April 6, 2021 – Physicspupil

When a pan of mass $m$ is put on top of a vertical spring scale of negligible weight, the downward displacement of the spring is observed to be $d$ . A ball of mass $M$ is now dropped from a height $h$ above the pan onto the pan. Suppose that the coefficient of restitution is equal to zero.

i. Show that the velocity of the pan immediately after the impact is

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ .

ii. From energy considerations, determine the maximum deflection of the pan after the ball has been dropped onto it.

Note: Force in spring scale is proportional to displacement.

Solution. (bad, if not wrong)

i.

Background.

1. If the coefficient of restitution is zero, the collision is perfectly inelastic.

2. Recall Hooke’s law:

$\begin{aligned} \mathbf{F_s} & =-k\mathbf{x} \\ F & = kx \\ U(x) & =\displaystyle{\frac{1}{2}}kx^2 \\ \end{aligned}$

Let $t$ be the time it takes for the ball to free-fall a vertical distance $h$ .

$\begin{aligned} s & =ut+\frac{1}{2}at^2 \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ h & =(0)t+\frac{1}{2}(g)t^2 \\ t & =\sqrt{\frac{2h}{g}} \end{aligned}$

The velocity of the ball right before the impact is given by

$\begin{aligned} v & =u+at \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ & = (0) + (g)\cdot\bigg(\sqrt{\frac{2h}{g}}\bigg) \\ & = \sqrt{2gh} \end{aligned}$

By conservation of linear momentum,

$\begin{aligned} M\cdot \sqrt{2gh} & = (M+m)v \\ v & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

is the velocity of the ball and the pan immediately after collision.

ii.

Setup.

From Hooke’s law, the spring force is given by $\mathbf{F}_s=-k\mathbf{x}$ , or $mg=kd$ . The spring constant, characteristic of the spring, is $k=mg/d$ . Let $x$ be the magnitude of downward displacement of the spring from its relaxed position in reaching the final state of equilibrium. Then,
$\begin{aligned} F & = kx \\ (M+m)g & = \bigg( \frac{mg}{d} \bigg) x \\ x & = \frac{(M+m)d}{m} \end{aligned}$

By the law of conservation of energy,

$\begin{aligned} \Delta\textrm{Mechanical Energy} & = 0 \\ \Delta\textrm{KE} + \Delta\textrm{PE} & = 0 \\ \Big( \textrm{KE}_{\textrm{final}} - \textrm{KE}_{\textrm{initial}}\Big) + \Big( \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}}\Big) & = 0 \end{aligned}$

$\begin{aligned} \textrm{KE}_{\textrm{initial}} & = \frac{1}{2}(M+m)\bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2 \\ & = \frac{M^2gh}{M+m} \\ \textrm{KE}_{\textrm{final}} & = 0 \\ \textrm{PE}_{\textrm{initial}} & = \frac{1}{2}kd^2\\ \textrm{PE}_{\textrm{final}} & = \frac{1}{2}k(x+d)^2 - (M+m)gx \end{aligned}$

$\begin{aligned} & \textrm{KE}_{\textrm{final}}-\textrm{KE}_{\textrm{initial}} \\ & = 0 - \frac{M^2gh}{M+m} \\ & = - \frac{M^2gh}{M+m} \\ & \\ & \textrm{PE}_{\textrm{final}}-\textrm{PE}_{\textrm{initial}} \\ = & \frac{1}{2}k(x+d)^2-(M+m)gx - \frac{1}{2}kd^2 \\ = & \frac{1}{2}kx^2+kxd - (M+m)gx \\ = & \frac{1}{2}\bigg( \frac{mg}{d} \bigg) x^2 + \bigg( \frac{mg}{d} \bigg) xd - (M+m)gx \\ = & \bigg( \frac{mg}{2d}\bigg) x^2 - Mgx \end{aligned}$

Thus,

$\begin{aligned} 0 & = -\frac{M^2gh}{M+m} + \frac{mgx^2}{2d} - Mgx \\ 0 & = \bigg( \frac{m}{2d} \bigg) x^2 + (-M)x + \bigg( -\frac{M^2h}{M+m} \bigg) \\ x & = \frac{-(-M)\pm\sqrt{(-M)^2-4(\frac{m}{2d})(-\frac{M^2h}{M+m})}}{2(\frac{m}{2d})} \\ x & = \frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}} \end{aligned}$

$\therefore$ The maximum deflection of the pan after the ball has been dropped onto it is

$\boxed{x = \displaystyle{\frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}}}}$ .

Done. Let’s overdo.

ii. (energy approach)

One can derive the equation of motion by considering energy conservation:

$\begin{aligned} \textrm{KE} & =\frac{1}{2}(M+m)\dot{x}^2 \\ \textrm{PE} & = \frac{1}{2}kx^2-(M+m)gx \\ 0 & = \frac{\mathrm{d}(\textrm{KE}+\textrm{PE})}{\mathrm{d}t} \\ 0 & = (M+m)\dot{x}\ddot{x} + kx\dot{x} - (m+M)g\dot{x} \\ 0 & = (M+m)\ddot{x} + kx - (m+M)g \\ \ddot{x} & = -\frac{k}{(M+m)}x + g \\ \end{aligned}$

or, by considering the Euler-Lagrange equation

where the Lagrangian $\mathcal{L}$ is defined

and then computing,

the Euler-Lagrange equation now reads

which agrees with the equation of motion derived previously.

ii. (force approach)

One can also derive the equation of motion by considering Newton’s second law:

$\begin{aligned} \dots\enspace \textrm{recall }& \textrm{Newton's second law }\dots \\ \textrm{Net }\mathbf{F} & = m\mathbf{a} \\ \dots\enspace \textrm{downward } & \textrm{taken to be positive \dots} \\ (M+m)\mathbf{g} - k\mathbf{x} & = (M+m)\ddot{\mathbf{x}} \\ \dots\enspace \textrm{consider } & \textrm{magnitude only }\dots \\ \ddot{x} & = -\frac{k}{M+m}x + g\\ \ddot{x} & = -\bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x+g \\ \ddot{x} & = -\omega^2x+g\qquad \textrm{where }\omega =\sqrt{\frac{k}{M+m}} \\ \dots \textrm{ go back } & \textrm{to an earlier line }\dots \\ (M+m)g - kx & = (M+m)\ddot{x} \\ \dots \textrm{let }x_{\textrm{eff}}\textrm{ be} & \textrm{ the effective displacement} \\ x_{\textrm{eff}} & = x-\frac{(M+m)g}{k} \\ \textrm{then\qquad} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \end{aligned}$

(to be continued)

Remark.

The effective displacement $x_{\textrm{eff}}$ is defined by

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)g}{k}}$ .

Substituting the spring constant $k$ with

$k=\displaystyle{\frac{mg}{d}}$ ,

$x_{\textrm{eff}}$ can be rewritten as

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)d}{m}}$ .

As seen in ii. Setup,

$\displaystyle{\frac{(M+m)d}{m}}$

is the equilibrium position.

$\therefore$ The effective displacement $x_{\textrm{eff}}$ is thus the distance extended or compressed with reference to the equilibrium position.

(continue)

$\begin{aligned} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & = - \bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & =-\omega^2x_{\textrm{eff}} \qquad \textrm{where }\omega = \sqrt{\frac{k}{M+m}} \\ \dots\textrm{ Solving it, }& \dots\\ x_{\textrm{eff}}(t) & = c_1\cos (\omega t) + c_2\sin (\omega t) \end{aligned}$

The constant coefficients (i.e., $c_1$ , $c_2$ ) should be obtained from the initial and boundary conditions.

Let $t=0$ be the instant the ball hits on the pan.

When $t=0$ ,

$\begin{aligned} x_{\textrm{eff}}(0) & = d - \frac{(M+m)d}{m} \\ c_1(1) + c_2(0) & = \frac{md-(M+m)d}{m} \\ c_1 & = -\bigg( \frac{M}{m} \bigg) d \\ \end{aligned}$

As the velocity of the pan immediately after the impact
was found in part (i) to be

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ ,

$\begin{aligned} \dot{x}_{\textrm{eff}}(t) & = -c_1\sin (\omega t)+c_2\cos (\omega t) \\ \dot{x}_{\textrm{eff}}(0) & = -c_1(0) + c_2(1) = \frac{M\sqrt{2gh}}{M+m} \\ c_2 & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

$\begin{aligned} \therefore x_{\textrm{eff}}(t) & = -\bigg( \frac{Md}{m} \bigg)\cos \Bigg( \bigg(\sqrt{\frac{k}{M+m}}\bigg) (t) \Bigg)\\ & \qquad + \bigg( \frac{M\sqrt{2gh}}{M+m}\bigg) \sin\Bigg( \bigg( \sqrt{\frac{k}{M+m}}\bigg) ( t)\Bigg) \\ \end{aligned}$

If there is no friction or none any other dissipation of energy, the spring will continue indefinitely and uninhibitedly its simple harmonic motion.

(to be continued)

Aside.

The angular frequency $\omega$ ( $=2\pi f$ ) is

$\omega =\displaystyle{\sqrt{\frac{k}{M+m}}}$ .

Hence the frequency $f$ is

$f=\displaystyle{\frac{\omega}{2\pi}}=\displaystyle{\frac{1}{2\pi}\sqrt{\frac{k}{M+m}}}$ ,

and the period $T$ is

$T=\displaystyle{\frac{1}{f}=2\pi\sqrt{\frac{M+m}{k}}}$ .

At time $t=\displaystyle{\frac{T}{2}}$ , the spring is at the equilibrium position (i.e., $x_{\textrm{eff}}(\frac{T}{2})$ ), whereas at time(s) $t=\displaystyle{\frac{T}{4},\, \frac{3T}{4}}$ , it reaches the extreme points (i.e., $x_{\textrm{eff}}(\frac{T}{4}),\, x_{\textrm{eff}}(\frac{3T}{4})$ ).

(continue)

Lemma.

The sinusoidal function

$x(t)=c_1\cos (\omega t)+c_2\sin (\omega t)$

can be written in the form

$x(t)=A\cos (\omega t-\varphi )$

where $A=\sqrt{c_1^2+c_2^2}$ and $\tan\varphi =\displaystyle{\frac{c_2}{c_1}}$ .

The amplitude (i.e., $A=\sqrt{c_1^2+c_2^2}$ above) is the maximum displacement of the spring from its equilibrium position.

$\begin{aligned} A & = \sqrt{c_1^2+c_2^2} \\ & = \sqrt{\bigg( -\frac{Md}{m} \bigg)^2 + \bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2} \\ & = \sqrt{\frac{M^2d^2}{m^2} + \frac{2M^2gh}{(M+m)^2}} \\ \end{aligned}$

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Day: April 6, 2021

202104060103 Solution to 1973-AL-AMATH-I-2