The post How Do You Calculate Gravitational Potential Energy and Force? appeared first on Dr Rom STEM Tuition.

]]>Use the following data to calculate the gravitational potential at the surface of the Moon. Mass of Earth = 81 × mass of Moon, radius of Earth = 3.7 × radius of Moon, gravitational potential at the surface of the Earth = –63 MJ per kg

Gravitational potential is a *scalar* which means it has a magnitude but no direction. This makes it easy to calculate. All we need to know is the size of all the masses in the neighbourhood and their distances. Then we add up the potential energy from each mass. If the distance of a mass \(M\) is \(r\) then its contribution to the gravitational potential is

where \(G\) is the gravitational constant which is *extremely small* as gravity is a very weak force. G is just \(6.672 \times 10^{-11}\) which is ten zeros after the decimal point before we get to 6: 0.00000000006672. Going back to the question, if we are working out the gravitational potential energy at the Earth’s surface it will be

where \(M_e\) is the Earth’s mass and \(r_e\) is the Earth’s radius. On the surface of the Moon the potential energy will be

\(\text{GPE}_m = – \frac{G M_m}{r_m}\)The question is asking us about the relationship between the gravitational potential energy on the Moon’s surface expressed in terms of the gravitational potential energy on the Earth’s surface. We know that

\(M_e = 81 M_m\)and that

\(r_e = 3.7 r_m\)Substituting these into the equation for the potential on the lunar surface

\( \begin{align*}\text{GPE}_m &= – \frac{G M_m}{r_m} \\

&= – \frac{G \frac{M_e}{81}}{\frac{r_e}{3.7}} \\

&= – \frac{3.7}{81} \frac{G M_e}{r_e} \\

&= \frac{3.7}{81} \text{GPE}_e \\

\end{align*}

\)

Use your previous answer to sketch the gravitational potential between the surface of the Earth and the surface of the moon.

From the previous answer, we know that the Earth sits in a deep potential well and the Moon sits in a much shallower well which is just 5% the depth of that of the Earth. The potential looks like this:

Use the graph above to calculate how much energy it would take to throw a 1 kg rock to the maximum potential point between the Earth and the Moon.

The potential difference is 3.86 – 1.28 megajoules, which is 2.6 megajoules.

What would be the kinetic energy of the rock when it hit the surface of the Earth?

The potential difference between the maximum energy point and the surface of the Earth is 62.55-1.28, which is 61.3 megajoules. That’s 24 times the energy it took to throw the rock to the tipping point (which is called the Lagrange point as described below). This is why a moon base would be a strategic threat to the Earth.

The gravitational force between two masses \(M\) and \(m\) is given by

\(F= -\frac{GMm}{r^2}\)Whereas gravitational potential energy scales as \(\frac{1}{r}\) the gravitational force scales as \(\frac{1}{r^2}\). That’s because gravitational potential energy is the amount of work done to bring a mass from an infinite distance to some point in space and when we integrate \(1/r^2\) we get a constant times \(1/r\).

Find the point between the Earth and the Moon at which the gravitational forces cancel one another out. Is this closer to the Moon or the Earth?

There is a point between the Earth and the Moon where the gravitational forces balance exactly called the *Lagrange point*. As the Moon has a lower mass than the Earth the Lagrange point is much closer to the Moon than the Earth. We can find this point by equating the gravitational force from the Earth and the Moon. The gravitational force between two masses \(M\) and \(m\) is given by

This scales as the square of the inverse of the separation of the masses, which means the gravitational force dwindles rapidly with distance. If we imagine a line from the centre of the Earth to the centre of the Moon the two forces we equate are

\( \begin{align*}F_e &= F_m \\

\frac{G M_e m}{r_e^2} &= \frac{G M_m m}{r_m^2} \\

\frac{M_e}{r_e^2} &= \frac{M_m}{r_m^2} \\

\frac{r_m^2}{r_e^2} &= \frac{M_m}{M_e} \\

\frac{r_m}{r_e} &= \sqrt{\frac{M_m}{M_e}}

\end{align*}

\)

As the ratio of the Earth’s mass to the Moon’s mass is about 81 the square root of that is 9. This means that the Earth-Moon Lagrange point is about 9 times as far from the Earth as from the Moon or about 11% of the way from the centre of the Moon to the centre of the Earth. If you do this calculation properly you have to correct for the fact that the whole system is in motion which increases the distance very slightly from 11% to 15% of the way from the Moon to Earth.

A satellite of mass \(m\) travels at angular speed \(\omega\) in a circular orbit at a height \(h\) above the surface of a planet of mass \(M\) and radius \(R\). Using these symbols, give an equation that relates the gravitational force on the satellite to the centripetal force.

We know that if we are in a circular orbit the acceleration is

\(a = \omega^2 r\)And the centripetal force due to gravity is

\(F = \frac{GMm}{r^2}\)However, in this example we have to remember that the satellite is orbiting at a distance of \(r = R+h\) i.e. the radius of the Earth plus the height above the surface of the Earth. Now we can equate mass times orbital acceleration with the gravitational force:

\( \begin{align*}F &= m a \\

\frac{GMm}{(R+h)^2} &= m \omega^2 (R+h) \\

\frac{GM}{(R+h)^2} &= \omega^2 (R+h)

\end{align*}

\)

Use the equation above to show that the orbital period, T, of the satellite is given by \(T^2=\frac{4 \pi^2 (R+h)^3}{GM}\)

We know that \(\omega = \frac{2 \pi}{T}\) so we can substitute for \(\omega\) in the equation above to get

\( \begin{align*}\frac{GM}{(R+h)^2} &= \left( \frac{2 \pi}{T} \right)^2 (R+h) \\

\frac{GM}{(R+h)^2} &= \left( \frac{4 \pi^2}{T^2} \right)^2 (R+h) \\

T^2 &= \frac{4 \pi^2 (R+h)^3}{GM} \\

\end{align*}

\)

Explain why the period of a satellite in orbit around the Earth cannot be less than 85

minutes. Your answer should include a calculation to justify this value. Mass of the Earth = \(6.00 \times 10^24\) kg, radius of the Earth = \(6.40 \times 10^6\) m.

The orbital height of the satellite must be positive and so the limiting case is when \(h=0\).

\( \begin{align*}T^2 &= \frac{4 \pi^2 (R+h)^3}{GM} \\

&= \frac{4 \pi^2 (6.40 \times 10^6)^3}{6.672 \times 10^{-11} \times 6.00 \times 10^24} \\

&= 2.6 \times 10^7 \\

T &= 5084

\end{align*}

\)

This means that the minimum orbital period when \(h=0\) is 5084 seconds, which is 85 minutes. As the height of the orbit increases so will the orbital period so 85 minutes is a minimum orbital period for a satellite.

The mass of the Earth \(6.00 × 10^24\) kg and its mean radius is \(6.40 \times 10^6\) m. Show that the radius of a geosynchronous orbit must be \(4.23 \times 10^7\) m.

We know that the orbital period for a geosynchronous satellite must be one day and that is enough to work out the radius of its orbit. For the orbit to be stable the Earth’s gravity must match the satellite’s mass times acceleration required for a circular orbit which is \(\omega^2 r\). We can use the relationship \(\omega = 2 \pi / T\) to substitute for the angular frequency where \(T\) is the length of a day in seconds which is \(24 \times 60 \times 60 = 86400\) seconds.

\(\begin{align*}\frac{GMm}{r^2} &= m \omega^2 r \\

GM &= \left ( \frac{2 \pi}{T} \right )^2 r^3 \\

GM &= \frac{4 \pi^2 r^3}{T^2} \\

r^3 &= \frac{GM T^2}{4 \pi^2} \\

r &= \sqrt[3]{\frac{GM T^2}{4 \pi^2}}

\end{align*}

\)

Putting in the numbers:

\(\begin{align*}r &= \sqrt[3]{\frac{6.672 \times 10^{-11} \times 6.00 × 10^24 \times 86400^2}{4 \pi^2}} \\

&= 4.23 \times 10^7 \\

\end{align*}

\)

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]]>The post Calculate Electric Potential and Field For Two Charges appeared first on Dr Rom STEM Tuition.

]]>We have a charge of +8 nanocoulombs and -4 nanocoulombs placed 60 millimetres apart. What is the electric potential at point P shown below. What is the size and direction of the electric field at point P?

This picture shows the size and position of the charges relative to point P.

Starting with the electric potential, it is a *scalar* which means it has a magnitude but no direction. This makes it easy to calculate. All we need to know is the size of all the charges in the neighbourhood and their distances. Then we add up the potential from each charge. If the distance of a charge from our point P is \(r\) then for each charge its contribution is

We have to remember to work in metres, and we can use Pythagoras’ theorem to work out the distances of the two charges. The +8 nC charge on the left is at a distance

\(\sqrt{(80 \times 10^-3)^2+(30 \times 10^-3)^2}=0.08544 \mbox{ m}\)and the -4 nC charge on the right is at the same distance from P. Now we can work out the electric potential from the +8 nC charge at P,

\(V_{\mbox{+8 nC}} = \frac{8 \times 10^{-9}}{4\pi \epsilon_0 0.08544}=841.5 \mbox{ V}\)and the contribution of the -4 nC charge is

\(V_{\mbox{-4 nC}} = \frac{-4 \times 10^{-9}}{4\pi \epsilon_0 0.08544}=-420.8 \mbox{ V}\)and as this is a scalar we can just add the potentials to give

\(V=V_{\mbox{+8 nC}} + V_{\mbox{-4 nC}} = 841.5-420.8 = 420.8 \mbox{ V}\)If we plot the electric potential it looks like this.

The positive charge creates the funnel that sticks up and the negative charge sticks down because its potential is negative. By convention the surface represents the work done, in joules, to drag one coulomb of positive charge from infinitely far away to each point in the 2d region.

Electric field is more difficult to calculate because it involves adding two vectors. The vectors we add are the fields from the +8 nC and -4 nC charges. The electric field at P is the force that would be felt by a +1 coulomb charge placed at P. Working out the magnitude of the field is easy because the formula for the force between charge \(q_1\) and \(q_2\) separated by distance \(r\) is

\(F=\frac{q_1 q_2}{4\pi \epsilon_0 r^2}\)We know that when we are calculating the field strength \(q_2\) is +1 coulombs, so starting with the +8 nC charge the magnitude of the field at P is

\(F_{\mbox{+8 nC}}=\frac{8 \times 10^{-9}}{4\pi \epsilon_0 0.08544^2}=9850 \mbox{ N C}^{-1}\)Because both charges, the +8 nanocoulomb charge and the “test” charge of +1 coulomb are positive the direction is directly away from the +8 nC charge. The magnitude of the field from the -4 nC charge at P is

\(F_{\mbox{-4 nC}}=\frac{-4 \times 10^{-9}}{4\pi \epsilon_0 0.08544^2}=4925 \mbox{ N C}^{-1}\)and the direction of this field is directly towards the -4 nC charge because the “test” charge of +1 coulomb and the -4 nC charge are opposite so the force between them is attractive.

To work out the direction of the force a diagram helps.

The two fields acting at P are shown in red for the 8 nC charge and blue for the -4 nC charge. To add the fields we simply lay them end-to-end. The resultant field is shown in green. To calculate the components of this field we can decompose the two forces into x and y components and add them. The angle the blue arrow is making with the vertical is

\(\tan^{-1} \left( \frac{30}{80} \right) = 20.6 \mbox{ degrees}\)The vertical component of the blue field is \(4925 \cos 20.6\) and the vertical component of the red field is \(-9850 \cos 20.6\) which adds up to give a vertical component of

\(F_{v} = 4925 \cos 20.6 = 4610 \mbox{ N C}^{-1}\)The horizontal components of the two fields are in the same direction (toward the right) so they add

\(F_{h} = 4925 \sin 20.6 + 9850 \sin 20.6 = 5198 \mbox{ N C}^{-1}\)The magnitude of this force is easy to calculate now we know its horizontal and vertical components using Pythagoras again

\(F = \sqrt{F_h^2 + F_v^2} = \sqrt{4610^2 + 5198^2} = 6948 \mbox{ N C}^{-1}\)This makes an angle 41.6 degrees below the horizontal

\(\tan^{-1} \left( \frac{4610}{5198} \right) = 41.6 \mbox{ degrees}\)The field at each point is the gradient of the potential. If we ignore magnitude and simply look at the direction of the field at each point it looks like this.

Because the convention is that the field is the force on a positive 1 coulomb charge it points away from the positive charge on the left and towards the negative charge on the right.

Fields and potential are always related in this way: field is the gradient of the potential. This is true for electric or gravitational potentials and forces. This is why the lines of equal potential (the contours) are always exactly at right angles with the force vectors.

If you found this explanation helpful and you want interactive one-to-one tuition then click on the button. It’s easy to book an hour and I can do online lessons too so you can learn from anywhere.

Book A Lesson With RaminIf you want to learn a bit of R and generate the cool interactive 3d visualisation of the potential here’s the code you need.

rm(list=ls()) mp <- 10000 # maximum potential avoids huge potential when r close to zero two.positive.charges <- function (x, y) { r1 <- sqrt((x+30e-3)^2+(y-80e-3)^2) # +8 nC charge at position x = -30 mm, y = +80 mm r2 <- sqrt((x-30e-3)^2+(y-80e-3)^2) # -4 nC charge at position x = +30 mm, y = +80 mm epsilon0 <- 8.854e-12 p1 <- 8e-9/(4*pi*epsilon0)/r1 # potential from +8 nC charge p2 <- -4e-9/(4*pi*epsilon0)/r2 # potential from -4 nC charge # truncate potential when it exceeds mp (huge positive or negative values) p1 <- pmin(p1,mp) p1 <- pmax(p1,-mp) p2 <- pmin(p2,mp) p2 <- pmax(p2,-mp) return (p1+p2) } # define a grid for the problem from -100 mm to +100 mm in x and from 0 to 200 mm in y # the point P in the question is at the origin x <- seq(-100e-3,100e-3,1e-3) y <- seq(0,200e-3,1e-3) grid <- expand.grid(x=x,y=y) # create an x-y grid with all combinations of x and y # outer calculates the potential at all x and y points in the grid # result stored in a data frame potential <- data.frame(x=grid$x ,y=grid$y ,potential=as.vector(outer(x,y,two.positive.charges))) # work out the maximum and minimum potential values plim <- range(potential$potential) plen <- plim[2] - plim[1] + 1 # build colour map based on range of potential, choose whichever you like best colorlut <- terrain.colors(plen) # height color lookup table colorlut <- heat.colors(plen) # height color lookup table colorlut <- rev(rainbow(plen)) # height color lookup table - we use this one col <- colorlut[ potential$potential - plim[1] + 1 ] # assign colors to heights for each point library(rgl) persp(x,y,outer(x,y,two.positive.charges),col=col) # 3d mesh plot of potential persp3d(x,y,outer(x,y,two.positive.charges),col=col) # interactive 3d plot in a new window library(ggplot2) library(reshape) # omit tails max.p <- as.numeric(quantile(potential$potential,probs = c(0.99))) # chop off anything above 99th percentile min.p <- as.numeric(quantile(potential$potential,probs = c(0.01))) # or below the 1st percentile potential <- potential[potential$potential > min.p,] potential <- potential[potential$potential < max.p,] library(RColorBrewer) cols <- brewer.pal(n = 10, name = "RdBu") library(scales) max.abs.potential <- max(abs(potential$potential)) g <- ggplot(potential, aes(x=x, y=y, z=potential)) + stat_contour(geom = "polygon", aes(fill = ..level..)) + geom_tile(aes(fill = potential)) + scale_fill_gradientn(colours = cols ,values=scales::rescale( c(max.abs.potential ,0 ,-max.abs.potential)) ,limits=c(-max.abs.potential,max.abs.potential)) + stat_contour(bins = 15) + xlab("x") + ylab("y") + guides(fill = guide_colorbar(title = "Electric Potential (volts)")) + theme_classic() + theme(legend.position="none") # define a coarse grid & potential for the arrows x <- seq(-100e-3,100e-3,10e-3) y <- seq(0,200e-3,10e-3) coarsegrid <- expand.grid(x=x,y=y) coarsepotential <- data.frame(x=coarsegrid$x ,y=coarsegrid$y ,potential=as.vector(outer(x,y,two.positive.charges))) # omit tails max.p <- as.numeric(quantile(coarsepotential$potential,probs = c(0.99))) min.p <- as.numeric(quantile(coarsepotential$potential,probs = c(0.01))) coarsepotential <- coarsepotential[coarsepotential$potential > min.p,] coarsepotential <- coarsepotential[coarsepotential$potential < max.p,] # calculate field by jiggling points and working out the gradient dx <- 1e-6 dy <- 1e-6 for (i.row in 1:nrow(coarsepotential)) { x <- coarsepotential[i.row,"x"] y <- coarsepotential[i.row,"y"] # x up-jiggle potential minus x down-jiggle divided by jiggle size is dV / dx coarsepotential[i.row,"dVdx"] <- (two.positive.charges(x+dx/2,y) - two.positive.charges(x-dx/2,y)) / dx # y up-jiggle potential minus y down-jiggle divided by jiggle size is dV / dy coarsepotential[i.row,"dVdy"] <- (two.positive.charges(x,y+dy/2) - two.positive.charges(x,y-dy/2)) / dy # scale all the field arrows to have length one scale <- sqrt(coarsepotential[i.row,"dVdx"]^2+coarsepotential[i.row,"dVdy"]^2) coarsepotential[i.row,"dVdx"] <- coarsepotential[i.row,"dVdx"] / scale coarsepotential[i.row,"dVdy"] <- coarsepotential[i.row,"dVdy"] / scale } scale <- 0.5e-2 # arrow scaling factor library(grid) g + geom_segment(data=coarsepotential ,aes(x=x,y=y,xend=x-dVdx*scale,yend=y-dVdy*scale) ,arrow = arrow(length = unit(0.3,"cm")) ,colour="red") # Now we answer the question in R... # What's the potential at point P (the origin)? two.positive.charges(x=0,y=0) # 420.7743 volts # what is the magnitude and direction of the field at P? x <- 0 y <- 0 # field is minus dV/dx because a charge moves down the potential slope like rolling down a hill # from a region of higher to lower potential dVdx <- -(two.positive.charges(x+dx/2,y) - two.positive.charges(x-dx/2,y)) / dx # -749.3976 dVdy <- -(two.positive.charges(x,y+dy/2) - two.positive.charges(x,y-dy/2)) / dy # 50.8635 # magnitude of field 6940.813 newtons per coulomb sqrt(dVdx^2+dVdy^2) # angle with horizontal -41.63354 degrees i.e. below horizontal pointing to the right 180/pi*atan(dVdy/dVdx) # R works in radians so multiply by 180/pi to convert to degrees # we can also do it the same way as in the text (theta <- 180/pi*atan(30/80)) # angle of both field vectors from vertical 20.55605 degrees # magnitude of force from +8 nC charge x <- -30e-3 y <- 80e-3 (r <- sqrt(x^2 + y^2)) # 0.08544004 metres (f8 <- 8e-9/(4*pi*8.854e-12*r^2)) # 9849.581 newtons per coulomb field strength for +8 nC charge # magnitude of force from -4 nC charge x <- 30e-3 y <- 80e-3 (r <- sqrt(x^2 + y^2)) # 0.08544004 metres, same as other charge thanks to symmetry (f4 <- -4e-9/(4*pi*8.854e-12*r^2)) # 4924.79 newtons per coulomb field strength for +8 nC charge # add components in horizontal and vertical directions (fv <- f4*cos(theta*pi/180) + f8*cos(theta*pi/180)) # 4611.225 newtons per coulomb (fh <- f4*sin(-theta*pi/180) + f8*sin(theta*pi/180)) # 9849.581 newtons per coulomb # magnitude of field 6940.813 newtons per coulomb sqrt(fv^2+fh^2) # direction of field 41.63354 degrees 180/pi*atan(fv/fh)

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]]>The post Particle Deflection in an Electric Field appeared first on Dr Rom STEM Tuition.

]]>Once we know the field strength it’s easy to calculate the force on a charge \(q\), we simply multiply by the field strength.

\(F = Eq = \frac{Vq}{d}\)A Level questions usually have an electron whizzing between the plates and ask about the force exerted on the electron by the plates, the resulting acceleration and the final velocity and direction of the electron. Let’s try solving a problem to see how this works.

Let’s say we have two plates separated by 20 millimetres with a voltage difference of 2000 volts. The length of the plates along which the electron travels is 30 millimetres. The horizontal velocity of the electron before it passes through the plates is \(1.5\times10^7 \mbox{ms}^{-1}\)

Firstly the strength of the electric field is

\(E=\frac{2000}{20\times10^{-3}}=10^5 \mbox{V m}^{-1}\)The force on the electron

\(F = Eq = 10^5 \times 1.6\times10^{-19}=1.6\times10^{-14} \mbox{N}\)And by Newton’s second law \(F=ma\) the acceleration of the electron toward the positive plate is

\(a = \frac{1.6\times10^{-14}}{9.11\times10^{-31}}=1.76\times10^{16} \mbox{ms}^{-2}\)The time spent between the plates, which we need to calculate the final vertical velocity of the electron is

\(\Delta t=\frac{30\times10^{-3}}{1.5\times10^7}= 2\times10^{-9} \mbox{s}\)In this two nanosecond period spent between the plates the vertical velocity increases by

\(\Delta v = a \Delta t = 1.76\times10^{16} \times 2\times10^{-9} = 3.51\times10^7 \mbox{ms}^{-1}\)We can calculate the angle the velocity of the electron makes with its original direction by working out the inverse tangent of the ration of the vertical and horizontal velocities

\(\theta=\tan^{-1}\frac{v_v}{v_h}=\tan^{-1}\frac{3.51\times10^7}{1.5\times10^7}=66.9 \mbox{ degrees}\)We can also work out the magnitude of the velocity by applying Pythagoras’ Theorem to the horizontal and vertical components of velocity

\(v=\sqrt{(3.51\times10^7)^2 + (1.5\times10^7)^2} = 3.82\times10^7 \mbox{ ms}^{-1} \)The post Particle Deflection in an Electric Field appeared first on Dr Rom STEM Tuition.

]]>The post Specific Heat Capacity and Specific Latent Heat of Vaporisation appeared first on Dr Rom STEM Tuition.

]]>We have a 2.2 kW kettle containing 0.5 kg of water at 20

^{o}C. The kettle is switched on for 3 minutes. What mass of water remains in the kettle at the end of this time? The specific heat capacity of water is 4186 J kg^{-1}K^{-1}and its specific latent heat of vaporisation is 2.26 MJ kg^{-1}.

It’s probably a good idea to draw a sketch of what’s happening. Notice that until it boils energy is flowing into the thermal energy store of the water. As the kinetic energy of the water molecules increases the temperature increases. But as the temperature reaches 100^{o}C it stops rising. Now the energy flows into breaking bonds between water molecules completely to change state from liquid water to water vapour which is a gas.

We split the problem into two parts. The first part is to calculate how long the temperature takes to reach 100^{o}C. The second part is to work out how much water is vaporised in the remaining time until 180 seconds.

We have to increase the temperature of the water from 20^{o}C to 100^{o}C. This is a temperature increase of 100^{o}C-20^{o}C=80^{o}C.

The energy required to raise the temperature of 1 kg of water by 1^{o}C is 4186 joules. To raise half a kilogram of water by 80 centigrade will require \(0.5 \times 80 \times 4186=167,440\) joules. We’re using the equation

where

\(\Delta E\) | change in energy | joules (J) |

\(m\) | mass | kilograms (kg) |

\(c\) | specific heat capacity | J kg^{-1} C^{-1} |

\(\Delta \theta\) | change in temperature | centigrade (C) or kelvin (K) |

The kettle’s heating element is delivering energy at a rate of 2,200 joules per second (watts). The time taken to deliver 167,440 joules is \(167440 \times 2200 = 76.1\) seconds. That’s because

\(t=\frac{E}{P}\)where

\(t\) | time to deliver energy | seconds (s) |

\(E\) | energy | joules (J) |

\(P\) | power | watts (W) or joules per second (J s^{-1}) |

The remaining time is spent boiling, and as 3 minutes is 180 seconds that is 180-76.1 or 103.9 seconds.

If we’re boiling the 0.5 kg of water for 103.9 seconds the energy delivered by the heating element in that time is \(103.9 \times 2200=228,560\) joules.

We know that it takes 2.26 megajoules to vaporise 1 kg of water. The amount of water that is vaporised by 228,560 joules is \(\frac{228560}{2.26 \times 10^6} = 0.101\) kg. As we began with 0.5 kg of water the remaining water has a mass of 0.5-0.101=0.399 kg. We’re using the equation

\(\Delta E=ml\)where

\(\Delta E\) | change in energy | joules (J) |

\(m\) | mass | kilograms (kg) |

\(l\) | latent heat of vaporisation | J kg^{-1} |

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]]>The post Integration Roadmap appeared first on Dr Rom STEM Tuition.

]]>Here’s the link: Integration Summary Sheet

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]]>The post How Variables Scale in Equations appeared first on Dr Rom STEM Tuition.

]]>and you know that mass is 1 kilogram and acceleration is 2 \(ms^{-2}\) then the force is \(1 \times 2 = 2\) newtons. But there’s something else you can do with equations that is incredibly useful and often crops up in exams, particularly in GCSE Mathematics and A-level Physics. This is spotting how variables *scale*.

What does scaling mean? Well, instead of putting in actual numbers into an equation you consider what would happen if the variables on the right-hand side of the equals sign are scaled up or down by a factor. Using Newton’s equation we can see that if you double the mass on the right-hand side you would double the force. Or if you double the acceleration you would double the force. This kind of scaling is the simplest type: it’s *linear*. The remaining examples in this post are more interesting because they aren’t linear.

The exam question that occurs all the time is scaling of area and volume with length. Let’s say we have a scaling gun that scales up animals. To create Godzilla we will breed a mini-Godzilla and then use the scaling gun to scale him up. Let’s think what will happen to four things:

- Godzilla’s height
- Godzilla’s surface area
- Godzilla’s volume
- Godzilla’s mass

Mini-Godzilla is a Tyrannosaurus rex, 10 metres high because we cloned him along the lines of Jurassic Park. Then we scale his height by a factor of ten. He is now 100 metres high!

Now the surface area of a solid depends on the *square* of its dimensions. For example, a regular cube has six sides and if each edge has length \(l\) then the surface area

Now if we increase \(l\) by a factor of ten the surface area \(A\) will increase by a factor of 100, which is 10 squared. The same is true of a sphere. If its radius is \(r\) then its surface area is

\(A = 4 \pi r^2\)which, if we increase the radius by a factor of ten, will increase by a factor of 100, which is 10 squared.

It’s true of a cube, a sphere, or Godzilla. If we increase an object’s dimensions by a factor of 10 its area will increase by a factor of 100.

What about Godzilla’s volume? If we think about our cube again its volume is

\(V=l^3\)which increases by a factor of 1000 if we increase the cube’s dimensions by ten. The same is true of a sphere with volume

\(V=\frac{4}{3} \pi r^3\)which, if we increase the radius by a factor of 10, has its volume increase by a factor of 1000. So Godzilla’s volume will also increase by a factor of 1000.

If the scaling gun doesn’t change the density of Godzilla then his mass will also scale up by a factor of 1000. This is a problem. Think of the stress on Godzilla’s legs. Stress is

\(\mbox{stress}=\frac{\mbox{force}}{\mbox{area}}\)The force is Godzilla’s weight and the area is the cross-sectional area of Godzilla’s legs. Weight goes up by a factor of 1000, the area goes up by a factor of 100 so the stress on Godzilla’s legs will increase by a factor of 10. His bones would shatter under his own weight!

If we think about his ability to radiate heat this would also be problematic. This is because he radiates heat through his surface, which has gone up by a factor of 100 but heat is generated throughout his body and that has scaled by a factor of 1000, so his surface area to volume ratio has decreased by a factor of 10. Godzilla’s legs would shatter and he would always be overheating. Maybe we shouldn’t use that scaling gun after all.

Exam questions often require you to scale from centimetres to metres but many people get confused when scaling areas from centimetres squared to metres squared or centimetres cubed to meters cubed. This works in exactly the same way as the Godzilla example above:

- Area:
- 1 m = 100 cm
- \(1 \mbox{m}^2 = 100^2\mbox{cm}^2 = 10,000 \mbox{cm}^2\)
- 1 m = 1000 mm
- \(1 \mbox{m}^2 = 1000^2 \mbox{mm}^2 =1,000,000 \mbox{mm}^2\)

- Volume
- 1 m = 100 cm
- \(1 \mbox{m}^3 = 100^3 \mbox{cm}^3 =1,000,000 \mbox{cm}^3\)
- 1 m = 1000 mm
- \(1 \mbox{m}^3 = 1000^3 \mbox{mm}^3 = 1,000,000,000 \mbox{mm}^3\)

To go from the **length** scaling factor to the **area** scaling factor you **square** the ratio.

To go from the **length** scaling factor to the **volume** scaling factor you **cube** the ratio.

To go from the **area** scaling factor to the **volume** scaling factor you first take the **square root** of the ratio (to turn it into a length ratio) then **cube** the ratio.

To go from the **volume** scaling factor to the **area** scaling factor you first take the **cube root** of the ratio (to turn it into a length ratio) then **square** the ratio.

Sometimes quantities scale down as another quantity scales up, a relationship called inverse scaling.

Travel time scales as the inverse of the speed

\(\mbox{Time}=\frac{\mbox{Distance}}{\mbox{Speed}}\)Density scales as the inverse of the volume

\(\mbox{Density}=\frac{\mbox{Mass}}{\mbox{Volume}}\)Gravitational potential energy scales as the inverse of the distance from a mass \(M\) with gravitational constant \(G\) is

\(\mbox{Gravitational Potential Energy}=-\frac{GM}{r}\)The power of a lens in diopters scales as the inverse of the focal length \(f\) of the lens

\(\mbox{P}=\frac{1}{f}\)There are lots of examples of equations where the scaling isn’t linear, it goes as the square of the scaling factor. Here are some examples:

Kinetic energy scales as the square of the velocity

\(\mbox{Kinetic Energy}=\frac{1}{2} m v^2\)Energy stored in a spring with spring constant \(k\) scales as the square of the extension \(x\) \(\mbox{Spring Energy} = \frac{1}{2} k x^2 \)

Maximum acceleration for a simple harmonic oscillator with amplitude \(A\) scales as the square of angular velocity \(\omega\) \(\mbox{Maximum Acceleration} = A \omega^2 \)

Power dissipated in a resistor of resistance \(R\) scales as the square of current \(I\) or voltage \(V\) \(\mbox{Power} = I^2 R = \frac{V^2}{R} \)

These seem trickier but instead of squaring you take the square root of the scaling factor.

The period of a pendulum with gravitational acceleration \(g\) scales as the square root of the pendulum length \(l\) \(T \approx 2\pi \sqrt{\frac{l}{g}}\)

The period of a mass on a spring \(g\) scales as the square root of the mass \(m\) \(T = \frac{1}{2\pi} \sqrt{\frac{m}{k}}\)

Escape velocity from a body with radius \(r\) scales as the square root of the mass of the planet \(M\) \(v_{\mbox{escape}} = \sqrt{\frac{2GM}{r}}\)

The root mean square velocity of gas molecules of mass \(m\) in an ideal gas scales as the square root of the temperature \(T\) \(v_{\mbox{rms}} = \sqrt{\frac{3kT}{m}}\)

Here are 11 questions that you can use to test your ability to find how equations scale:

The post How Variables Scale in Equations appeared first on Dr Rom STEM Tuition.

]]>The post Radioactive Decay and Neanderthal Art appeared first on Dr Rom STEM Tuition.

]]>The age of calcium carbonate deposits on top of cave paintings in the North, Centre and South of Spain were found to be 64,800 years when the inhabitants of what is now Spain were Neanderthals. Modern humans are thought to have arrived in Spain 20,000 years later. The images show what looks to be symbolic art: for example at La Pasiega in Cantabria you can see a ladder with dots and paintings of animals inside the panels of the ladder. The white crust you see above the painting, which consists of calcite deposits, was used for uranium-thorium dating.

This dating method involves two decays rather than one. Naturally occurring water contains the uranium isotope \({}^{234}_{92}U\) which decays into thorium which does not normally appear in groundwater. This thorium then decays to radium.

\({}^{234}_{92}U \rightarrow {}^{230}_{90}Th + {}^4_2\alpha \rightarrow {}^{226}_{88}Ra + {}^4_2\alpha\)The method relies on the differing half-lives of the two reactions.

- 245,000 years for \({}^{234}U \rightarrow {}^{230}Th\) decay
- 75,000 years for \({}^{230}Th \rightarrow {}^{226}Ra\) decay

The half-life of the thorium isotope is three times shorter than the uranium isotope. After a certain amount of time the level of thorium builds up and its rate of decay eventually approaches its rate of production so that the total amount of thorium relative to the amount of uranium reaches equilibrium. We know that the rate of decay of uranium if we have \(N_U\) atoms and the decay constant is \(\lambda_U\), is

\(\frac{dN_U}{dt}=-\lambda_U N_U\)

The decay rate of uranium is the production rate of thorium, but thorium also decays at rate \(-\lambda_T N_T\) which means the rate of change of the amount of thorium is

\(\frac{dN_T}{dt}=\lambda_U N_U – \lambda_T N_T\)The system reaches equilibrium, which means that \(\frac{dN_T}{dt}=0\) when

\(\lambda_U N_U = \lambda_T N_T\)or rearranging

\(\frac{\lambda_U}{\lambda_T} = \frac{N_T}{N_U}\)We can convert the half-lives of the isotopes into decay constants using the formula \(\lambda=\frac{\ln(2)}{t_\frac{1}{2}}\).

\(\lambda_U=\frac{\ln(2)}{245000}=2.83\times 10^-6\) \(\lambda_T=\frac{\ln(2)}{75000}=9.24\times 10^-6\)At equilibrium

\(\frac{N_T}{N_U} = \frac{2.83\times 10^-6}{9.24\times 10^-6} = 0.31\)It turns out that we can solve these equations using a nice little integration trick which you may find useful. It relies on the product rule:

\(d(uv)=v du + u dv\)It’s as if we have water draining slowly out of one tank (uranium) into another tank (thorium) which drains quickly. Let’s start out with the simple bit which is the decay of uranium which feeds the production of thorium.

\(\frac{dN_U}{dt} = -\lambda_U N_U\)We can separate the two integrals by moving \(N_U\) over to the left hand side and integrating

\(\int_{N_U^0}^{N_U^t}\frac{dN_U}{N_U} = -\lambda_U \int_0^t dt\)We can solve these two easy integrals

\(\left.\ln N_U\right|_{N_0}^{N_t} = \left.-\lambda_U t \right|_{0}^t\)and the limits give

\(\ln \frac{N_U^t}{N_U^0} = -\lambda_U t \)Take exponentials of both sides

\(\frac{N_U^t}{N_U^0} = e^{-\lambda_U t}\)And we come up with the familiar equation for radiactive decay

\(N_U^t = N_U^0 e^{-\lambda_U t}\)Now the hard bit which is solving both the uranium and thorium decay equations. Let’s start with the rate equation for thorium, remember the first term is the production of thorium (positive) through decay of uranium and the second term (negative) is the consequent decay of thorium

\(\frac{dN_T}{dt}=\lambda_U N_U – \lambda_T N_T\)We solved the first term above so we can substitute that here

\(\frac{dN_T}{dt}=\lambda_U N_U^0 e^{-\lambda_U t} – \lambda_T N_T\)We rearrange to put all the thorium terms on the left hand side and all the uranium terms on the right hand side

\(\frac{dN_T}{dt} + \lambda_T N_T=\lambda_U N_U^0 e^{-\lambda_U t}\)The game here is to make the left hand side look like the derivative of a product. We multiply everything by \(e^{\lambda_T t}\) to get

\(\frac{dN_T}{dt} e^{\lambda_T t} + \lambda_T N_T e^{\lambda_T t} = \lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t}\)Magically the left hand side looks like the derivative of a product which gives us the left hand side integral straight away

\(\frac{d}{dt} N_T e^{\lambda_T t} = \frac{dN_T}{dt} e^{\lambda_T t} + \lambda_T N_T e^{\lambda_T t}\)The integral on the right hand side is easy so we put them together

\( N_T e^{\lambda_T t} = \int \lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t} = \frac{\lambda_U N_U^0 e^{-\lambda_U t + \lambda_T t}}{\lambda_T – \lambda_U} + C\)We can get rid of the \(e^{\lambda_T t}\) term by multiplying by \(e^{-\lambda_T t}\) to give

\( N_T = \frac{\lambda_U N_U^0 e^{-\lambda_U t}}{\lambda_T – \lambda_U} + C e^{-\lambda_T t}\)The integration constant \(C\) can be found by setting \(t=0\) when we know that the amount of thorium is some initial value \(N_T = N_T^0\) and

\( N_T^0 = \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U} + C \)This can be rearranged

\( C = N_T^0 – \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U}\)Substituting this back into our equation for \(N_T\) gives us

\( N_T = \frac{\lambda_U N_U^0 e^{-\lambda_U t}}{\lambda_T – \lambda_U} + N_T^0 e^{-\lambda_T t} – \frac{\lambda_U N_U^0 }{\lambda_T – \lambda_U} e^{-\lambda_T t}\)Which can be rearranged as

\( N_T = \frac{\lambda_U N_U^0 (e^{-\lambda_U t} – e^{-\lambda_T t})}{\lambda_T – \lambda_U} + N_T^0 e^{-\lambda_T t} \)The second term is simply the decay of the initial amount of thorium, which is usually zero because thorium is not usually present in groundwater. The first term is the interesting one.

If we look at the abundance of thorium and uranium numerically we can see that uranium decays exponentially as we would expect. Thorium is not present initially but is created as uranium decays, its abundance peaks then falls as the abundance of its parent decreases and slows its production rate and decay into radium exceeds its rate of production.

The ratio of thorium to uranium resulting from decay is a steadily increasing amount that stabilises at 0.44.

Notice the equilibrium value we calculated (0.31) is the value of this ratio when the amount of thorium reaches its maximum. It’s easy to find when the amount of thorium peaks by differentiating the equation for thorium assuming the initial amount of thorium is zero

\( \frac{d N_T}{dt} = \frac{\lambda_U N_U^0 (-\lambda_U e^{-\lambda_U t} + \lambda_T e^{-\lambda_T t})}{\lambda_T – \lambda_U} \)The numerator is zero when

\( \lambda_U e^{-\lambda_U t} = \lambda_T e^{-\lambda_T t} \)If we rearrange

\( e^{(\lambda_T – \lambda_U) t} = \frac{\lambda_T}{\lambda_U} \)Take logs

\( (\lambda_T – \lambda_U) t = \ln \frac{\lambda_T}{\lambda_U} \)And

\( t = \frac{\ln \frac{\lambda_T}{\lambda_U}}{\lambda_T – \lambda_U} \)Using the decay constants for uranium and thorium the amount of thorium peaks after about 185,000 years. We can also see from the graph above that the ratio is sensitive up to around 250,000 years which means it works well for archaeological remains from hominids.

The post Radioactive Decay and Neanderthal Art appeared first on Dr Rom STEM Tuition.

]]>The post Circle Theorems appeared first on Dr Rom STEM Tuition.

]]>Below is a circle with a red line cutting through it joining two points on the circle: A and B. This line is called a ** chord** and the little bit of the circle that it chops off at the top is the

- If we join any point on the circle to A and B this forms a triangle. The angle subtended by the lines from A and B to any point will
*always*be equal.

To illustrate this I’ve drawn two triangles below. See how they make the shape of a butterfly? That’s the first theorem: *Angles in the same segment are equal*

If we move the two points so the butterfly gets a bit squidged notice that the angles are exactly the same, 45 degrees. Bear in mind that in some questions where this theorem applies the chord AB won’t be drawn, but you should still be able to see the butterfly shape.

Slide the two points at the bottom of the circle around and see whether the angles change. Then try sliding those points above the chord into the minor segment at the top. Does the angle change now? Why is that? Are we in another segment? Now try moving the points on the red arc. What’s happened to the angles now?

To spot this pattern the key ingredients are:

- A chord
- The centre of the circle
- A point on the circumference

I’m a Star Trek fan and to me this pattern looks like a Starfleet symbol:

Slide the two points at the bottom of the circle around and see whether the angles change. What happens if you move the point at the top? To see another theorem slide one of the points so that the blue lines make a diameter and the angle alpha is 180 degrees. What happens to the angle beta between the red lines?

If you see a quadrilateral where all the corners lie on the circle’s circumference, this is called a * cyclic quadrilateral*, the opposite angles of the quadrilateral add up to 180 degrees.

Slide the two points at the top of the circle around and see whether the angles change. See how the red and blue angles always match?

This is probably the most tricky of the circle theorems. It is easy to spot if you recognise three ingredients:

- A circle
- A tangent to the circle
- A triangle inside the circle with one corner lying on the tangent

This combination means you can relate the angles of the triangle and the angles to the tangent on the opposite side. To see what this means try dragging the points around in the picture below.

Slide the two points at the top of the circle around and see whether the angles change. See how the red and blue angles always match?

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]]>The post Simultaneous Equations appeared first on Dr Rom STEM Tuition.

]]>Simultaneous equations are easy if you follow these simple steps. Rearrange the first equation to get an expression for x, substitute that into the second equation and solve it for the value of y. Then once you have a number for y you can find the value of x by substituting into either equation.

An equation like y=mx + c defines a set of points for which a relationship between the x value and y value are true. For example, if we have the line y=x

- Point (1,1) lies on the line because the equation is true for that point (y=x=1).
- Point (2,1) doesn’t lie on the line because y=1 which is not equal to the x value 2.

In this way * all points on a line are solutions of the line’s equation*.

A set of simultaneous equations defines a set of points for which *multiple* statements are true. For example if both y=x *and* y=2x+3 then only a single point in the plane can be true for both equations. The solution is now a single point and in this article we’ll see how to calculate that point. We’ll also see how to calculate the solutions for more complex curves and lines.

Let’s start simple and work our way up to more complex examples.

\(\begin{align} y &= x \\\\ y &= 2x + 3 \end{align}\)If both statements are true then we can solve this problem in two ways, but both use the same idea. We are trying to get rid of one variable, either x or y, so that we can solve for the other variable. First, let’s do it by substitution. We know that \(y=x\) so we can substitute that into \(y=2x+3\) and solve. Let’s try it.

- Label your equations (1) and (2) so you know what you’ve got
- Choose one equation, say (1), and rearrange it to get just y on the left hand side
- Substitute for y in equation (2), now you have an equation with just y’s and numbers
- Rearrange this equation to solve for the value of y
- Substitute the numerical value of y into equation (1) and rearrange to get the value of x

All we did was substitute \(x=y\) in \(y=2x+3\) then move all the y’s onto one side of the equals sign and all the numbers onto the other side. Now we can find x easily because we know \(y=x\) so the solution is (-3,-3). We can easily check this is a solution for both equations because \(-3 = -3\) (from the equation \(y=x\)) and \(-3=2 \times -3 + 3=-6 + 3\).

- Label your equations (1) and (2) so you know what you’ve got
- Choose to get rid of x or y. Let’s say you choose to get rid of y.
- Multiply (1) and/or (2) by some number so you have the same number of y’s
- Subtract the multiplied up version of (1) from (2)
- Rearrange this equation to solve for the value of x
- Substitute the numerical value of x into equation (1) and rearrange to get the value of y

The other method is to subtract one equation from the other to get rid of x or y. Then we solve for x or y just as we did by substitution. We have a y on its own in both equations so by subtracting the bottom equation from the top

\(\begin{align}y – y &= (2x+3) – x \\\\ 0 &= x + 3 \end{align}\)We know \(x+3=0\) so \(x=-3\) and as before we know \(y=x\) so the solution is (-3,-3).

Two lines can only ever cross at one point (or no point if they’re parallel). A quadratic and a line are more complicated because they can cross at two points, one point or no point. Two solutions must be reminding you of something… yes to solve these problems you will have to solve a quadratic equation!

Let’s visualise what’s going on here. A quadratic equation is a parabola and the solutions are the points where the line cuts the parabola. The equations of the line and the parabola are

\(\begin{align} y &= x-2 \\\\ y &= x^2 + x – 6 \end{align}\)

To solve this we just substitute to get rid of one of the variables, like this.

\(\begin{align} y &= x^2 + x – 6 \\\\ y &= (y+2)^2 + (y-2) – 6 \\\\ y &= (y+2)(y+2) + y – 4 \\\\ y &= y^2 + 4y + 4 + y – 4 \\\\ 0 &= y^2 + 4y \\\\ 0 &= y(y+4) \end{align}\)This must have two solutions: either y+4=0, in which case y=-4 or y=0. That’s because we’re multiplying two numbers together and they’re producing zero so either one or the other must be zero. By substituting we can then work out the values of x for the these two solutions. We know that x=y+2, so the solutions are (-2,-4) and (2,0). You can also see these are the solutions by looking at the graph above.

The equation of a circle with radius r is

\(x^2+y^2=r^2\)If we have a pair of simultaneous equations that consist of a circle and a line then, just as with a parabola, there are three possibilities:

- Two solutions because the line meets the circumference of the circle in two places
- One solution where the line is a tangent to the circle at a single point
- No solutions if the line does not cut the circle

This means the equation we end up solving is a quadratic because this has two, one or zero solutions. Here’s an example.

The equations of the line and the circle are:

\(\begin{align} y &= -2x + 5 \\\\ x^2 + y^2 & = 10 \end{align}\)Let’s solve this by substitution. We can replace the y in the circle equation with -2x + 5 from the line equation then it becomes a quadratic like so:

\(\begin{align} x^2 + (-2x + 5)^2 &= 10 \\\\ x^2 + 4x^2 – 10x – 10x + 25 &= 10 \\\\ 5x^2 – 20x + 15 &= 0 \\\\ x^2 – 4x + 3 &= 0 \\\\ (x-1)(x-3) &= 0 \end{align}\)So the solutions are x=1 and x=3 and substituting into the line equation we can work out the two points where the line intersects the circle are (1,3) and (3,-1).

Here is a worksheet to give you a bit more practice:

The post Simultaneous Equations appeared first on Dr Rom STEM Tuition.

]]>The post Using Maths to Make Crispier Roast Potatoes appeared first on Dr Rom STEM Tuition.

]]>I was intrigued by a Tweet from Edge Hotel School in Essex about making roast potatoes crispier.

Dear Heston Blumenthal, we are not trying to roast your technique but we have a challenge for you..... @TheHestonTeam @TastyPotatoes @IoH_Online @TheCrownatBray pic.twitter.com/zYTnPgkwgK

— Edge Hotel School (@EdgeHotelSchool) January 15, 2018

Normally people quarter their potatoes before roasting them so that the four pieces are about the same size. What the students suggested was making diagonal cuts at thirty degrees to the long axis of the potato. Their calculations show that this increases the surface area of the slices by 65%. Here's their diagram.

First let's check their sums. A quick rifle around my vegetable basket showed that a typical potato was 12 cm long, 9 cm wide and 8 cm thick. We can assume that the potato is roughly elliptical in cross-section to make the sums easier because the area of an ellipse is pi times its radius in either axis.

The long axis and short axis of an ellipse are called the major and minor axes, and so an ellipse has two radii: the semi-major axis (labelled "a" below) and semi-minor axis (labelled "b"). For a circle the values of a and b are equal and so we get a single radius r=a=b.

This isn't just about potatoes; if you study planetary orbits, which are elliptical, you will also see these terms used. Orbits which are very squashed are defined as being very "eccentric" where eccentricity is zero for circles and approaches 1 for extremely elongated orbits,

\(e=\sqrt{1-\frac{b^2}{a^2}}\)

Earth's orbit is very nearly circular, it's eccentricity is 0.02, Mercury's a bit higher at 0.2. For really high eccentricity you have to look at comets that drift into the inner solar system and have extremely elongated orbits. The orbit of Halley's Comet has an eccentricity of 0.97.

Moving from space back to Earth, let's slice the potato traditionally. First we slice along the long dimension. The major axis is 12 cm and the minor axis is 8 cm so the "radii" are 6 cm and 4 cm. We end up with the potato cut in half and the cut surface area of one piece is

\(6 \times 4 \times \pi = 24 \pi\)

The cut surface area of both pieces is twice that. Then we slice again through the waist of the potato to make 4 pieces. This slice has a cross-section with semi-major axis 4.5 cm (half of 9 cm width) and semi-minor axis of 4 cm (half of 8 cm thickness). Each face of this cut has area

\(4 \times 4.5 \times \pi = 18 \pi\)

Now let's try the new method. The long slice is the same as before but now we have two diagonal cuts at 30 degrees to that first cut. Here's a 3d model:

If you slice a 3d ellipsoid the slice surfaces are ellipses. This cut produces two side wedges each of which has two faces of area 13.8 pi. In total that's:

**Traditional**: 24 pi + 18 pi = 42 pi**New Method**: 24 pi + 2 x 13.8 pi = 52 pi

The diagonal slices according to these calculations produce cut surfaces which are **24% larger in area** than the traditional method. That's well below the 65% increase that's in the tweet from Edge Hotel School, but is still an improvement. The precise benefit depends on the elongation of your potato with longer potatoes having a larger surface area. In astronomical terms we'd say the more eccentric your potato the crisper your tatties.

The post Using Maths to Make Crispier Roast Potatoes appeared first on Dr Rom STEM Tuition.

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