Author Archives: Engineering Math

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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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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