Author Archives: Engineering Math

Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of $0.50\ \text{km/s}$ . We are asked to verify this value if we are given a radius of $r=0.100\ \text{m}$ and angular velocity of $\omega = 50000 \ \text{rev/min}$ . We are going to use the formula

v = r \omega

Since we are given a linear speed in $\text{km/s}$ , we are going to convert the radius to $\text{km}$ , and the angular velocity to $\text{rad/sec}$

r=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}

\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

\begin{align*} v & = r \omega \\ \\ v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\ v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

Parent	Satellite	Average orbital radius r(km)	Period T(y)	r³ / T² (km³ / y²)
Sun	Earth	$1.496 \times 10^{8}$	1	$3.35 \times 10^{24}$

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496 \times10^{8} \ \text{km}

The angular velocity is

\begin{align*} \omega & = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ \omega & = 1.9910 \times 10^{-7}\ \text{rad/sec} \end{align*}

The linear velocity is

\begin{align*} v & = r \omega \\ \\ v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\ v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The linear velocity is about 30 km/s.

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Problem 6-17: The acceleration due to gravity at the position of a satellite located above the Earth

What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

Solution:

The acceleration due to gravity of a body and the Earth is given by the formula

g= G \frac{M}{r^2}

where $G$ is the gravitational constant, $M$ is the mass of the Earth, and $r$ is the distance of the object to the center of the Earth. We know that the approximate radius of the Earth is $r=6.3781 \times 10^6 \ \text{m}$ .

The percentage of the acceleration at 300 km above the Earth of the acceleration due to gravity at Earth’s surface is

\displaystyle \frac{\left( \frac{GM}{r^2} \right)_2}{\left( \frac{GM}{r^2} \right)_1} \times 100\%

Note that the subscript 2 indicates the satellite located 300 km above the Earth, and the subscript 1 indicates the object at the Earth’s surface. Also, from the expression above, we can cancel $G$ and $M$ from the numerator and denominator because these are constants. So, we are down to

\frac{\left( \frac{1}{r^2} \right)_2}{\left( \frac{1}{r^2} \right)_1} \times 100\% = \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\%

Substituting the values, we have

\begin{align*} \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\% & = \frac{\left( 6.3781 \times 10^6 \ \text{m} \right)^{2}}{\left( 6.3781 \times 10^6 \ \text{m}+300 \times 10^{3} \ \text{m} \right)^{2}} \times 100\% \\ \\ & = 91.2172\% \\ \\ & = 91.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The percentage of the acceleration at the Earth’s surface of the acceleration due to gravity at the position of a satellite located 300 km above the Earth is about 91.2%.

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Problem 6-16: Calculating the centripetal acceleration of an ice skater’s nose

Olympic ice skaters are able to spin at about 5.00 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9.00 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

Solution:

We are given an angular velocity, $\omega = 5 \ \text{rev/sec}$

Part A

For this part, we are asked to convert the angular velocity to units of radians per second.

\begin{align*} \omega & = \frac{5.00\ \text{rev}}{\text{sec}}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \\ \\ \omega & = 31.4159 \ \text{rad/sec} \\ \\ \omega & = 31.4 \ \text{rad/sec}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part, we are asked to solve for the centripetal acceleration. We are going to use the formula $a_{c} = r \omega ^2$ given $r=0.120\ \text{m}$ and $\omega = 31.4159 \ \text{rad/s}$ .

\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.120 \ \text{m} \right) \left( 31.4159 \ \text{rad/s} \right)^2 \\ \\ a_{c} & = 118.4350 \ \text{m/s}^2 \\ \\ a_{c} & = 118 \ \text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

For this part, we are going to directly solve the centripetal acceleration.

\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.120 \ \text{m} \right)\left( \frac{9\ \text{rev}}{\text{s}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}}\right)^2 \\ \\ a_{c} & = 383.7302 \ \text{m/s}^2 \\ \\ a_{c} & = 384 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That is quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity! It is no wonder that he ruptured small blood vessels in his spins.

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Curing the Compulsive Gambler: Challenging Probability Problem

Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)

Solution:

If Mr. Brown wins once in 36 turns, he is even with the casino. His probability of losing all 36 times is $\displaystyle \left( \frac{37}{38} \right)^{36} \approx 0.383$ . In a single turn, his expectation is

35\left( \frac{1}{38} \right)-1\left( \frac{37}{38} \right) = - \frac{2}{38}\ \text{dollars}

and in 36 turns

-\frac{2}{38}\left( 36 \right) = -1.89 \ \text{dollars}

Against Kind Friend, Mr. Brown has an expectation of

+20\left( 0.617 \right)-20\left( 0.383 \right)\approx +4.68 \ \text{dollars}

And so, all told, Mr. Brown gains +4.68 – 1.89 = + 2.79 dollars per 36 trials; he is finally making money at roulette. Possibly Kind Friend will be cured first. Of course, when Brown loses all 36, he is out $56, which may jolt him a bit.

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Chuck-a-Luck: Challenging Probability Problem

Chuck-a-Luck is a gambling game often played at carnivals and gambling houses. A player may bet on anyone of the numbers 1, 2, 3, 4, 5, 6. Three dice are rolled. If the player’s number appears on one, two, or three of the dice, he receives respectively one, two, or three times his original stake plus his own money back; otherwise, he loses his stake. What is the player’s expected loss per unit stake? (Actually, the player may distribute stakes on several numbers, but each such stake can be regarded as a separate bet.)

Solution:

Let us compute the losses incurred (a) when the numbers on the three dice are different, (b) when exactly two are alike, and (c) when all three are alike. An easy attack is to suppose that you place a unit stake on each of the six numbers, thus betting six units in all. Suppose the roll produces three different numbers, say 1, 2, 3. Then the house takes the three unit stakes on the losing numbers 4, 5, 6 and pays off the three winning numbers 1, 2, 3. The house won nothing, and you won nothing. That result would be the same for any roll of three different numbers.

Next suppose the roll of the dice results in two of one number and one of a second, say 1, 1, 2. Then the house can use the stakes on numbers 3 and 4 to payoff the stake on number 1, and the stake on number 5 to payoff that on number 2. This leaves the stake on number 6 for the house. The house won one unit, you lost one unit, or per unit stake you lost 1/6.

Suppose the three dice roll the same number, for example, 1, 1, 1. Then the house can pay the triple odds from the stakes placed on 2, 3, 4 leaving those on 5 and 6 as house winnings. The loss per unit stake then is 2/6. Note that when a roll produces a multiple payoff the players are losing the most on the average.

To find the expected loss per unit stake in the whole game, we need to weight the three kinds of outcomes by their probabilities. If we regard the three dice as distinguishable –say red, green, and blue — there are $6 \times 6 \times 6= 216$ ways for them to fall.

In how many ways do we get three different numbers? If we take them in order, 6 possibilities for the red, then for each of these, 5 for the green since it must not match the red, and for each red-green pair, 4 ways for the blue since it must not match either of the others, we get $6 \times 5 \times 4 = 120$ ways.

For a moment skip the case where exactly two dice are alike and go on to three alike. There are just 6 ways because there are 6 ways for the red to fall and only 1 way for each of the others since they must match the red.

This means that there are $216 - 126 = 90$ ways for them to fall two alike and one different. Let us check that directly. There are three main patterns that give two alike: red-green alike, red-blue alike, or green-blue alike. Count the number of ways for one of these, say red-green alike, and then multiply by three. The red can be thrown 6 ways, then the green 1 way to match, and the blue 5 ways to fail to match, or 30 ways. All told then we have $3 \times 30 = 90$ ways, checking the result we got by subtraction.

We get the expected loss by weighting each loss by its probability and summing as follows:

\underbrace{\frac{120}{216}\times 0}_\text{none alike} + \underbrace{\frac{90}{216}\times \frac{1}{6}}_\text{2 alike}+\underbrace{\frac{6}{216}\times \frac{2}{6}}_\text{3 alike} = \frac{17}{216} \approx 0.079

Thus, you lose about 8% per play. Considering that a play might take half a minute and that government bonds pay you less than 4% interest for a year, the attrition can be regarded as fierce.

This calculation is for regular dice. Sometimes a spinning wheel with a pointer is used with sets of three numbers painted in segments around the edge of the wheel. The sets do not correspond perfectly to the frequencies given by the dice. In such wheels I have observed that the multiple payoffs are more frequent than for the dice, and therefore the expected loss to the bettor greater.

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Coin in Square: Challenging Probability Problem

In a common carnival game, a player tosses a penny from a distance of about 5 feet onto the surface of a table ruled in 1-inch squares. If the penny (3/4 inch in diameter) falls entirely inside a square, the player receives 5 cents but does not get his penny back; otherwise, he loses his penny. If the penny lands on the table, what is his chance to win?

Solution:

When we toss the coin onto the table, some positions for the center of the coin are more likely than others, but over a very small square we can regard the probability distribution as uniform. This means that the probability that the center falls into any
region of a square is proportional to the area of the region, indeed, is the area of the region divided by the area of the square. Since the coin is 3/8 inch in radius, its center must not land within 3/8 inch of any edge if the player is to win. This restriction generates a square of side 1/4 inch within which the center of the coin must lie for the coin to be in the square. Since the probabilities are proportional to areas, the probability of winning is $\displaystyle \left( \frac{1}{4} \right)^2 = \frac{1}{16}$ . Of course, since there is a chance that the coin falls off the table altogether, the total probability of winning is smaller still. Also, the squares can be made smaller by merely thickening the lines. If the lines are 1/16 inch wide, the winning central area reduces the probability to $\displaystyle \left( \frac{3}{16} \right)^{2} = \frac{9}{256}$ or less than $\displaystyle \frac{1}{28}$ .

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Trials until First Success: Challenging Probability Problem

On the average, how many times must a die be thrown until one gets a 6?

Solution:

Let $p$ be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let $q = 1 - p$ ):

Trial	Probability of success on trial
1	$p$
2	$pq$
3	$pq ^2$
.	.
.	.
.	.

The sum of the probabilities is

\begin{align*} p+pq+pq^2+\ldots & = p\left( 1+q+q^2+\ldots \right) \\ \\ & = \frac{p}{1-q} \\ \\ & = \frac{p}{p} \\ \\ & = 1 \end{align*}

The mean number of trials, $m$ , is by definition,

m = p + 2pq + 3pq^2 + 4pq^3+ \ldots

Note that our usual trick for summing a geometric series works:

qm = pq + 2pq^2+3pq^3 + \ldots

Subtracting the second expression from the first gives

m-qm=p+pq+pq^2+\ldots

m\left( 1-q \right) = 1

Consequently,

mp=1

and

m=1/p

We see that $p=1/6$ , and so $m=6$ .

On the average, a die must be thrown 6 times until one gets a 6.

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The Flippant Juror: Challenging Probability Problem

A three-man jury has two members each of whom independently has probability p of making the correct decision and a third member who flips a coin for each decision (majority rules). A one-man jury has probability p of making the correct decision. Which jury has the better probability of making the correct decision?

Solution:

The two juries have the same chance of a correct decision. In the three-man jury, the two serious jurors agree on the correct decision in the fraction $p \times p = p^2$ of the cases, and for these cases the vote of the joker with the coin does not matter. In the other correct decisions by the three-man jury, the serious jurors vote oppositely, and the joker votes with the “correct” juror. The chance that the serious jurors split is $p\left( 1-p \right)+\left( 1-p \right)p$ or $2p\left( 1-p \right)$ . Halve this because the coin favors the correct side half the time. Finally, the total probability of a correct decision by the three-man jury is $p^{2}+p\left( 1-p \right) =p^{2}+p-p^{2}=p$ , which is identical with the probability given for the one-man jury.

The two options have equal probability of making the correct decision.

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Problem 6-13: The motion of the WWII fighter plane propeller

The propeller of a World War II fighter plane is 2.30 m in diameter.

(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?

(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?

(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

Solution:

Part A

We are converting the angular velocity $\omega = 1200\ \text{rev/min}$ into radians per second.

\begin{align*} \omega = & \frac{1200\ \text{rev}}{\text{min}}\times \frac{2\pi \ \text{radian}}{1\ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{sec}} \\ \\ \omega = & 125.6637 \ \text{radians/sec} \\ \\ \omega = & 126 \ \text{radians/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are now solving the linear speed of the tip of the propeller by relating the angular velocity to linear velocity using the formula $v = r \omega$ . The radius is half the diameter, so $r= \frac{2.30\ \text{m}}{2} = 1.15 \ \text{m}$ .

\begin{align*} v & = r \omega \\ \\ v & = \left( 1.15 \ \text{m} \right)\left( 125.6637 \ \text{radians/sec} \right) \\ \\ v & = 144.5132 \ \text{m/s} \\ \\ v & = 145 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

From the computed linear speed and the given radius of the propeller, we can now compute for the centripetal acceleration $a_{c}$ using the formula

a_{c} = \frac{v^2}{r}

If we substitute the given values, we have

\begin{align*} a_{c} & = \frac{v^2}{r} \\ \\ a_{c} & = \frac{\left( 144.5132 \ \text{m/s} \right)^2}{1.15 \ \text{m}} \\ \\ a_{c} & = 18160.0565 \ \text{m/s}^2 \\ \\ a_{c} & = 1.82\times 10^{4} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can convert this value in multiples of $g$

\begin{align*} a_{c} & = 18160.0565 \ \text{m/s}^2 \times \frac{g}{9.81 \ \text{m/s}^2} \\ \\ a_{c} & = 1851.1780 g \\ \\ a_{c} & = 1.85\times 10^{3} \ g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-12: The approximate total distance traveled by planet Earth since its birth

Taking the age of Earth to be about 4×10⁹ years and assuming its orbital radius of 1.5 ×10¹¹ m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).

Solution:

First, we need to compute for the linear velocity of the Earth using the formula below knowing that the Earth has 1 full revolution in 1 year

v=r\omega

where $r=1.5\times 10^{11} \ \text{m}$ and $\omega = 2\pi \ \text{rad/year}$ . Substituting these values, we have

\begin{align*} v & = r \omega \\ \\ v & = \left( 1.5\times 10^{11} \ \text{m} \right)\left( 2 \pi \ \text{rad/year} \right) \\ \\ v & = 9.4248\times 10^{11} \ \text{m/year} \end{align*}

Knowing the linear velocity, we can compute for the total distance using the formula

\Delta x = v \Delta t

We can now substitute the given values: $v = 9.4248\times 10^{11} \ \text{m/year}$ and $\Delta t = 4\times 10^{9} \ \text{years}$ .

\begin{align*} \Delta x & = v \Delta t \\ \\ \Delta x & = \left( 9.4248\times 10^{11} \ \text{m/year} \right) \left( 4\times 10^{9} \ \text{years} \right) \\ \\ \Delta x & = 3.7699 \times 10^{21} \ \text{m} \\ \\ \Delta x & = 4 \times 10^{21} \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

Part B

What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

Solution:

Olympic ice skaters are able to spin at about 5.00 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9.00 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

Solution:

Part A

Part B

Part C

Part D

Mr. Brown always bets a dollar on the number 13 at roulette against the advice of Kind Friend. To help cure Mr. Brown of playing roulette, Kind Friend always bets Brown $20 at even money that Brown will be behind at the end of 36 plays. How is the cure working?

(Most American roulette wheels have 38 equally likely numbers. If the player’s number comes up, he is paid 35 times his stake and gets his original stake back; otherwise, he loses his stake)

Solution:

Solution:

Solution:

On the average, how many times must a die be thrown until one gets a 6?

Solution:

Solution:

The propeller of a World War II fighter plane is 2.30 m in diameter.

(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?

(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?

(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

Solution:

Part A

Part B

Part C

Taking the age of Earth to be about 4×109 years and assuming its orbital radius of 1.5 ×1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).

Solution:

Taking the age of Earth to be about 4×10⁹ years and assuming its orbital radius of 1.5 ×10¹¹ m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).