Author Archives: Engineering Math

Problem 6-5: Calculating the angular velocity of a baseball pitcher’s forearm during a pitch

A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?

Solution:

We are given the linear velocity of the ball in the pitcher’s hand, v=35.0\ \text{m/s}, and the radius of the curvature, r=0.300 \ \text{m}. Linear velocity v and angular velocity \omega are related by

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

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

\begin{align*}
\omega & = \frac{v}{r} \\
\\
\omega & = \frac{35.0 \ \text{m/s}}{0.300 \ \text{m}} \\
\\
\omega & = 116.6667 \ \text{rad/s} \\ 
\\
\omega & = 117 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The angular velocity of the forearm is about 117 radians per second.

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Problem 6-4: Period, angular velocity, and linear velocity of the Earth

(a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth’s surface?

Solution:

Part A

The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is

\begin{align*}
\text{Period} & = 24 \ \text{hours} \\
\\
\text{Period} & = 24 \ \text{hours} \times \frac{3600 \ \text{seconds}}{1 \ \text{hour}} \\
\\
\text{Period} & = 86400 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The angular velocity \omega is the rate of change of an angle,

\omega = \frac{\Delta \theta}{\Delta t},

where a rotation \Delta \theta takes place in a time \Delta t.

From the given problem, we are given the following: \Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution}, and \Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds}. Therefore, the angular velocity is

\begin{align*}
\omega & = \frac{\Delta\theta}{\Delta t} \\
\\
\omega & = \frac{1 \ \text{revolution}}{1440 \ \text{minutes}}\\
\\
\omega & = 6.94 \times 10^{-4}\ \text{rpm}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can also express the angular velocity in units of radians per second. That is

\begin{align*}
\omega & = \frac{\Delta\theta}{\Delta t} \\
\\
\omega & = \frac{2\pi \ \text{radian}}{86400 \ \text{seconds}}\\
\\
\omega & = 7.27 \times 10^{-5}\ \text{radians/second}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

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

v = r \omega

From the given problem, we are given the following values: r=6.4 \times 10^{6} \ \text{meters}, and \omega = 7.27 \times 10^{-5}\ \text{radians/second}. Therefore, the linear velocity at the surface of the earth is

\begin{align*}
v & =r \omega \\
\\
v & = \left( 6.4 \times 10^{6} \ \text{meters} \right)\left( 7.27 \times 10^{-5}\ \text{radians/second} \right) \\
\\
v & = 465.28 \  \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-3: Calculating the number of revolutions given the tires radius and distance traveled

An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?

Solution:

The rotation angle \Delta \theta is defined as the ratio of the arc length to the radius of curvature:

\Delta \theta = \frac{\Delta s}{r}

where arc length \Delta s is distance traveled along a circular path and r is the radius of curvature of the circular path.

From the given problem, we are given the following quantities: r=0.260 \ \text{m}, and \Delta s = 80000 \ \text{km}.

\begin{align*}
\Delta \theta & = \frac{\Delta s}{r} \\
\\
\Delta \theta & = \frac{80000 \ \text{km} \times \frac{1000 \ \text{m}}{1 \ \text{km}}}{0.260 \ \text{m}} \\
\\
\Delta \theta & = 307.6923077 \times 10^{6} \ \text{radians} \times\frac{1 \ \text{rev}}{2\pi \  \text{radians}} \\
\\
\Delta \theta & = 48970751.72 \ \text{revolutions}  \\
\\
\Delta \theta & = 4.90 \times 10^{7} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-2: Conversion of units from rpm to revolutions per second and radians per second

Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?

Solution:

This is a problem on conversion of units. We are given a rotation in revolutions per minute and asked to convert this to revolutions per second and radians per second.

For the first part, we are asked to convert 6 rev/min to revolutions per second. 

\begin{align*}
\frac{6 \ \text{rev}}{\text{minute}} & =  \frac{6 \ \text{rev}}{\bcancel{\text{minute}}} \times \frac{1 \ \bcancel{\text{minute}}}{60 \ \text{seconds}} \\ \\
& = 0.1 \ \text{rev/second} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the next part, we are going to convert 6 rev/min to radians per second.

\begin{align*}
\frac{6 \ \text{rev}}{\text{minute}} & =  \frac{6 \ \bcancel{\text{rev}}}{\bcancel{\text{minute}}} \times \frac{2\pi \ \text{radians}}{1 \ \bcancel{\text{rev}}} \times \frac{1 \ \bcancel{\text{minute}}}{60 \ \text{seconds}} \\ \\
& = 0.6283 \  \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Finding the value/s of x for which a function is discontinuous – Problem 1.5.1

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PROBLEM:

Find the value or values of x for which the function is discontinuous.

\large \displaystyle f\left( x \right)=\frac{3x}{x-5}
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Solution:

A function \displaystyle f\left( x \right) is continuous at \displaystyle x=a if \displaystyle \lim_{x \to a} f\left( x \right)=f\left( a \right), which implies these three conditions:

  1. \displaystyle f\left( a \right) is defined.
  2. \displaystyle \lim_{x \to a} f\left( x \right)=L exists, and
  3. \displaystyle L=f\left( a \right)

We are given a rational function. A rational function is not defined when the denominator is equal to zero. If we equate the denominator to zero, we can compute the value/s of \displaystyle x where the function is discontinuous.

\begin{align*}
x-5 & = 0 \\
x & = 5 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The function is not continuous at \displaystyle x=5.

The graph of the function \displaystyle f\left( x \right)=\frac{3x}{x-5} is drawn below. It can be seen that there is an infinite discontinuity at \displaystyle x=5.

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Strength of Materials Problem 101 – Stress in each section of a composite bar

A composite bar consists of an aluminum section rigidly fastened between a bronze section and a steel section as shown in Fig. 1-8a. Axial loads are applied at the positions indicated. Determine the stress in each section.

Strength of Materials by Andrew Pytel and Ferdinand Singer Problem 101
Figure 1.8a

Solution:

We must first determine the axial load in each section to calculate the stresses. The free-body diagrams have been drawn by isolating the portion of the bar lying to the left of imaginary cutting planes. Identical results would be obtained if portions lying to the right of the cutting planes had been considered.

Solve for the internal axial load of the bronze

Free body diagram for the internal axial load of the bronze section for Problem 101 of Strength of Materials by Ferdinand Singer and Andrew Pytel
The free-body diagram of the bronze section
\begin{align*}
\sum_{}^{}F_x & = 0  \to  \\
-4000\ \text{lb}+P_{br} & = 0 \\
P_{br} & = 4000 \ \text{lb} \ \text{(tension)}
\end{align*}

Solve for the internal axial load of the aluminum

Free-body diagram of the aluminum section for problem 101 of Strength of materials by Andrew Pytel and Ferdinand Singer
The free-body diagram of the aluminum section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000 \ \text{lb} + 9000 \ \text{lb} - P_{al} & = 0 \\
P_{al} & = 5000 \ \text{lb} \ \text{(Compression)}
\end{align*}

Solve for the internal axial load of the aluminum

The free-body diagram of the steel section
\begin{align*}
\sum_{}^{}F_x & = 0 \\
-4000\ \text{lb} + 9000 \ \text{lb} + 2000\ \text{lb} - P_{st} & =0 \\
P_{st} & = 7000 \ \text{lb} \ \text{(Compression)}
\end{align*}

We can now solve the stresses in each section.

For the bronze

\begin{align*}
\sigma_{br} & = \frac{P_{br}}{A_{br}} \\
& = \frac{4000\ \text{lb}}{1.2 \ \text{in}^2} \\
& = 3330 \ \text{psi}\ \text{(Tension)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the aluminum

\begin{align*}
\sigma_{al} & = \frac{P_{br}}{A_{al}} \\
& = \frac{5000\ \text{lb}}{1.8 \ \text{in}^2} \\
& = 2780 \ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

For the steel

\begin{align*}
\sigma_{st} & = \frac{P_{st}}{A_{st}} \\
& = \frac{7000\ \text{lb}}{1.6 \ \text{in}^2} \\
& = 4380\ \text{psi}\ \text{(Compression)} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 4 Problem 8

What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (Such deceleration caused one test subject to black out and have temporary blindness.)

Solution:

We are given the following: v_{0}=1000 \ \text{km/h}, v_{f}=0 \ \text{km/h}, \Delta t = 1.1 \ \text{s}.

The acceleration is computed as the change in velocity divided by the change in time.

\begin{align*}
a & = \frac{\Delta v}{\Delta t} \\
a & = \frac{v_{f}-v_{o}}{\Delta t} \\
a & = \frac{\left( 0\ \text{km/h}-1000 \ \text{km/h} \right)\left( \frac{1000 \ \text{m}}{1\ \text{km}} \right) \left( \frac{1\ \text{h}}{3600\ \text{s}} \right)}{1.1\ \text{s}} \\
a & = -252.5\ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}
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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2

According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 20 pieces of cotton fiber was taken and the absorbency on each piece was measured. The following are the absorbency values:

18.7121.4120.7221.8119.2922.4320.17
23.7119.4420.5018.9220.3323.0022.85
19.2521.7722.1119.7718.0421.12

(a) Calculate the sample mean and median for the above sample values.
(b) Compute the 10% trimmed mean.
(c) Do a dot plot of the absorbency data.
(d) Using only the values of the mean, median, and trimmed mean, do you have evidence of outliers in the data?

Solution:

Part A. The sample mean is computed as follows:

\begin{align*}
\bar x & = \frac{\Sigma x_{i}}{n} \\
\bar x & = \frac{18.71+21.41+20.72+\cdots +21.12}{20} \\
\bar x & = 20.77 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can solve for the sample median by arranging the data in increasing order first.

18.04, \
18.71, \
18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85, \
23.00, \
23.71

Since there are 20 measurements (even), the middle measurements are the (20/2) 10th and the (20/2 + 1) 11th measurement. The 10th measurement is 20.50 and the 11th measurement is 20.72. The median is the average of these two measurements. 

\begin{align*}
\tilde x & = \frac{20.50+20.72}{2} \\
\tilde x & = 20.61 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The 10% trimmed mean is calculated by removing 10% of the lowest data and 10% of the highest data. That is, removing the 2 lowest and 2 highest data. We are left with the following:

18.92, \
19.25, \
19.29, \
19.44, \
19.77, \
20.17, \
20.33, \
20.50 \\
20.72, \
21.12, \
21.41, \
21.77, \
21.81, \
22.11, \
22.43, \
22.85

The 10% trimmed mean, \bar x _{tr10} is

\begin{align*}
\bar x _{tr10} & = \frac{\Sigma x_{i}}{n} \\
\bar x _{tr10} & = \frac{18.92+19.25+\cdots+22.85}{16} \\
\bar x _{tr10} & = 20.74 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The dot plot is shown

dot plot for Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.2

Part D. Since the values of the mean, median, and trimmed mean are not actually far from each other, we can conclude that there are no outliers in the given measurements.

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College Physics by Openstax Chapter 4 Problem 7

(a) If the rocket sled shown in Figure 4.31 starts with only one rocket burning, what is the magnitude of its acceleration? Assume that the mass of the system is 2100 kg, the thrust T is 2.4 \times 10^{4} N, and the force of friction opposing the motion is known to be 650 N. (b) Why is the acceleration not one-fourth of what it is with all rockets burning?

Solution:

Considering the direction of motion as the positive direction, we are given the following: T=2.4 \times 10^4 \ \text{N}, f=-650 \ \text{N}, and mass, m=2100 \ \text{kg}.

Part A. The magnitude of the acceleration can be computed using Newton’s Second Law of Motion.

\begin{align*}
\Sigma F & =ma \\
2.4\times 10^4 \ \text{N}-650 \ \text{N} & = 2100 \ \text{kg}\times a \\
23350 & = 2100 a \\
\frac{23350}{2100} & = \frac{\cancel{2100} a}{\cancel{2100}} \\
a & = \frac{23350}{2100} \\
a & = 11 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B. The acceleration is not one-fourth of what it was with all rockets burning because the frictional force is still as large as it was with all rockets burning. \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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Probability & Statistics for Engineers & Scientists Ninth Edition by Walpole Exercise 1.1

The following measurements were recorded for the drying time, in hours, of a certain brand of latex paint.

3.42.54.82.93.6
2.83.35.63.72.8
4.44.05.23.04.8

Assume that the measurements are a simple random sample.
(a) What is the sample size for the above sample?
(b) Calculate the sample mean for these data.
(c) Calculate the sample median.
(d) Plot the data by way of a dot plot.
(e) Compute the 20% trimmed mean for the above data set.
(f) Is the sample mean for these data more or less descriptive as a center of location than the trimmed mean?

Solution:

Part A. Sample size, n is the total number of measurements.

n=15 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B. The sample mean, \bar x is computed as follows:

\begin{align*}
\bar x & = \sum_{i=1}^{n}\frac{x_{i}}{n} \\
& = \frac{3.4+2.5+4.8+2.9+3.6+2.8+3.3+5.6+3.7+2.8+4.4+4.0+5.2+3.0+4.8}{15} \\
& = \frac{56.8}{15} \\
& = 3.79 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C. The sample median, \tilde x is the number at the middle of the arranged measurements in increasing magnitude. There are 15 measurements, n=15. If we arranged the data in increasing magnitude, the median is the measurement in the middle.

2.5, \ 2.8, \ 2.8, \ 2.9, \ 3.0, \ 3.3, \ 3.4, \ \underset{\color{Blue} \text{middle number}}{3.6}, \ 3.7, \ 4.0, \ 4.4, \ 4.8, \ 4.8, \ 5.2, \ 5.6

The middle number is 3.6. That is

\tilde x= 3.6 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D. This dot plot has been created using the Statistical Software Rguroo

a dot plot for the data: 3.4, 2.5, 4.8, 2.9, 3.6, 2.8, 3.3, 5.6, 3.7, 2.8, 4.4, 4.0, 5.2, 3.0, 4.8. this dot plot was made possible through Rgurro at https://www.rguroo.com/

Part E. The 20% trimmed mean means the average of the measurements left after removing 20% highest and 20% lowest data. This means we remove the 3 highest and 3 lowest numbers. Therefore, the data becomes

\ 2.9, \ 3.0, \ 3.3, \ 3.4, \ 3.6,  \ 3.7, \ 4.0, \ 4.4, \ 4.8,

The 20% trimmed mean, \bar x _{tr20} is

\begin{align*}
\bar x _{tr20} & = \frac{2.9+3.0+ \cdots+4.8}{9} \\
\bar x _{tr20} & = \frac{33.1}{9} \\
\bar x _{tr20} & = 3.678 \ \text{hours} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part F. The sample mean for these data is \bar x = 3.79 \ \text{hours} while the 20% trimmed mean is \bar x _{tr20} = 3.678 \ \text{hours}. Seems like the two means are not really far from each other, but because of the elimination of the extreme values, we can treat the trimmed mean as a better descriptive mean.

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