Continuum Mechanics * – Page 4

September 28, 2022September 29, 2022

202209281002 Exercise 14.2 (Q24)

Analysis of waves in shallow water (depth much less than wavelength) yields the following wave equation:

$\displaystyle{\frac{\partial^2y}{\partial x^2}=\frac{1}{gh}\frac{\partial^2y}{\partial t^2}}$

where $h$ is the water depth and $g$ the gravitational acceleration. Give an expression for the wave speed.

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

(top-down approach)

Write a traveling sinusoidal wave in the form

$y(x,t)=A\cos (kx\pm \omega t)$

the wave speed $v$ being such as

$\displaystyle{v=\frac{\lambda}{T}=\frac{2\pi /k}{2\pi /\omega}=\frac{\omega}{k}}$ .

$\begin{aligned} \textrm{LHS} & = \frac{\partial^2y}{\partial x^2} \\ & = \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}\big( A\cos (kx\pm\omega t)\big) \bigg) \\ & = \frac{\partial}{\partial x}\big( kA\sin (kx\pm\omega t)\big) \\ & = -k^2A\cos (kx\pm\omega t) \\ \textrm{RHS} & = \frac{1}{gh}\frac{\partial^2y}{\partial t^2} \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\bigg(\frac{\partial}{\partial t}\big( A\cos (kx\pm\omega t)\big)\bigg) \\ & = \frac{1}{gh}\frac{\partial}{\partial t}\big(\omega A\sin (kx\pm\omega t)\big) \\ & = \frac{1}{gh}\big( -\omega^2A\cos (kx\pm\omega t)\big) \\ \textrm{LHS} & = \textrm{RHS} \\ \Rightarrow v = \frac{\omega}{k} & = \sqrt{gh} \\ \end{aligned}$

$\therefore$ The wave speed $v$ is given by $v=\sqrt{gh}$ .

September 27, 2022October 24, 2022

202209271040 Exercise 8.4 (Q53)

Neglecting the Earth’s rotation, show that the energy needed to launch a satellite of mass $m$ into circular orbit at altitude $h$ is

$\displaystyle{\bigg( \frac{GM_\textrm{E}m}{R_{\textrm{E}}}\bigg)\bigg(\frac{R_{\textrm{E}}+2h}{2(R_{\textrm{E}}+h)}\bigg)}$ .

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Abortive attempt.

(energy/work-done approach)

By conservation of mechanical energy,

$\begin{aligned} \Delta (\textrm{KE}+\textrm{PE}) & = 0 \\ (\textrm{KE}_f-\textrm{KE}_i) + (\textrm{PE}_f-\textrm{PE}_i) & = 0 \\ \bigg(\frac{1}{2}mv^2 - \frac{1}{2}mu^2\bigg) + \big(mg_f(R_\textrm{E}+h)-mg_iR_\textrm{E}\big) & = 0 \\ \end{aligned}$

$\begin{aligned} g_f & = G\frac{M_\textrm{E}}{(R_\textrm{E}+h)^2} \\ g_i & = G\frac{M_\textrm{E}}{(R_\textrm{E})^2} \\ \end{aligned}$

On the one hand, the gravitational pull provides the centripetal force for revolving at the orbital speed $v$ :

$\begin{aligned} \text{}_MF_m & = \text{}_mF_M \\ \frac{mv^2}{R_{\textrm{E}}+h} & = G\frac{mM_\textrm{E}}{(R_{\textrm{E}}+h)^2} \\ v & = \sqrt{\frac{GM_\textrm{E}}{R_{\textrm{E}}+h}} \\ \end{aligned}$

On the other hand, the escape speed $u$ of the satellite is

$\begin{aligned} \frac{1}{2}mu^2 & = G\frac{mM_\textrm{E}}{R_\textrm{E}} \\ u & = \sqrt{\frac{2GM_\textrm{E}}{R_\textrm{E}}} \\ \end{aligned}$

but what is this question asking for?

$\Delta\textrm{KE}=\displaystyle{\frac{m(v^2-u^2)}{2}}$

(to be continued)

September 27, 2022October 24, 2022

202209271004 Exercise 4.5 (Q31)

An elevator accelerates downward at $2.4\,\mathrm{m\, s^{-2}}$ . What force does the elevator’s floor exert on a $52\,\mathrm{kg}$ passenger?

_{Extracted from R. Wolfson. (2016). Essential University Physics.}

Roughwork.

Take downward positive, $\downarrow\textrm{+ve}$ . By Newton’s 2^nd Law,

$\mathbf{F}_{\textrm{net}}=m\mathbf{a}$ .

Given $m=52\,\mathrm{kg}$ . Let $F$ be the unknown magnitude of force:

$\begin{aligned} \mathbf{a} & = 2.4\,\hat{\mathbf{k}} \\ \mathbf{F}_\textrm{net} & = -F\,\hat{\mathbf{k}}+mg\,\hat{\mathbf{k}} \\ & = (-F+mg)\,\hat{\mathbf{k}} \\ \end{aligned}$

Then,

$\begin{aligned} -F+mg & = 2.4m \\ F & = (g-2.4)m \\ & = (9.81-2.4)(52) \\ & = 385\,\mathrm{N}\qquad \textrm{(3 s.f.)} \end{aligned}$

$\therefore$ The force the elevator’s floor exerts on the passenger is $385\,\mathrm{N}$ upward.

September 27, 2022October 24, 2022

202209270930 Exercise 5.3.B (Q1-Q6)

1. What is the potential energy of a $10\,\mathrm{kg}$ mass $25$ metres above the ground?
2. A $3\,\mathrm{kg}$ mass is $20\,\mathrm{m}$ above the ground. How much potential energy does it have?
3. How high must you raise a $25\,\mathrm{kg}$ mass before it has a potential energy of $8500\,\mathrm{J}$ ?
4. How high must you raise a $2\,\mathrm{kg}$ mass before it has a potential energy of $10\,\mathrm{J}$ ?
5. A mass has a potential energy of $1000\,\mathrm{J}$ when it is $40\,\mathrm{m}$ above ground level. What is the mass?
6. A mass has a potential energy of $20\,\mathrm{J}$ when it is $50\,\mathrm{cm}$ above ground level. What is the mass?

_{Extracted from B. Kennedy. (1999). Progressive Problems for ‘S’ Grade Physics.}

Background.

$\textrm{PE}=mgh$

Roughwork.

1.

$\begin{aligned} \textrm{PE} & = mgh \\ & = (10)(9.81)(25) \\ & = 2450\,\mathrm{J}\qquad \textrm{(3 s.f.)} \\ \end{aligned}$

2.

$\begin{aligned} \textrm{PE} & = mgh \\ & = (3)(9.81)(20) \\ & = 589\,\mathrm{J}\qquad \textrm{(3 s.f.)} \end{aligned}$

3.

$\begin{aligned} \textrm{PE} & = mgh \\ 8500 & = (25)(9.81)(h) \\ h & = 34.7\,\mathrm{m}\qquad \textrm{(3 s.f.)} \end{aligned}$

4.

$\begin{aligned} \textrm{PE} & = mgh \\ 10 & = (2)(9.81)(h) \\ h & = 51.0\,\mathrm{cm}\qquad \textrm{(3 s.f.)} \end{aligned}$

5.

$\begin{aligned} \textrm{PE} & = mgh \\ 1000 & = (m)(9.81)(40) \\ m & = 2.55\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}$

6.

$\begin{aligned} \textrm{PE} & = mgh \\ 20 & = (m)(9.81)(0.5) \\ m & = 4.08\,\mathrm{kg}\qquad \textrm{(3 s.f.)} \end{aligned}$

July 8, 2022October 24, 2022

202207081631 Statics Figures (Elementary) Q1

In a lift, the panel screen displays signs on an array of seven bars of light-emitting diodes (LEDs):

such that the numeric digits $0$ , $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ and $9$ when lit up are as follow:

For each configuration of lit-up cells, assuming that the mass distribution is uniform and the plate thickness negligible, find the centre of mass (CM), and hence the centre of gravity (CG).

Intuition.

Try arguing $\textrm{\scriptsize{NOT}}$ by symmetry in mathematics, $\textrm{\scriptsize{BUT}}$ by moment in physics.

We state without proof the centre of mass (CM) in each cell is as shown below:

For $N$ particles in 2-D:

$\begin{aligned} x_{\textrm{CM}} & = \frac{m_1x_1+m_2x_2+\cdots +m_Nx_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_ix_i}{M} \\ y_{\textrm{CM}} & = \frac{m_1y_1+m_2y_2+\cdots +m_Ny_N}{m_1+m_2+\cdots +m_N} = \frac{\sum_{i=1}^{N}m_iy_i}{M} \\ \end{aligned}$

Define the center at the origin $O(0,0)$ , and let $\mathbf{r}_1$ , $\mathbf{r}_2$ , $\mathbf{r}_3$ , $\mathbf{r}_4$ , $\mathbf{r}_5$ , and $\mathbf{r}_6$ be the vectors pointing to the CM of each cell:

such that

$\begin{aligned} \mathbf{r}_1 & = 2\,\hat{\mathbf{j}} \\ \mathbf{r}_2 & = -1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_3 & = 1\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \mathbf{r}_4 & = -1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \mathbf{r}_5 & = -2\,\hat{\mathbf{j}} \\ \mathbf{r}_6 & = 1\,\hat{\mathbf{i}} -1\,\hat{\mathbf{j}} \\ \end{aligned}$

To illustrate how to get the CM for the number signs, we first begin with $1$ :

$\mathbf{1}_{\textrm{CM}} = (x_{\textrm{CM,1}},y_{\textrm{CM,1}})$ .

$\begin{aligned} \mathbf{1}_{\textrm{CM}} & = \frac{m\mathbf{r}_3+m\mathbf{r}_6}{m+m} \\ & = \frac{1}{2}\mathbf{r_3} + \frac{1}{2}\mathbf{r_6} \\ & = \frac{1}{2}(1,1) + \frac{1}{2}(1,-1) \\ & = (1,0) \\ \end{aligned}$

The remaining are left the reader as an exercise.

May 31, 2022October 24, 2022

202205311107 Problem 4.23

Boxes $A$ and $B$ are in contact on a horizontal, frictionless (i.e. $f=0$ ) surface, as shown in the Figure below. Box $A$ has mass $20.0\,\mathrm{kg}$ and box $B$ has mass $5.0\,\mathrm{kg}$ . A horizontal force of $100\,\mathrm{N}$ is exerted on box $A$ . What is the magnitude of the force that box $A$ exerts on box $B$ ?

_{extracted from Problem 4.23, Sears and Zemansky’s University Physics}

Steps.

Draw the free-body diagrams of $A$ , $B$ , and $A+B$ .

Apply Newton’s $2^\textrm{nd}$ law $\textrm{Net }F=ma$ :

$\begin{aligned} F_A- {}_{B}F_A - f_A & = m_Aa_A \\ {}_{A}F_B - f_B & = m_Ba_B \\ F_A-f_{A+B} & = m_{A+B}a_{A+B} \\ \end{aligned}$

Conditioning the equations of motion:

$\begin{aligned} a_A=a_B & =a_{A+B} \\ f_A=f_B=f_{A+B} & =0 \\ {}_{B}F_A & ={}_{A}F_{B} \\ m_A+m_B & =m_{A+B} \\ \end{aligned}$

and substituting numbers for symbols, write:

$\begin{aligned} 100 - {}_{A}F_{B} & = 20a \\ {}_{A}F_{B} & = 5a \\ 100 & =25a \\ \end{aligned}$

$\therefore$ The magnitude ${}_{A}F_{B}$ of the force that box $A$ exerts on box $B$ is $20\,\mathrm{N}$ .

February 8, 2022October 24, 2022

202202081217 Dynamics Figures (Elementary) Q6

13. Two identical $2\,\mathrm{kg}$ trolleys are connected by a light string and pulled by a $12\,\mathrm{N}$ force as shown. Assuming the surface to be frictionless,

(a) calculate the acceleration of the two trolleys;

(b) find the tension $T$ in the string.

_{Extracted from B. Kennedy. (2001). Higher Physics Progressive Problems.}

Solution.

We draw three free-body diagrams of i. $m_a$ ; ii. $m_b$ ; and iii. $m_a+m_b$ :

In respective diagrams there are three equations of motion.

By Newton’s second law,

$\textrm{Net }\mathbf{F}=m\mathbf{a}$ :

we have

$\begin{aligned} T & = m_aa \\ 12-T & = m_ba \\ 12 & = (m_a+m_b)a \\ \end{aligned}$

Answers. (a) acceleration $a=3\,\mathrm{m\, s^{-2}}$ ; (b) tension $T=6\,\mathrm{N}$ .

February 8, 2022January 19, 2023

202202080958 Dynamics Figures (Elementary) Q5

2. A man walking with a speed $v$ constant in magnitude and direction passes under a lantern hanging at a height $H$ above the ground. Find the velocity which the edge of the shadow of the man’s head moves over the ground with if his height is $h$ .

_{Extracted from B. Bukhovtsev et al. (1978). Problems in Elementary Physics.}

Solution.

Let $x=0$ be the position of the lantern; let the man walk in the positive $x$ -direction; and let the position of the man be $x_m(t)$ and that of the shadow of his head $x_s(t)$ . So the length $s$ of his shadow is $|x_s-x_m|$ .

By comparing similar triangles, we have

$\displaystyle{\frac{H}{x_s} = \frac{h}{s}}$ .

Thus,

$\begin{aligned} \frac{H}{x_s} & = \frac{h}{x_s-x_m} \\ x_s & = \bigg(\frac{H}{H-h}\bigg) x_m \\ \dot{x}_s & = \bigg(\frac{H}{H-h}\bigg) \dot{x}_m \\ \dot{x}_s & = \bigg(\frac{H}{H-h}\bigg) v \\ \end{aligned}$

$\therefore$ The edge of the shadow of the man’s head moves with a velocity $(\frac{H}{H-h}) v\,\hat{\mathbf{i}}$ over the ground.

February 7, 2022October 24, 2022

202202071621 Dynamics Figures (Elementary) Q4

1.3.10. A $2\,\mathrm{kg}$ mass and a $3\,\mathrm{kg}$ mass are linked by a light string passed over a frictionless pulley. Calculate the acceleration of the system.

_{Extracted from B. Kennedy. (2001). Higher Physics Progressive Problems.}

Solution.

At first glance, as mass $m_2$ is heavier than mass $m_1$ , one can expect that the heavier mass will fall down and the lighter mass will move up.

Draw the free body diagram of each mass below:

Write the equation of motion for both masses:

$\begin{aligned} T-m_1g & = m_1a \\ m_2g-T & = m_2a \\ & \\ T-2g & = 2a \\ 3g-T & = 3a \\ & \\ a & = \frac{g}{5} \\ T & = \frac{12g}{5} \\ \end{aligned}$

$\therefore$ Both masses $m_1$ and $m_2$ will accelerate with a magnitude of $g/5=1.96\,\mathrm{m\, s^{-2}}$ .

February 7, 2022October 24, 2022

202202071437 Dynamics Figures (Elementary) Q3

1.17. A stationary object explodes into two fragments of relative mass $1:100$ . At the instant of break-up, the larger mass has a velocity of $10\,\mathrm{m\, s^{-1}}$ . Calculate i. the velocity of the smaller mass, ii. the ratio of their kinetic energies at this instant.

_{Extracted from M. Nelkon. (1971). Graded Exercises and Worked Examples in Physics.}

Solution.

Provided that $u=0$ , $m_1:m_2=1:100$ , and $v_2=10\,\mathrm{m\, s^{-1}}$ .

By the law of conservation of linear momentum,

$\begin{aligned} M\mathbf{u} & = m_1\mathbf{v}_1+m_2\mathbf{v}_2 \\ (M)(0\,\hat{\mathbf{i}}) & = \bigg(\frac{1}{101}M\bigg) (-v_1\,\hat{\mathbf{i}})+\bigg(\frac{100}{101}M\bigg) (+10\,\hat{\mathbf{i}}) \\ v_1 & = 1000 \\ \mathbf{v}_1 & = -1000\,\hat{\mathbf{i}}\\ \end{aligned}$

The ratio of their kinetic energies is given by

$\begin{aligned} &\quad \textrm{KE}_1:\textrm{KE}_2 \\ & = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} \\ & = \frac{(\frac{1}{2})(\frac{1}{101}M)(1000)^2}{(\frac{1}{2})(\frac{100}{101}M)(10)^2} \\ & = 100:1 \\ \end{aligned}$