Apply independently

Given

load $m = 2.0\,\text{t} = 2000\,\text{kg}$, jib $m_\text{p} = 4.0\,\text{t}$ and $r_\text{p} = 12\,\text{m}$, counterweight $m_\text{v} = 9.0\,\text{t}$ and $r_\text{v} = 10\,\text{m}$

Find

the distance $r$ of the trolley from the axis of the tower when the total moment of the weights is zero

Principle

Three weights act on the jib: the weight of the load at the trolley (through the cable), the weight of the jib itself at its centre of mass, and the weight of the counterweight. The axis of rotation is chosen to be the axis of the tower, about which the jib tends to tip. The lever arm of each weight is the perpendicular distance of its line of action from the axis, that is, the horizontal distance. The load and the weight of the jib turn the jib towards the side of the load, the counterweight in the opposite direction. In equilibrium with respect to rotation the sum of the moments is zero.

Formula

$m_\text{p} g r_\text{p} + m g r - m_\text{v} g r_\text{v} = 0$. The acceleration of free fall cancels, so

Conversions

$1\,\text{t} = 1000\,\text{kg}$, so $m_\text{p} = 4000\,\text{kg}$ and $m_\text{v} = 9000\,\text{kg}$.

Substitution

Answer

$r \approx 21\,\text{m}$

Two significant figures like the distances and masses.

Reasonableness

The moment of the counterweight is about twice the moment of the jib, so the load has to make up the difference. A light load has to be far out, and the point is within the $40\,\text{m}$ length of the jib.

Given

the data of part a, trolley at the tip of the jib $r_\text{max} = 40\,\text{m}$, permitted total moment $M_\text{max} = 600\,\text{kNm}$ and $g = 9.81\,\text{m/s}^2$

Find

the greatest load $m$ at the tip of the jib

Principle

When the load is at the tip, the total moment points towards the side of the load, and with the greatest permitted load it is equal to the limit. The moments are calculated about the axis of the tower as in part a, with the positive direction of rotation towards the side of the load.

Formula

so

Conversions

Substitution

The moment of the load may therefore be

so

Answer

$m \approx 2.6\,\text{t}$

Two significant figures like the data of the crane.

Reasonableness

Without a load the total moment is $882.9\,\text{kNm} - 470.9\,\text{kNm} = 412\,\text{kNm}$ towards the counterweight, so the limit is not exceeded even by the empty crane. The load at the tip is smaller than the load the crane carries closer to the tower, as in the load charts of cranes.

Find

whether the counterweight can be increased by a couple of tonnes so that heavier loads can be lifted from the tip

Principle

The moment of the counterweight turns the jib in the opposite direction to the load. The limit applies to the total moment in both directions, so both the fully loaded and the empty crane must be checked.

Model answer

The loaded crane. A larger counterweight reduces the total moment on the side of the load, so a heavier load could be lifted at the tip. The foreman is right about this.

The empty crane. Without a load the total moment points towards the counterweight: now it is $412\,\text{kNm}$. An addition of two tonnes increases it by $2000\,\text{kg} \cdot 9.81\,\text{m/s}^2 \cdot 10\,\text{m} = 196.2\,\text{kNm}$, so the total moment becomes $608\,\text{kNm}$. This exceeds the limit of $600\,\text{kNm}$, and the tower is overloaded on the side of the counterweight whenever there is no load on the hook.

How much could be added. For the empty crane there is room of $600\,\text{kNm} - 412\,\text{kNm} = 188\,\text{kNm}$ below the limit, so the counterweight can be increased by at most $188\,000\,\text{Nm} : (9.81\,\text{m/s}^2 \cdot 10\,\text{m}) \approx 1.9\,\text{t}$. Then the greatest load at the tip would be $1200\,\text{kNm} : 392.4\,\text{m}^2/\text{s}^2 \approx 3.1\,\text{t}$.

Conclusion. The suggestion does not work as it stands. The counterweight is dimensioned so that the limit is not exceeded with the empty or the fully loaded crane. Instead of two tonnes, the counterweight could be increased by at most about $1.9\,\text{t}$, and the lifting capacity cannot be increased further without a sturdier tower.

Find

the free-body diagram of the load while the jib turns steadily, and an evaluation of the operator's explanation

Principle

The load is chosen as the object. The bodies interacting with it are the Earth (weight) and the cable (tension); nothing else touches the load, and air resistance is small at low speed. When the jib turns steadily, the load moves in a horizontal circle about the axis of the tower, so it has a centripetal acceleration towards the axis. The net force must then point towards the axis.

Model answer

The load is drawn as a rectangle tilted along the cable.
- The weight $\overline{G}$ starts from the centre and points straight down.
- The tension $\overline{S}$ of the cable starts from the upper surface where the cable is attached and points along the cable upwards and slightly towards the axis of the tower, because the load is farther out than the trolley.

In the vertical direction the load does not accelerate, so the vertical component of the tension is equal to the weight, and the tension is slightly larger than the weight. The horizontal component of the tension is the net force, and it points towards the axis of the tower like the centripetal acceleration $\overline{a}_\text{n}$, which is drawn separately in its own colour. In the model picture the angle of the cable has been drawn much larger than in the problem so that the arrows can be told apart.

The operator's explanation. There is no centrifugal force: no body pulls the load outwards, and there is no outward force in the free-body diagram. When the jib starts to turn, the load tends to continue straight ahead because of its inertia. That is why it stays on the outer edge of the circle and the cable tilts, until the horizontal component of the tension is enough to turn the motion of the load onto a circular path. So the load is not thrown outwards but lags behind its straight path only by as much as the cable needs to pull it towards the centre.

Given

the radius $r = 21\,\text{m}$ of part a, period $T = 90\,\text{s}$, cable length $15\,\text{m}$ and $g = 9.81\,\text{m/s}^2$

Find

the angle $\theta$ between the cable and the vertical, and whether it matters

Principle

The load moves in uniform circular motion, so its acceleration is the centripetal acceleration $a_\text{n} = \tfrac{v^2}{r}$ towards the axis of the tower. The speed is obtained from the period. Newton's second law is written for the free-body diagram of part d by components: axis $x$ horizontally towards the axis of the tower and $y$ upwards. The tilt moves the load only by centimetres, so the distance of the trolley, $21\,\text{m}$, can be used as the radius of the circle.

Formula

$x$: $S\sin\theta = m\dfrac{v^2}{r}$ and $y$: $S\cos\theta - mg = 0$. Dividing the equations by each other gives $\tan\theta = \dfrac{v^2}{rg}$, where $v = \dfrac{2\pi r}{T}$. The mass cancels.

Conversions

Not needed.

Substitution

so $\theta = 0.598^\circ$

Answer

$\theta \approx 0.60^\circ$

Two significant figures like the period $90\,\text{s}$. The load has moved outwards from the trolley by $15\,\text{m} \cdot 0.01043 \approx 0.16\,\text{m}$.

Reasonableness

The centripetal acceleration

is only about one percent of the acceleration of free fall, so the angle is small and the tension is practically equal to the weight. Steady turning therefore hardly matters in lifting work. A sudden start or stop, on the other hand, makes the load swing, because the cable cannot immediately give the load the same acceleration as the trolley.

Find

a graph whose points, according to the circular-orbit model, fit a straight line through the origin

Graph

In the model, the moon is kept in a circular orbit by the gravity of Jupiter, which is the only significant force acting on the moon. It is the centripetal force of the circular motion of the moon: $\gamma\dfrac{Mm}{r^2} = m\dfrac{v^2}{r}$, so $v^2 = \gamma M \cdot \dfrac{1}{r}$. The orbital speed is calculated from the period: $v = \dfrac{2\pi r}{T}$, with the radius in metres and the period in seconds ($1\,\text{d} = 86\,400\,\text{s}$). For example, for Io $v = \dfrac{2\pi \cdot 421.8 \cdot 10^{6}\,\text{m}}{1.769 \cdot 86\,400\,\text{s}} = 17\,340\,\text{m/s}$, so $v^2 = 3.01 \cdot 10^{8}\,\text{m}^2/\text{s}^2$ and $\tfrac{1}{r} = 2.37 \cdot 10^{-9}\,\tfrac{1}{\text{m}}$.

$\tfrac{1}{r}$ is chosen for the horizontal axis and $v^2$ for the vertical axis.

Model answer

All five points are on the same straight line through the origin, so the circular-orbit model, in which gravity is the centripetal force, describes the motion of the moons well. The line passes through the origin because according to the model the orbital speed would be zero infinitely far away ($\tfrac{1}{r} = 0$). According to the model, the slope of the line is $\gamma M$, which is the same for all the moons, although the masses of the moons are very different.

Given

the graph of part a and $\gamma = 6.674 \cdot 10^{-11}\,\tfrac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$

Find

the mass $M$ of Jupiter

Principle

According to the model $v^2 = \gamma M \cdot \tfrac{1}{r}$, so the slope $k$ of the line is $\gamma M$. The slope is read from a point of the line far from the origin, so that the relative error of the reading is small.

Reading the graph

The line passes through the origin and the point $(5.0 \cdot 10^{-9}\,\tfrac{1}{\text{m}}$; $6.35 \cdot 10^{8}\,\text{m}^2/\text{s}^2)$, so

Formula

$k = \gamma M$, so $M = \dfrac{k}{\gamma}$

Conversions

because $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.

Substitution

Answer

$M \approx 1.9 \cdot 10^{27}\,\text{kg}$

Two significant figures, because the slope is read from the graph to an accuracy of about two or three figures.

Reasonableness

The value in the data book is $1.898 \cdot 10^{27}\,\text{kg}$, about $320$ Earth masses, so the result is correct. Jupiter is the most massive planet in the Solar System, and its mass is about a thousandth of the mass of the Sun.

Given

the rotation period of Jupiter $T = 9.93\,\text{h}$ and $\gamma M = 1.27 \cdot 10^{17}\,\text{m}^3/\text{s}^2$ from part b

Find

the radius $r$ of the orbit of the probe

Principle

The probe stays above the same cloud formation only if it orbits Jupiter in the same direction and in the same time as Jupiter rotates about its axis. The centripetal force of the circular motion of the probe is the gravity of Jupiter, as for the moons, so $v^2 = \dfrac{\gamma M}{r}$, and the orbital speed is $v = \dfrac{2\pi r}{T}$.

Formula

so

Conversions

Substitution

so $r = 1.60 \cdot 10^{8}\,\text{m}$

Answer

$r \approx 1.6 \cdot 10^{8}\,\text{m} = 160\,000\,\text{km}$

From the centre of Jupiter. Two significant figures like the result of part b.

Reasonableness

The period $9.93\,\text{h}$ is shorter than that of Amalthea, $0.4982\,\text{d} \approx 12\,\text{h}$, so the orbit must be inside the orbit of Amalthea ($181\,400\,\text{km}$), as it is. The radius of Jupiter is about $71\,500\,\text{km}$, so the probe would orbit at a height of about $89\,000\,\text{km}$ above the clouds.

Find

an evaluation of the claim that Ganymede would orbit faster than Io on the orbit of Io, and the assumptions of the result of part b

Principle

The orbital speed is obtained from the condition that gravity is the centripetal force of the circular motion of the moon: $\gamma\dfrac{Mm}{r^2} = m\dfrac{v^2}{r}$, where $m$ is the mass of the moon.

Model answer

The claim. It is true that at the same distance a larger gravitational force acts on a more massive moon, because the force is proportional to the mass $m$ of the moon. However, a more massive moon needs a correspondingly larger centripetal force to stay on the same circular orbit, because the required force $m\tfrac{v^2}{r}$ is also proportional to the mass. The mass of the moon cancels from the equation, leaving $v^2 = \dfrac{\gamma M}{r}$: the orbital speed depends only on the mass $M$ of Jupiter and the radius of the orbit. On the orbit of Io, Ganymede would therefore orbit with the same speed and in the same time as Io. The claim is not correct. The same can be seen from the graph of part a: moons of very different masses are on the same line.

The assumptions. The result of part b requires that
- the orbits of the moons are circles whose centre is the centre of Jupiter; in reality Jupiter and the moon orbit their common centre of mass, but the masses of the moons are so small compared with the mass of Jupiter that the difference is insignificant
- only the gravity of Jupiter acts on the moon; the attractions of the other moons and the Sun are small in comparison
- the mass of Jupiter pulls as if it were concentrated at its centre, which holds well for a spherically symmetric body.

Find

a graph of the oscillation measurements whose points fit a straight line, and the model fitted to it

Graph

The period of a spring and a mass is $T = 2\pi\sqrt{\dfrac{m}{k}}$. Squaring the equation gives $T^2 = \dfrac{4\pi^2}{k}\,m$: the square of the period is proportional to the mass. The points are therefore drawn in a $(m, T^2)$ coordinate system. The period is a tenth of the measured time, for example $T = 4.74\,\text{s} : 10 = 0.474\,\text{s}$ and $T^2 = 0.2247\,\text{s}^2$.

Model answer

The points are almost on a straight line, so the square of the period grows steadily with the mass, as the model predicts. A straight line is fitted to the points so that it passes between them; it is not forced through the origin, because the measured points do not point there. When the line is extended to the vertical axis, it meets the axis at about $0.028\,\text{s}^2$, slightly above the origin.

Given

the graph of part a

Find

the spring constant $k$ of the spring

Principle

According to the model $T^2 = \dfrac{4\pi^2}{k}\,m$ the slope of the line is $\dfrac{4\pi^2}{k}$. The intercept on the vertical axis does not affect the slope, so the spring constant is obtained from the slope although the line does not pass through the origin. The slope is read from two points of the line far apart from each other.

Reading the graph

The line passes through the points $(0.100\,\text{kg}$; $0.226\,\text{s}^2)$ and $(0.350\,\text{kg}$; $0.722\,\text{s}^2)$, so the slope is

Formula

so

Conversions

$1\,\text{kg/s}^2 = 1\,\text{N/m}$, because $1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2$.

Substitution

Answer

$k \approx 19.9\,\text{N/m}$

Three significant figures, because the masses and times have been measured to three figures and the points are very close to the line.

Reasonableness

The spring stretches about $10\,\text{cm}$ under a $200\,\text{g}$ mass, which corresponds to a spring constant of about $20\,\text{N/m}$. Such a spring is typical in school oscillation experiments.

Find

what the deviation of the line from the origin can be due to and how reliable the spring constant of part b is

Principle

Random error scatters the points on both sides of the line. A systematic error or a deficiency of the model shifts the results in the same direction. The result is reliable when different methods give the same value within the uncertainties.

Model answer

Uncertainty. In hand timing the uncertainty due to reaction time is about $0.2\,\text{s}$. Spread over ten periods, it is only $0.02\,\text{s}$ in the period, which is the reason for measuring ten periods. The points deviate from the line by at most about $0.005\,\text{s}^2$, so the random variation is small and does not explain the intercept of $0.028\,\text{s}^2$ on the vertical axis.

The cause of the deviation. In the model $T = 2\pi\sqrt{\tfrac{m}{k}}$ only the mass oscillates and the spring is massless. The mass of the spring itself, $45\,\text{g}$, however, moves along: its lower end moves with the mass and its upper end is at rest. The oscillating mass is therefore larger than the hanging mass alone, which shifts the whole line upwards but does not change its slope. From the intercept the extra mass is $0.028\,\text{s}^2 : 1.984\,\text{s}^2/\text{kg} \approx 0.014\,\text{kg}$, about a third of the mass of the spring. The result makes sense, because the lower part of the spring moves almost as much as the mass and the upper part hardly at all.

Comparison. From the measurement at rest $k = \dfrac{mg}{x} = \dfrac{0.200\,\text{kg} \cdot 9.81\,\text{m/s}^2}{0.098\,\text{m}} \approx 20.0\,\text{N/m}$. It differs from the result of part b, $19.9\,\text{N/m}$, by less than one percent, so two independent methods support each other.

Conclusion. There is no contradiction: the spring constant is reliable, because it is obtained from the slope, which the mass of the spring does not affect. If the line had been forced through the origin, the spring constant would have been too small, about $18.9\,\text{N/m}$.

Given

$m = 0.250\,\text{kg}$, amplitude $A = 5.0\,\text{cm}$ and $k = 19.9\,\text{N/m}$ from part b

Find

the greatest speed $v_\text{max}$ of the mass

Principle

The energy of the oscillation is conserved when friction and air resistance are small. At an extreme position the mass is momentarily at rest, and the whole energy $\tfrac{1}{2}kA^2$ is potential energy. At the equilibrium position the displacement is zero, so the whole energy is kinetic energy and the speed is greatest. For a vertical spring the weight only shifts the equilibrium position, so the displacement is measured from the equilibrium position.

Formula

$\tfrac{1}{2}mv_\text{max}^2 = \tfrac{1}{2}kA^2$, so $v_\text{max} = A\sqrt{\dfrac{k}{m}}$

Conversions

$A = 5.0\,\text{cm} = 0.050\,\text{m}$

Substitution

Answer

$v_\text{max} \approx 0.45\,\text{m/s}$

Two significant figures like the amplitude $5.0\,\text{cm}$.

Reasonableness

The period is about $0.73\,\text{s}$, and during a period the mass travels $4A = 20\,\text{cm}$, so its average speed is about $0.28\,\text{m/s}$. The greatest speed is larger than this, as it should be. The mass of the spring itself reduces the real greatest speed slightly, because the same energy is also shared with the spring.

Given

the graph of the signal of the microphone over the interval $0\text{–}20\,\text{ms}$

Find

the frequency $f$ of the sound

Principle

The period of a periodic motion is the time after which the whole waveform starts to repeat in the same form. The signal is not a sine wave: each period has a high and a low peak, so the distance between two adjacent peaks is not the period. The reading is more accurate when the duration of several periods is read.

Reading the graph

The high peaks are at $1\,\text{ms}$, $5\,\text{ms}$, $9\,\text{ms}$, $13\,\text{ms}$ and $17\,\text{ms}$. The graph has five whole periods in a time of $20.0\,\text{ms}$, so

Formula

$f = \dfrac{1}{T}$

Conversions

$T = 4.0\,\text{ms} = 0.0040\,\text{s}$

Substitution

Answer

$f = 250\,\text{Hz}$

Two or three significant figures: the duration of five periods can be read accurately from the guide lines.

Reasonableness

The frequency is typical of the sound of an instrument, close to middle C. The shape of the signal shows that in addition to the fundamental frequency it contains double the frequency, $500\,\text{Hz}$: the distance between a high and a low peak is half a period. If the distance between peaks, $2.0\,\text{ms}$, is taken as the period, $500\,\text{Hz}$ is wrongly obtained.

Given

the fundamental frequency $f = 250\,\text{Hz}$ and the speed of sound $v = 343\,\text{m/s}$; the pipe is open at both ends

Find

the length $L$ of the pipe

Principle

A standing wave forms in the air column of the pipe when the sound wave is reflected from the ends of the tube. At an open end the air can move freely, so there is an antinode of the oscillation there. When both ends are open, in the fundamental mode the tube has two antinodes at the ends and one node in the middle, so the length of the tube is half the wavelength. The wavelength is obtained from the basic equation of waves $v = f\lambda$.

Formula

$L = \dfrac{\lambda}{2}$ and $\lambda = \dfrac{v}{f}$, so $L = \dfrac{v}{2f}$

Conversions

Not needed.

Substitution

Answer

$L \approx 0.686\,\text{m}$

Three significant figures like the speed of sound.

Reasonableness

The wavelength is $1.37\,\text{m}$, and the pipe is half of it, just under seventy centimetres, which is a usual size for an organ pipe of this pitch. A real pipe is slightly shorter, because the antinode of the standing wave is a little outside the end of the tube.

Given

$L = 0.686\,\text{m}$, $v = 343\,\text{m/s}$ at a temperature of $20.0\,{}^\circ\text{C}$, the speed of sound decreases by about $0.6\,\text{m/s}$ per degree, winter temperature $8.0\,{}^\circ\text{C}$

Find

the change $\Delta f$ of the fundamental frequency

Principle

The wavelength of the standing wave is determined by the length of the pipe, $\lambda = 2L$, which practically does not change. In cold air sound travels more slowly, so with the same wavelength the frequency $f = \tfrac{v}{\lambda}$ decreases in the same proportion as the speed of sound.

Formula

$v' = v - 0.6\,\tfrac{\text{m/s}}{{}^\circ\text{C}} \cdot \Delta T$, $f' = \dfrac{v'}{2L}$ and $\Delta f = f - f'$

Conversions

The temperature falls by $20.0\,{}^\circ\text{C} - 8.0\,{}^\circ\text{C} = 12.0\,{}^\circ\text{C}$.

Substitution

so

Answer

The fundamental frequency decreases by about $5\,\text{Hz}$ (more precisely $5.2\,\text{Hz}$) to about $245\,\text{Hz}$. The change is known to one significant figure, because the change of speed $0.6\,\text{m/s}$ per degree is given to one figure.

Reasonableness

The speed of sound decreases by about $2\,\%$, so the frequency also decreases by about $2\,\%$, and $2\,\%$ of $250\,\text{Hz}$ is $5\,\text{Hz}$. A semitone at this pitch is about $15\,\text{Hz}$, so in winter the organ sounds about a third of a semitone lower. That is why organs are tuned at the temperature at which they are usually played.

Given

the sound intensity level $L_1 = 88\,\text{dB}$ at a distance $r_1 = 4.0\,\text{m}$, the distance of the organist $r_2 = 2.0\,\text{m}$, practice time $3\,\text{h}$; upper action value $L_0 = 85\,\text{dB}$ and the corresponding time $t_0 = 8\,\text{h}$

Find

the permitted daily exposure time $t_\text{perm}$ of the organist and whether it is enough

Principle

When a source radiates uniformly in all directions, the same power spreads over the surface of a sphere $4\pi r^2$, so the intensity is $I = \dfrac{P}{4\pi r^2}$. When the distance is halved, the intensity quadruples. The change in the intensity level is obtained from the ratio of the intensities. The permitted exposure time halves for every $3\,\text{dB}$ rise.

Formula

where

$n = \dfrac{L_2 - L_0}{3\,\text{dB}}$ and $t_\text{perm} = \dfrac{t_0}{2^{\,n}}$

Conversions

Not needed.

Substitution

so

Answer

$t_\text{perm} \approx 1\,\text{h}$

The time is not enough: three hours of practice is about three times the permitted time. The organist should use hearing protection or practise with quieter stops.

Reasonableness

A level of $94\,\text{dB}$ is of the same order as the noise of a lawn mower at the ear of the user, when hearing protection is recommended. In the point-source model the sound power of the organ would be

which is plausible for an instrument.

Find

an evaluation of the assumptions of the model of part d, the direction of the real exposure, and a reliable way to find it out

Principle

The point-source model assumes that sound starts from a single point and travels without obstacles or reflections. For each assumption we consider whether the level at the organist would in reality be larger or smaller than the model gives.

Model answer

The organ is not a point source. The pipes are spread over an area several metres wide and high, and the distance of the organist from most of the pipes is much more than $2\,\text{m}$. Near an extended source the level does not rise by $6\,\text{dB}$ every time the distance is halved, so the $94\,\text{dB}$ of part d is probably too large an estimate.

Reflections. The walls, floor and vaults of the church reflect sound, and reverberation is heard throughout the hall. Part of the level measured at a distance of $4.0\,\text{m}$ is reverberation, which does not get louder closer up. This also makes the estimate too large.

Playing. During practice the organ is seldom played at full power all the time, so the average level is lower. On the other hand, in some churches the position of the pipes and the console may direct sound straight at the organist.

A reliable way. A sound level meter measures the A-weighted level at the organist's ear throughout a normal practice and its average is calculated and compared with the action value $85\,\text{dB}$. The exposure is probably smaller than the estimate of part d, but only a measurement tells whether hearing protection is needed.