Tag Archives: Physics PDF solutions

Problem 6-13: The motion of the WWII fighter plane propeller

The propeller of a World War II fighter plane is 2.30 m in diameter.

(a) What is its angular velocity in radians per second if it spins at 1200 rev/min?

(b) What is the linear speed of its tip at this angular velocity if the plane is stationary on the tarmac?

(c) What is the centripetal acceleration of the propeller tip under these conditions? Calculate it in meters per second squared and convert to multiples of g.

Solution:

Part A

We are converting the angular velocity \omega = 1200\ \text{rev/min} into radians per second.

\begin{align*}
\omega = & \frac{1200\ \text{rev}}{\text{min}}\times \frac{2\pi \ \text{radian}}{1\ \text{rev}} \times \frac{1 \ \text{min}}{60 \ \text{sec}} \\ \\
\omega = & 125.6637 \ \text{radians/sec} \\ \\
\omega = & 126 \ \text{radians/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are now solving the linear speed of the tip of the propeller by relating the angular velocity to linear velocity using the formula v = r \omega . The radius is half the diameter, so r= \frac{2.30\ \text{m}}{2} = 1.15 \ \text{m} .

\begin{align*}
v & = r \omega \\ \\
v & = \left( 1.15 \ \text{m} \right)\left( 125.6637 \ \text{radians/sec} \right) \\ \\
v & = 144.5132 \ \text{m/s} \\ \\
v & = 145 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

From the computed linear speed and the given radius of the propeller, we can now compute for the centripetal acceleration a_{c} using the formula 

a_{c} = \frac{v^2}{r}

If we substitute the given values, we have

\begin{align*}
a_{c} & = \frac{v^2}{r} \\ \\
a_{c} & = \frac{\left( 144.5132 \ \text{m/s} \right)^2}{1.15 \ \text{m}} \\ \\
a_{c} & = 18160.0565 \ \text{m/s}^2 \\ \\
a_{c} & = 1.82\times 10^{4} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

We can convert this value in multiples of g 

\begin{align*}
a_{c} & = 18160.0565 \ \text{m/s}^2 \times \frac{g}{9.81 \ \text{m/s}^2} \\ \\
a_{c} & = 1851.1780 g \\ \\
a_{c} & = 1.85\times 10^{3} \ g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-12: The approximate total distance traveled by planet Earth since its birth

Taking the age of Earth to be about 4×109 years and assuming its orbital radius of 1.5 ×1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun).

Solution:

First, we need to compute for the linear velocity of the Earth using the formula below knowing that the Earth has 1 full revolution in 1 year

v=r\omega

where r=1.5\times 10^{11} \ \text{m} and \omega = 2\pi \ \text{rad/year} . Substituting these values, we have

\begin{align*}
v & = r \omega \\ \\
v & = \left( 1.5\times 10^{11} \ \text{m} \right)\left( 2 \pi \ \text{rad/year} \right) \\ \\
v & = 9.4248\times 10^{11} \ \text{m/year}
\end{align*}

Knowing the linear velocity, we can compute for the total distance using the formula

\Delta x = v \Delta t

We can now substitute the given values: v = 9.4248\times 10^{11} \ \text{m/year} and \Delta t = 4\times 10^{9} \ \text{years} .

\begin{align*}
\Delta x & = v \Delta t \\ \\
\Delta x & = \left( 9.4248\times 10^{11} \ \text{m/year}  \right) \left( 4\times 10^{9} \ \text{years} \right) \\ \\
\Delta x & = 3.7699 \times 10^{21} \ \text{m} \\ \\
\Delta x & = 4 \times 10^{21} \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-11: Calculating the centripetal acceleration of a runner in a circular track

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If the runner completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of their centripetal acceleration as they run the curved portion of the track?

Solution:

Centripetal acceleration a_{c} is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity v and has the magnitude

a_{c}=\frac{v^{2}}{r}

We can solve for the constant speed of the runner using the formula

v=\frac{\Delta x}{\Delta t}

We are given the distance \Delta x = 200 \ \text{m} , and the total time \Delta t = 23.2\ \text{s} . Therefore, the velocity is

\begin{align*}
v & =\frac{\Delta x}{\Delta t} \\ \\ 
v & = \frac{200\ \text{m}}{23.2\ \text{s}} \\ \\
v & = 8.6207\ \text{m/s}
\end{align*}

From the given problem, we are given the following values: r=30\ \text{m} . We now have the details to solve for the centripetal acceleration.

\begin{align*}
a_{c} & = \frac{v^{2}}{r} \\ \\
a_{c} & = \frac{\left( 8.6207\ \text{m/s} \right)^2}{30\ \text{m}} \\ \\
a_{c} & = 2.4772\ \text{m/s}^{2} \\ \\
a_{c} & = 2.5\  \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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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 6-8: 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 6-4: 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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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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