Category Archives: Sciences

Includes Engineering Sciences such as Statics, Dynamics, Strength of Materials, Physics, Fluid Mechanics, Chemistry

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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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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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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Torsion Featured Image: Chapter 3 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 3: Torsion

Problem 301

Problem 302

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Simple Strain Featured Image: Chapter 2 of the book of Andrew Pytel and Ferdinand Singer Strength of Materials 4th Edition

Chapter 2: Simple Strain

Problem 201

Problem 202

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