Author Archives: Engineering Math

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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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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Problem 6-9: A construct your own problem on College Physics involving angular motion

Construct Your Own Problem

Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. Construct a problem in which you calculate the necessary angular velocity that assures the riders will not slide down the wall. Include a free body diagram of a single rider. Among the variables to consider are the radius of the cylinder and the coefficients of friction between the riders’ clothing and the wall.

Problem 6-8: An integrated problem involving circular motion, momentum, and projectile motion

Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

From the given problem, we are given the following values: $v=35.0\ \text{m/s}$ and $r=1.05\ \text{m}$ . We are required to solve for the angular velocity $\omega$ .

The linear velocity, $v$ and the angular velocity, $\omega$ are related by the equation

v=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into the formula, we can directly solve for the value of the angular velocity. That is,

\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{35.0\ \text{m/s}}{1.05\ \text{m}} \\ \\ \omega & = 33.3333\ \text{rad/sec} \\ \\ \omega & = 33.3 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part of the problem, we are going to use Newton’s second law of motion in term of linear momentum which states that the net external force equals the change in momentum of a system divided by the time over which it changes. That is

F_{net} = \frac{\Delta p}{\Delta t} = \frac{m\left( v_f - v_i \right)}{t}

For this problem, we are given the following values: $m=0.500\ \text{kg}$ , $t=20.0\times 10^{-3} \ \text{s}$ , $v_{f}=20.0\ \text{m/s}$ , and $v_{i}=0$ . Substituting all these values into the equation, we can solve directly for the value of the net external force.

\begin{align*} F_{net} & = \frac{\left( 0.500\ \text{kg} \right)\left( 20.0\ \text{m/s}-0\ \text{m/s} \right)}{20.0\times 10^{-3}\ \text{s}} \\ \\ F_{net} & = 500\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

This is a problem on projectile motion. In this particular case, we are solving for the range of the projectile. The formula for the range of a projectile is

R=\frac{v_{0}^2 \sin 2\theta}{g}

We are asked to solve for the maximum range, and we know that the maximum range happens when the angle $\theta$ is $45^\circ$ .

\begin{align*} R & = \frac{\left( 20.0\ \text{m/s} \right)^{2} \sin \left( 2\left( 45^\circ \right) \right)}{9.81 \ \text{m/s}^2} \\ \\ R & = 40.7747\ \text{m} \\ \\ R & = 40.8 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-7: Calculating the angular velocity of a truck’s rotating tires

A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?

Solution:

The linear velocity, $v$ and the angular velocity $\omega$ are related by the equation

v=r\omega \ \text{or} \ \omega=\frac{v}{r}

From the given problem, we are given the following values: $r=0.420 \ \text{m}$ and $v=32.0 \ \text{m/s}$ . Substituting these values into the formula, we can directly solve for the angular velocity.

\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{32.0 \ \text{m/s}}{0.420 \ \text{m}} \\ \\ \omega & = 76.1905 \ \text{rad/s} \\ \\ \omega & = 76.2 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Then, we can convert this into units of revolutions per minute:

\begin{align*} \omega & = 76.1905 \ \frac{\bcancel{\text{rad}}}{\bcancel{\text{sec}}}\times \frac{1 \ \text{rev}}{2\pi\ \bcancel{\text{rad}}}\times \frac{60\ \bcancel{\text{sec}}}{1\ \text{min}} \\ \\ \omega & = 727.5657\ \text{rev/min} \\ \\ \omega & = 728\ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-6: Calculating the linear velocity of the lacrosse ball with the given angular velocity

In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?

Solution:

The linear velocity, $v$ and the angular velocity, $\omega$ of a rotating object are related by the equation

v=r\omega

From the given problem, we have the following values: $\omega=30.0 \ \text{rad/s}$ and $r=1.30 \ \text{m}$ . Substituting these values in the formula, we can directly solve for the linear velocity.

\begin{align*} v & =r\omega \\ \\ v & = \left( 1.30 \ \text{m} \right)\left( 30.0 \ \text{rad/s} \right) \\ \\ v & = 39.0 \ \text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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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:

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:

Construct Your Own Problem

Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.

Solution:

Part A

Part B

Part C

A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?

Solution:

In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?

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).