2.3 Centre of Mass and Torque

It might help to know that $-x^2+2x+1 = 2-(x-1)^2$.

$(1,1)$

The centroid of a region doesn't have to be a point in the region.

$(0,0)$

Read over the very beginning of Section 2.3 in the CLP-2 text, specifically Equation 2.3.1.

In general, false.

Use Equation 2.3.1 in the CLP-2 text.

$3.5$ metres from the left end

Imagine cutting out the shape and setting it on top of a pencil, so that the pencil lines up with the vertical line $x=a$. Will the figure balance, or fall to one side? Which side?

(a) to the left (b) to the left (c) not enough information
(d) along the line $x=a$ (e) to the right

You can find the heights of the centres of mass using symmetry.

$\displaystyle\frac{39200\pi}{9}(12-\pi)\approx 121,212~\text{J}$

Think about whether your answers should have repetition.

(a), (b) $\dfrac{1}{x}\,\dee{x}$ (c), (d) $\log 3$ (e), (f) $\dfrac{2}{\log 3}$

The definition of a definite integral (Definition 1.1.9 in the CLP-2 text) will tell you how to convert your limits of sums into integrals.

(a) $\displaystyle\frac{\sum\limits_{i=1}^n\left[\frac{b-a}{n}\rho\left(\textcolor{red}{a+\left(i-\tfrac12\right)(\tfrac{b-a}{n})}\right)\times\left( \textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)\right]}{\sum\limits_{i=1}^n\frac{b-a}{n}\rho\left(\textcolor{red}{ a+(i-\tfrac12)\left(\tfrac{b-a}{n}\right) } \right)}$ (b) $\displaystyle\bar x=\frac{\int_a^b x\rho(x)\,\dee{x}}{\int_a^b\rho(x)\,\dee{x}}$

In (a), the slices all have the same width, so the area of the slices is larger (and hence the density of $R$ is higher) where $T(x)-B(x)$ is larger.

(a)

(b) $(T(x)-B(x))\,\dee{x}$ (c) $T(x)-B(x)$ (d) $\bar x=\dfrac{\int_a^b x({T(x)-B(x)})\,\dee{x}}{\int_a^b({T(x)-B(x)})\,\dee{x}}$

Part (a) is a significantly different model from the last question.

(a) The strips between $x=a$ and $x=a'$ at the left end of the figure all have the same centre of mass, which is the $y$-value where $T(x)=B(x)$, $x<0$. So, there should be multiple weights of different mass piled up at that $y$-value.

Similarly, the strips between $x=b'$ and $x=b$ at the right end of the figure all have the same centre of mass, which is the $y$-value where $T(x)=B(x)$, $x>0$. So, there should be a second pile of weights of different mass, at that (higher) $y$-value.

Between these two piles, there are a collection of weights with identical mass distributed fairly evenly. The top and bottom ends of $R$ (above the uppermost pile, and below the lowermost pile) have no weights.

One possible answer (using twelve slices):

(b) The area of the strip is $(T(x)-B(x))\,\dee{x}$, and its centre of mass is at height $\dfrac{T(x)+B(x)}{2}$.
(c) $\displaystyle\bar y=\frac{\int_a^b \big(T(x)^2-B(x)^2\big)\,\dee{x}}{2\int_a^b\big( T(x)-B(x)\big)\,\dee{x}}$

Which method involves more work: horizontal strips or vertical strips?

$\displaystyle\bar x = -\frac{1}{3} \int_{-1}^0 6x^2\ \dee{x}$

This is a straightforward application of Equation 2.3.2 in the CLP-2 text.

$\bar x=\dfrac{14}{3}$

Remember the derivative of arctangent is $\frac{1}{1+x^2}$

$\displaystyle \bar x=\frac{\log 10.1}{2(\arctan 10 + \arctan(3))}\approx 0.43$

This is a straightforward application of Equations 2.3.3 and 2.3.4 in the

CLP-2 text. Note that you're only asked for the $y$-coordinate of the centroid.

$\displaystyle\bar y = \frac{3}{4e}-\frac{e}{4}$

You can use a trigonometric substitution to find the area, then a partial fraction decomposition to find the $y$-coordinate of the centroid. Remember $\sin(1/2)=\pi/6$.

(a)

(b) $\dfrac{3\log 3}{8\pi}$

Vertical slices will be easier than horizontal. An integration by parts might be helpful to find $\bar x$, while trigonometric identities are important to finding $\bar y$.

$\displaystyle\bar x=\frac{\frac{\pi}{4}\sqrt{2}-1}{\sqrt{2}-1}$ and $\displaystyle\bar y=\frac{1}{4(\sqrt{2}-1)}$

No trigonometric substitution is necessary if you're clever with your $u$-substitutions, and remember the derivative of arctangent.

(a) $\displaystyle\bar x = \frac{k}{A}\big[\sqrt{2}-1\big]$, $\displaystyle\bar y = \frac{k^2\pi}{8A}$ (b) $\displaystyle k=\frac{8}{\pi}\big[\sqrt{2}-1\big]$

In $R$, the top function is $x-x^2$, and the bottom function is $x^2-3x$.

(a)

(b) $\dfrac{8}{3}$ (c) $1$

Remember $\diff{}{x}\{\arctan x\} = \frac{1}{1+x^2}$.

$\dfrac{2}{\pi}\log 2\approx 0.44127$

You can save quite a bit of work by, firstly, exploiting symmetry and, secondly, thinking about whether it is more efficient to use vertical strips or horizontal strips.

$\bar x=0$ and $\bar y= \dfrac{12}{24+9\pi}$

Sketch the region, being careful the domain of $\sqrt{9-4x^2}$. You can save quite a bit of work by exploiting symmetry.

(a) $\dfrac{9}{4}\pi$ (b) $\bar x = 0$ and $\bar y = \dfrac{4}{\pi}$

Horizontal slices will be easier than vertical.

$(\bar x, \bar y) = \left( 1,-\dfrac{2}{\pi}\right)$

Start with a picture: whether you use vertical slices or horizontal, you'll need to break your integral into multiple pieces.

$\displaystyle\left(\frac{e^2-3/2}{e^2-5/2},\frac{e^4-7}{4e^2-10}\right)\approx (1.2,2.4)$

For practice, do the computation twice — once with horizontal strips and once with vertical strips. Watch for improper integrals.

$\bar y = \dfrac{8}{5}$

Draw a sketch. In part (b) be careful about the equation of the right hand boundary of $A$.

(a) $\displaystyle \bar x = \frac{8}{11}$, $\displaystyle\bar y = \frac{166}{55}$ (b) $\displaystyle\pi \int_0^4 y\,\dee{y} + \pi \int_4^6 (6-y)^2\,\dee{y}$

Draw a sketch. Rotating about a horizontal line is similar to rotating about the $x$-axis, but for the radius of a slice, you'll need to know $|y-(-1)|$: the distance from the outer edge of the region (the boundary function's $y$-value) to $y=-1$.

(a) $\displaystyle\bar y = \frac{e}{4} - \frac{3}{4e}$ (b) $\displaystyle\pi\left(\frac{e^{2}}{2}+2e -\frac{3}{2}\right)$

Go back to the derivation of Equation 2.3.3 in the CLP-2 text (centroid for a region) to figure out what to do when your surface does not have uniform density. We will consider a rod $R$ that reaches from $x=0$ to $x=4$, and the mass of the section of the rod along $[a,b]$ is equal to the mass of the strip of our rectangle along $[a,b]$.

$(3,1.5)$

Horizontal slices will help you, where symmetry doesn't, to set up a rod $R$ whose centre of mass is the same as one coordinate of the centre of mass of the circle. When you're integrating, trigonometric substitutions are sometimes the easiest way, and sometimes not.

The equation of a circle of radius 3, centred at $(0,3)$, is $x^2+(y-3)^2=9$.

$(0,3.45)$

The model in the question gives you the setup to solve this problem. You know how to find the centre of mass of a rod–that's Equation 2.3.2 in the CLP-2 text — so all you need to find is $\rho(y)$, the density of the rod at position $y$. To find this, consider a thin slice of the cone at position $y$ with thickness $\dee{y}$. Its volume $V(y)$ is the same as the mass of the small section of the rod at position $y$ with thickness $\dee{y}$. So, the density of the rod at position $y$ is $\rho(y)=\frac{V(y)}{\dee{y}}$.

(a) $\dfrac{h}{4}$ (b) $\displaystyle\frac{\frac{1}{2}h^2k - \frac{2}{3}hk^2+\frac{1}{4}k^3}{h^2-hk+\frac{1}{3}k^2}$

Use similar triangles to show that the shape of the lower (also upper) half of the hourglass is a truncated cone, where the untruncated cone would have had a height 10 cm.

To calculate the centre of mass of the upturned sand using the result of Question 29, you should find $h=9.8$ (not $h=10$–think carefully about our model from Question 29) and $k=8.8$. For the centre of mass of the sand before turning, $h=10$ and $k=6$.

about 0.833 N

The techniques of Section 2.1 get pretty complicated here, so it's easiest to use the techniques we developed in Questions 6, 29, and 30 in this section. That is, (1) find the height of the centre of mass of the water in its starting and ending positions, and then (2) model the work done as the work moving a point mass with the weight of the water from the first centre of mass to the second.

The height change of the centre of mass is all that matters to calculate the work done against gravity, so you only have to worry about the height of the centres of mass.

(a) 17,150$\pi$ J (b) $\displaystyle\frac{2450}{9}\pi\left(8\pi-9\right)\approx 13,797\text{ J}$ (c) about 74%

The area of $R$ is precisely one, so the error in your approximation is the error involved in approximating $\int_0^{\sqrt{\pi/2}}2x^2\sin(x^2)\,\dee{x}$.

$\displaystyle \bar x =\frac{\pi}{162}\sqrt{\frac{\pi}{2}}\bigg[ \sin\left(\frac{\pi}{72}\right)+2\sin\left(\frac{\pi}{18}\right)+9\sin\left(\frac{\pi}{8}\right)+ 8\sin\left(\frac{2\pi}{9}\right)+25\sin\left(\frac{25\pi}{72}\right) +9\bigg]\approx 0.976$