Tag Archives: Rotation Angle

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, vv and the angular velocity ω\omega are related by the equation

v=rω or ω=vrv=r\omega \ \text{or} \ \omega=\frac{v}{r}

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

ω=vrω=32.0 m/s0.420 mω=76.1905 rad/sω=76.2 rad/s  (Answer)\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:

ω=76.1905 radsec×1 rev2π rad×60 sec1 minω=727.5657 rev/minω=728 rev/min  (Answer)\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 6-5: 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 m/sv=35.0\ \text{m/s}, and the radius of the curvature, r=0.300 mr=0.300 \ \text{m}. Linear velocity vv and angular velocity ω\omega are related by

v=rω or ω=vrv=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,

ω=vrω=35.0 m/s0.300 mω=116.6667 rad/sω=117 rad/s  (Answer)\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 6-3: 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 Δθ\Delta \theta is defined as the ratio of the arc length to the radius of curvature:

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

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

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

Δθ=ΔsrΔθ=80000 km×1000 m1 km0.260 mΔθ=307.6923077×106 radians×1 rev2π radiansΔθ=48970751.72 revolutionsΔθ=4.90×107 revolutions  (Answer)\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

Problem 1

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Problem 3

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Problem 6

Problem 7

Problem 8

Problem 9

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Centripetal Acceleration

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

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Centripetal Force

Problem 23

Problem 24

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Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

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Newton’s Universal Law of Gravitation

Problem 33

Problem 34

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Problem 37

Problem 38

Problem 39

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Problem 42

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Satellites and Kepler’s Laws: An Argument for Simplicity

Problem 43

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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

Δs=Δθ×r\Delta s=\Delta \theta \times r

Therefore, the total distance traveled is

Δs=(200000 rotations ×2π radian1 rotation)(1.15 m2)Δs=722566.3103 mΔs=722.6 km  (Answer)\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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