Tag Archives: College Physics Complete Solution Manual

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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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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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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College Physics by Openstax Chapter 4 Problem 6

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

Solution:

Since the rockets are off, the only force acting on the sled is the friction $f$ . This force is against the direction of motion. By using Newton’s Second Law of Motion, we have.

\begin{align*} \Sigma F & = ma \\ -f & = ma \\ -f & = \left( 2100 \ \text{kg} \right)\left( -196 \ \text{m/s}^{2} \right) \\ -f & = -411600 \ \text{N} \\ f & = 411600 \ \text{N} \\ f & = 411.6 \ \text{kN} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The force necessary to produce the given deceleration is 411.6 kN.

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College Physics by Openstax Chapter 4 Problem 5

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force F (in newtons) is the person exerting on the mower? Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

Solution:

We can isolate the mower and expose the forces acting on it. This is the free-body diagram.

The free-body diagram of the mower: the force F exerted by the person and the friction force f exerted by the ground on the mower.

There are two forces acting on the mower in the horizontal directions:

$\textbf{F}:$ This is the force exerted by the person on the mower, and it is going to the right. This is the first unknown in the problem. We treat this as a positive force since it is directed to the right. The value of this force is $F=51 \ \text{N}$
$\textbf{f}:$ This is the friction force directed opposite the motion of the mower. We treat this as a negative force because it is directed to the left. The value of this force is $f=24 \ \text{N}$ .

Part A. The third force in the figure is the net force, $F_{net}$ . This is the vector sum of the forces $F$ and $f$ . That is

\begin{align*} F_{net} & =F-f \\ 51 \ \text{N} & = F-24 \ \text{N} \\ F & = 51 \ \text{N} +24 \ \text{N} \\ F & = 75 \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The person should exert a force of 75 N to produce a net force of 51 N.

Part B. When the force F is removed, the friction is now the only force acting on the mower. The friction is acting opposite the direction of motion. The direction of motion is indicated by the blue arrow in the figure below, and the friction force is the red arrow.

The free-body diagram of the mower when force F is removed.

Using Newton’s Second Law, we can solve for the deceleration of the mower.

\begin{align*} \Sigma F & = ma \\ -24 \ \text{N} & = \left( 24 \ \text{kg} \right) \ a \\ \frac{-24 \ \text{N}}{24 \ \text{kg}} & = \frac{\cancel{24 \ \text{kg}}\ \ a}{\cancel{24 \ \text{kg}}} \\ a & = \frac{-24 \ \text{N}}{24 \ \text{kg}} \\ a & = -1 \ \text{m/s}^{2} \\ \end{align*}

Using this deceleration computed above, we can solve for the distance traveled by the mower before coming to stop.

\begin{align*} \left( v_{f} \right)^{2} & = \left( v_{o} \right)^{2}+2a\Delta x \\ \left( 0 \ \text{m/s} \right)^{2} & = \left( 1.5 \ \text{m/s} \right)^{2}+2\left( -1 \ \text{m/s}^{2} \right)\Delta x \\ 0 & = 2.25-2 \Delta x \\ 2 \Delta x & = 2.25 \\ \frac{\cancel{2}\Delta x}{\cancel{2}} & =\frac{2.25}{2} \\ \Delta x & = 1.125 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 4 Problem 4

Since astronauts in orbit are apparently weightless, a clever method of measuring their masses is needed to monitor their mass gains or losses to adjust diets. One way to do this is to exert a known force on an astronaut and measure the acceleration produced. Suppose a net external force of 50.0 N is exerted and the astronaut’s acceleration is measured to be 0.893 m/s². (a) Calculate her mass. (b) By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. Discuss how this would affect the measurement of the astronaut’s acceleration. Propose a method in which recoil of the vehicle is avoided.

Solution:

We are given the following: $\sum F = 50.0 \ \text{N}$ , and $a=0.893 \ \text{m/s}^{2}$ .

Part A. We can solve for the mass, $m$ by using Newton’s second law of motion.

\begin{align*} \sum F & = ma \\ 50.0 \ \text{N} & = m \left( 0.893 \ \text{m/s}^{2} \right) \\ m & = \frac{50.0 \ \text{N}}{0.893 \ \text{m/s}^{2}} \\ m & = 56.0 \ \text{kg}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B. The measured acceleration is equal to the sum of the accelerations of the astronauts and the ship. That is

a_{measured}=a_{astronaut}+a_{ship}

If a force acting on the astronaut came from something other than the spaceship, the spaceship would not undergo a recoil. $\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)$

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College Physics by Openstax Chapter 4 Problem 3

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

Solution:

From Newton’s Second Law of Motion, $\sum F =ma$ . Substituting the given values, we have

\begin{align*} \sum F & = ma \\ 60.0 \ \text{N} & = \left( 4.50 \ \text{kg} \right) \ a \\ a & = \frac{60.0 \ \text{N}}{4.50 \ \text{kg}} \\ a & = 13.3 \ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Tag Archives: College Physics Complete Solution Manual

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:

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

Part B

Part C

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:

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:

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.)

College Physics by Openstax Chapter 4 Problem 7

College Physics by Openstax Chapter 4 Problem 6

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

College Physics by Openstax Chapter 4 Problem 5

College Physics by Openstax Chapter 4 Problem 4

College Physics by Openstax Chapter 4 Problem 3

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

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:

(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

Part B

Part C

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:

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:

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.)

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s2. What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.

A cleaner pushes a 4.50-kg laundry cart in such a way that the net external force on it is 60.0 N. Calculate its acceleration.

(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?

The same rocket sled drawn in Figure 4.30 is decelerated at a rate of 196 m/s². What force is necessary to produce this deceleration? Assume that the rockets are off. The mass of the system is 2100 kg.