Tag Archives: Angular Velocity

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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Solution Guides to College Physics by Openstax Chapter 6 Banner

Chapter 6: Uniform Circular Motion and Gravitation

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Rotation Angle and Angular Velocity

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Centripetal Acceleration

Problem 22

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Centripetal Force

Problem 29

Problem 30

Problem 31

Problem 32

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Newton’s Universal Law of Gravitation

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

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Satellites and Kepler’s Laws: An Argument for Simplicity

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

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Problem 6-1: Odometer reading based on the number of wheel revolutions


Semi-trailer trucks have an odometer on one hub of a trailer wheel. The hub is weighted so that it does not rotate, but it contains gears to count the number of wheel revolutions—it then calculates the distance traveled. If the wheel has a 1.15 m diameter and goes through 200,000 rotations, how many kilometers should the odometer read?


Solution:

The formula for the total distance traveled is

\Delta s=\Delta \theta \times r

Therefore, the total distance traveled is

\begin{align*}
\Delta s & =\left(200000\:\text{rotations}\:\times \frac{2\pi \:\text{radian}}{1\:\text{rotation}}\right)\left(\frac{1.15\:\text{m}}{2}\right) \\
\Delta s & =722566.3103\:\text{m} \\
\Delta s & =722.6\:\text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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