Category Archives: Engineering Mathematics Blog

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 6-15: The centripetal acceleration at the tip of a helicopter blade

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.

(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

Solution:

Part A

We are given the following values: $r=4.00\ \text{m}$ , and $\omega = 300 \ \text{rev/min}$ .

Let us convert the angular velocity to unit of radians per second.

\omega = 300 \ \frac{\text{rev}}{\text{min}} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}}\times \frac{1\ \text{min}}{60 \ \text{sec}} = 31.4159 \ \text{rad/sec}

The centripetal acceleration at the tip of the helicopter blade can be computed using the formula

a_{c} = r \omega ^2

If we substitute the given values into the formula, we have

\begin{align*} a_{c} & = r \omega^2 \\ \\ a_{c} & = \left( 4.00\ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 3947.8351 \ \text{m/s}^2 \\ \\ a_{c} & = 3.95 \times10^3 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are asked to solve for the linear velocity of the blade’s tip. We are going to use the formula

v=r \omega

We just needed to substitute the given values into the formula.

\begin{align*} v & = r \omega \\ \\ v & = \left( 4.00 \ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right) \\ \\ v & = 125.6636 \ \text{m/s} \\ \\ v & = 126 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Let us compare this with the speed of light which is 340 m/s.

\frac{125.6636 \ \text{m/s}}{340\ \text{m/s}} \times 100 \%= 36.9599 \% =37.0\%

The linear velocity of the blades tip is 37.0% of the speed of light.

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Problem 6-14: The centripetal acceleration and a linear speed of a point on an edge of an ordinary workshop grindstone

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.

(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.

(b) What is the linear speed of a point on its edge?

Solution:

We are given the following values: $r=7.50\ \text{cm}$ , and $\omega = 6500\ \text{rev/min}$ . We need to convert these values into appropriate units so that we can come up with sensical units when we solve for the centripetal acceleration.

r = 7.50 \ \text{cm} = 0.075 \ \text{m}

\omega = 6500 \ \text{rev/min} \times\frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{min}}{60\ \text{sec}} = 680.6784 \ \text{rad/sec}

Part A

We are asked to solve for the centripetal acceleration $a_{c}$ . Basing on the given data, we are going to use the formula

a_{c} = r \omega ^{2}

Substituting the given values, we have

\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.075 \ \text{m} \right) \left( 680.6784 \ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 34749.2313 \ \text{m/s}^2 \\ \\ a_{c} & = 3.47 \times 10^{4} \ \text{m/s} ^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Now, we can convert the centripetal acceleration in multiples of g.

\begin{align*} a_{c} & = 34749.2313 \ \text{m/s}^2 \times \frac{g}{9.81 \ \text{m/s}^2}\\ \\ a_{c} & =3542.2254g \\ \\ a_{c} & = 3.54\times 10^3 g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are then asked for the linear speed, $v$ of the point on the edge. So, we can use the given values to find the linear speed. We are going to use the formula

v=r\omega

If we substitute the given values, we have

\begin{align*} v & = r \omega \\ \\ v & = \left( 0.075 \ \text{m} \right)\left( 680.6784\ \text{rad/sec} \right) \ \ \\ \\ v & = 51.0509 \ \text{m/s} \\ \\ v & = 51.1 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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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Problem 6-11: Calculating the centripetal acceleration of a runner in a circular track

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If the runner completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of their centripetal acceleration as they run the curved portion of the track?

Solution:

Centripetal acceleration $a_{c}$ is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity $v$ and has the magnitude

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

We can solve for the constant speed of the runner using the formula

v=\frac{\Delta x}{\Delta t}

We are given the distance $\Delta x = 200 \ \text{m}$ , and the total time $\Delta t = 23.2\ \text{s}$ . Therefore, the velocity is

\begin{align*} v & =\frac{\Delta x}{\Delta t} \\ \\ v & = \frac{200\ \text{m}}{23.2\ \text{s}} \\ \\ v & = 8.6207\ \text{m/s} \end{align*}

From the given problem, we are given the following values: $r=30\ \text{m}$ . We now have the details to solve for the centripetal acceleration.

\begin{align*} a_{c} & = \frac{v^{2}}{r} \\ \\ a_{c} & = \frac{\left( 8.6207\ \text{m/s} \right)^2}{30\ \text{m}} \\ \\ a_{c} & = 2.4772\ \text{m/s}^{2} \\ \\ a_{c} & = 2.5\ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-10: The angular velocity of a person in a circular fairground ride

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

Solution:

Centripetal acceleration $a_{c}$ is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The relationship between the centripetal acceleration $a_{c}$ and the angular velocity $\omega$ is given by the formula

a_{c}=r\omega^{2}

Now, taking the formula and solving for the angular velocity:

\omega = \sqrt{\frac{a_{c}}{r}}

From the given problem, we are given the following values: $r=8.00\ \text{m}$ and $a_{c}=1.50\times 9.81 \ \text{m/s}^2=14.715\ \text{m/s}^2$ . If we substitute these values in the formula, we can solve for the angular velocity.

\begin{align*} \omega & = \sqrt{\frac{a_{c}}{r}} \\ \\ \omega & = \sqrt{\frac{14.715\ \text{m/s}^2}{8.00\ \text{m}}} \\ \\ \omega & = 1.3561\ \text{rad/sec} \\ \\ \end{align*}

Then, we can convert this value into its corresponding value at the unit of revolutions per minute.

\begin{align*} \omega & = 1.3561\ \frac{\text{rad}}{\text{sec}} \times \frac{60\ \text{sec}}{1\ \text{min}}\times \frac{1\ \text{rev}}{2\pi \ \text{rad}} \\ \\ \omega & = 12.9498\ \text{rev/min} \\ \\ \omega & = 13.0 \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Solution:

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

Solution:

Solution:

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.

(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

Solution:

Part A

Part B

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.

(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.

(b) What is the linear speed of a point on its edge?

Solution:

Part A

Part B

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:

Solution:

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow has an 8.00 m radius, at how many revolutions per minute will the riders be subjected to a centripetal acceleration whose magnitude is 1.50 times that due to gravity?

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).