (in two-body analysis) as the reduced mass of the box/hanging mass system, and is the "effective" mass of the combined box/hanging mass when viewing its motion as if it were a single body.

Note that whereas (4) contains only the external gravitational force, (13) contains only the internal Hooke’s Law force. The solution to (13) represents the relative displacement of the box and the mass as they fall tethered together via the spring; it is given by

$(15)\,\,\,\,\,z(t)=A\cos(\omega t +\phi)$

where

$(16)\,\,\,\,\,\omega=\sqrt{\frac{k}{\mu}}$

and A and $\phi$ are determined by the initial conditions (i.e., the values of z and dz/dt at t = 0).

From (3), (7), and (8a), we find that z(0) = d; and since neither the box nor the mass has an initial velocity—both $y_{M}'(t=0)$ and $y_{m}'(t=0)$ are zero—we find that A = d and $\phi=0$ , so that

$(17)\,\,\,\,\,z(t)=d\cos(\omega t)$ .

The equation of motion of the hanging mass can be extracted from the equations of the motion of the Center of Mass of the combined box/hanging mass and of the spring’s extension: By using (5), (10), and (12) to obtain a relationship between ${y_{m}(t)}$ and ${y_{M}(t)}$ , and (3) and (17) to obtain another such relationship, the equation of motion of the hanging mass can be shown to be

$(18)\,\,\,\,{y_{m}(t)}=\frac{1}{2}gt^2+l+\frac{md+Md\cos(\omega t)}{m+M}$ .

Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

#1 2009-05-19 14:02:11

Spring falling

#2 2009-06-25 13:33:32

Re: Spring falling

#3 2009-07-02 04:06:14

Re: Spring falling

myvalq wrote:

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