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College Physics by Openstax Chapter 6 Problem 26

The Ideal Speed on a Banked Curve

Problem:

What is the ideal speed to take a 100 m radius curve banked at a 20.0° angle?

Solution:

The formula for the ideal speed on a banked curve can be derived from the formula of the ideal angle. That is, starting from \tan \theta = \frac{v^2}{rg}, we can solve for v.

v = \sqrt{rg \tan \theta}

For this problem, we are given the following values:

  • radius of curvature, r=100\ \text{m}
  • acceleration due to gravity, g=9.81\ \text{m/s}^2
  • banking angle, \theta = 20.0 ^\circ

If we substitute the given values into our formula, we have

\begin{align*}
v = & \sqrt{rg \tan \theta} \\ \\
v = & \sqrt{\left( 100\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right) \left( \tan 20.0 ^\circ \right) } \\ \\
v = & 18.8959\ \text{m/s} \\ \\
v = & 18.9\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The ideal speed for the given banked curve is about 18.9\ \text{m/s}.

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College Physics by Openstax Chapter 6 Problem 25

The ideal banking angle of a curve on a highway

Problem:

What is the ideal banking angle for a gentle turn of 1.20 km radius on a highway with a 105 km/h speed limit (about 65 mi/h), assuming everyone travels at the limit?

Solution:

The ideal banking angle (meaning there is no involved friction) of a car on a curve is given by the formula:

\theta = \tan^{-1} \left( \frac{v^2}{rg} \right)

We are given the following values:

  • radius of curvature, \displaystyle r = 1.20\ \text{km} \times \frac{1000\ \text{m}}{1\ \text{km}} = 1200\ \text{m}
  • linear velocity, \displaystyle v=105\ \text{km/h}\times \frac{1000\ \text{m}}{1\ \text{km}} \times \frac{1\ \text{h}}{3600\ \text{s}} = 29.1667\ \text{m/s}
  • acceleration due to gravity, \displaystyle g = 9.81\ \text{m/s}^2

If we substitute these values into our formula, we come up with

\begin{align*}
\theta & = \tan^{-1} \left( \frac{v^2}{rg} \right) \\ \\
\theta & = \tan^{-1} \left[ \frac{\left( 29.1667\ \text{m/s} \right)^2}{\left( 1200\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\
\theta & = 4.1333 ^\circ \\ \\
\theta & = 4.13 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The ideal banking angle for the given highway is about 4.13 ^\circ.

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College Physics by Openstax Chapter 6 Problem 24

Centripetal Force of a Rotating Wind Turbine Blade

Problem:

Calculate the centripetal force on the end of a 100 m (radius) wind turbine blade that is rotating at 0.5 rev/s. Assume the mass is 4 kg.

Solution:

We are given the following values:

  • radius, r=100\ \text{m}
  • angular velocity, \omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4\ \text{kg}

Centripetal force F_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude F_c = m a_c which can also be expressed as

F_c = m \frac{v^2}{r} \quad \text{or} \quad \ F_c = mr \omega^2

For this particular problem, we are going to use the formula F_c = mr \omega^2. If we substitute the given values, we have

\begin{align*}
F_c & =mr \omega^2 \\ \\
F_c & = \left( 4\ \text{kg} \right)\left( 100\ \text{m} \right)\left( 3.1416\ \text{rad/sec} \right)^2 \\ \\
F_c & = 3947.8602\ \text{N} \\ \\
F_c & = 4 \times 10^3\ \text{N}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The centripetal force on the end of the wind turbine blade is approximately 4 \times 10^3\ \text{N}.

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College Physics by Openstax Chapter 6 Problem 23

The centripetal force of a child riding a merry-go-round

Problem:

(a) A 22.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force must she exert to stay on if she is 1.25 m from its center?

(b) What centripetal force does she need to stay on an amusement park merry-go-round that rotates at 3.00 rev/min if she is 8.00 m from its center?

(c) Compare each force with her weight.

Solution:

Part A

We are given the following values: m=22.0\ \text{kg}, \omega = 40.0\ \text{rev/min}, and r=1.25\ \text{m}. We are asked to solve for the centripetal force, F_c.

Centripetal force F_c is any force causing uniform circular motion. It is a “center-seeking” force that always points toward the center of rotation. It is perpendicular to linear velocity v and has magnitude F_c=m a_c, which can also be expressed as F_c = m \frac{v^2}{r} or F_c = m r \omega ^2. Basing from the given values, we are going to solve the problem using the formula

F_c = m r \omega ^2

First, we need to convert the angular velocity \omega to \text{rad/sec} for unit homogeneity. 

40\ \text{rev/min} \times \frac{2\pi\ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} = 4.1888\ \text{rad/sec}

Now, we can substitute the given values into our formula.

\begin{align*}
F_c & = m r \omega ^2 \\ \\
F_c & = \left( 22.0\ \text{kg}\right) \left( 1.25\ \text{m}\right) \left( 4.1888\ \text{rad/sec}\right)^2 \\ \\
F_c & = 482.5162\ \text{N} \\ \\
F_c & = 483\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

Let us convert the angular velocity to radians per second.

3.00\ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}}=0.3142 \ \text{rad/sec}

Now, we can substitute the given values

\begin{align*}
F_c & = m r \omega ^2 \\ \\
F_c & = \left( 22.0\ \text{kg}\right) \left( 8.00\ \text{m}\right) \left( 0.3142\ \text{rad/sec}\right)^2 \\ \\
F_c & = 17.3750\ \text{N} \\ \\
F_c & = 17.4\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For the first centripetal force we solved in Part A, 

\frac{F_c}{w} = \frac{483\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 2.24

The centripetal force is 2.24 times the weight of the child.

For the centripetal force we solved in Part B, we have

\frac{F_c}{w} = \frac{17.4\ \text{N}}{\left( 22\ \text{kg} \right)\left( 9.81\ \text{m/s}^2 \right)} = 0.0806

The centripetal force is only about 8% of the child’s weight.

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College Physics by Openstax Chapter 6 Problem 21

Centripetal acceleration of an amusement park ride shaped like a Viking ship

Problem:

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. The speed at the bottom of the arc is 23.4 m/s.

(a) What is the centripetal acceleration at the bottom of the arc?

(b) Draw a free-body diagram of the forces acting on a rider at the bottom of the arc.

(c) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

(d) Discuss whether the answer seems reasonable.

Solution:

Problem 6-20: The centripetal acceleration of the commercial jet’s tires, and the force of a determined bacterium in it

At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m.

(a) At how many rev/min are the tires rotating?

(b) What is the centripetal acceleration at the edge of the tire?

(c) With what force must a determined 1.00×10−15 kg bacterium cling to the rim?

(d) Take the ratio of this force to the bacterium’s weight.

Solution:

We are given the following quantities: linear speed, v=60.0 \ \text{m/s}, radius is half the diameter, r=0.425 \ \text{m}.

Part A

We can compute the angular velocity based on the given using the formula, \displaystyle \omega = \frac{v}{r}. 

\begin{align*}
\omega & = \frac{v}{r} \\ \\
\omega & = \frac{60.0 \ \text{m/s}}{0.425 \ \text{m}} \\ \\
\omega & = 141.1765 \ \text{rad/sec}
\end{align*}

Now, we can convert this into the required unit of rev/min.

\begin{align*}
\omega & = 141.1765\ \frac{\text{rad}}{\text{sec}} \times \frac{1\ \text{rev}}{2\pi\ \text{rad}} \times \frac{60\ \text{sec}}{1\ \text{min}} \\ \\
\omega & = 1348.1363 \ \text{rev/min} \\ \\
\omega & = 1.35 \times 10^{3} \ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The centripetal acceleration at the edge of the tire can be computed using the formula, a_{c} = r \omega ^{2}.

\begin{align*}
a_{c} & = r \omega ^2 \\ \\
a_{c} & = \left( 0.425\ \text{m} \right) \left(141.1765\ \text{rad/sec} \right)^2 \\ \\
a_{c} & = 8470.5918 \ \text{m/s}^2 \\ \\ 
a_{c} & = 8.47 \times 10 ^{3} \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

From the second law of motion, the force is equal to the product of the mass and the acceleration. In this case, we are going to use the formula, F_c = m a_c . We are given the mass to be m=1.00 \times 10 ^{-15}\ \text{kg} , and the centripetal acceleration is solved in Part B.

\begin{align*}
F_c & = ma_c \\ \\
F_c & = \left( 1 \times 10^{-15}\ \text{kg}\right) \left(8470.5918 \ \text{m/s}^2\right) \\ \\
F_c & = 8.4705918 \times 10 ^{-12}\ \text{kg m/s}^2 \\ \\
F_c & = 8.47 \times 10^{-12} \ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

The ratio of this force, F_c to the weight of the bacterium is

\begin{align*}
\frac{F_c}{mg} & = \frac{8.4705819 \times 10 ^{-12}\ \text{N}}{\left( 1 \times 10^{-15} \text{kg}  \right)\left(9.81 \ \text{m/s}^2 \right)} \\ \\
\frac{F_c}{mg} & = 863.4640 \\ \\
\frac{F_c}{mg} & = 863 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 6-18: The linear speed of an ultracentrifuge and Earth in its orbit

Verify that the linear speed of an ultracentrifuge is about 0.50 km/s, and Earth in its orbit is about 30 km/s by calculating:

(a) The linear speed of a point on an ultracentrifuge 0.100 m from its center, rotating at 50,000 rev/min.

(b) The linear speed of Earth in its orbit about the Sun (use data from the text on the radius of Earth’s orbit and approximate it as being circular).

Solution:

Part A

We are given a linear speed of an ultracentrifuge of 0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100\ \text{m} and angular velocity of \omega = 50000 \ \text{rev/min}. We are going to use the formula

v = r \omega

Since we are given a linear speed in \text{km/s}, we are going to convert the radius to \text{km}, and the angular velocity to \text{rad/sec}

r=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km} 
\omega = 50000 \ \text{rev/min} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1\ \text{min}}{60\ \text{sec}} =5235.9878\ \text{rad/sec}

Now, we can substitute these into the formula

\begin{align*}
v & = r \omega \\ \\
v & = \left( 0.0001 \ \text{km} \right)\left( 5235.9878 \ \text{rad/sec} \right) \\ \\
v & = 0.5236 \ \text{km/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This value is about 0.500 km/s.

Part B

From Table 6.2 of the book

ParentSatelliteAverage orbital radius r(km)Period T(y)r3 / T2 (km3 / y2)
SunEarth1.496 \times 10^{8} 13.35 \times 10^{24}

Using the same formulas we used in Part A, we can solve for the linear velocity of the Earth around the sun. The radius is

r=1.496 \times10^{8} \ \text{km}

The angular velocity is

\begin{align*}
\omega &  = 1 \ \frac{\text{rev}}{\text{year}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{year}}{365.25 \ \text{days}} \times \frac{1\ \text{day}}{24\ \text{hours}}\times \frac{1\ \text{hour}}{3600\ \text{sec}} \\ \\ 
\omega &  = 1.9910 \times 10^{-7}\ \text{rad/sec}
\end{align*}

The linear velocity is

\begin{align*}
v & = r \omega \\ \\
v & = \left( 1.496\times 10^{8}\ \text{km} \right)\left( 1.9910 \times 10 ^ {-7} \right) \ \text{rad/sec}\\ \\
v & = 29.7854\ \text{km/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The linear velocity is about 30 km/s.

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Problem 6-17: The acceleration due to gravity at the position of a satellite located above the Earth

What percentage of the acceleration at Earth’s surface is the acceleration due to gravity at the position of a satellite located 300 km above Earth?

Solution:

The acceleration due to gravity of a body and the Earth is given by the formula

g= G \frac{M}{r^2}

where G is the gravitational constant, M is the mass of the Earth, and r is the distance of the object to the center of the Earth. We know that the approximate radius of the Earth is r=6.3781 \times 10^6 \ \text{m} .

The percentage of the acceleration at 300 km above the Earth of the acceleration due to gravity at Earth’s surface is

\displaystyle \frac{\left( \frac{GM}{r^2} \right)_2}{\left( \frac{GM}{r^2} \right)_1} \times 100\%

Note that the subscript 2 indicates the satellite located 300 km above the Earth, and the subscript 1 indicates the object at the Earth’s surface. Also, from the expression above, we can cancel G and M from the numerator and denominator because these are constants. So, we are down to

\frac{\left( \frac{1}{r^2} \right)_2}{\left( \frac{1}{r^2} \right)_1} \times 100\% = \frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\%

Substituting the values, we have

\begin{align*}
\frac{\left( r^2 \right)_1}{\left( r^2 \right)_2} \times 100\% & = \frac{\left( 6.3781 \times 10^6 \ \text{m} \right)^{2}}{\left( 6.3781 \times 10^6 \ \text{m}+300 \times 10^{3} \ \text{m} \right)^{2}} \times 100\% \\ \\
& = 91.2172\% \\ \\
& = 91.2\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The percentage of the acceleration at the Earth’s surface of the acceleration due to gravity at the position of a satellite located 300 km above the Earth is about 91.2%.

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Problem 6-16: Calculating the centripetal acceleration of an ice skater’s nose

Olympic ice skaters are able to spin at about 5.00 rev/s.

(a) What is their angular velocity in radians per second?

(b) What is the centripetal acceleration of the skater’s nose if it is 0.120 m from the axis of rotation?

(c) An exceptional skater named Dick Button was able to spin much faster in the 1950s than anyone since—at about 9.00 rev/s. What was the centripetal acceleration of the tip of his nose, assuming it is at 0.120 m radius?

(d) Comment on the magnitudes of the accelerations found. It is reputed that Button ruptured small blood vessels during his spins.

Solution:

We are given an angular velocity, \omega = 5 \ \text{rev/sec}

Part A

For this part, we are asked to convert the angular velocity to units of radians per second. 

\begin{align*}
\omega & = \frac{5.00\ \text{rev}}{\text{sec}}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} \\ \\
\omega & = 31.4159 \ \text{rad/sec} \\ \\
\omega & = 31.4 \ \text{rad/sec}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

For this part, we are asked to solve for the centripetal acceleration. We are going to use the formula a_{c} = r \omega ^2 given r=0.120\ \text{m} and \omega = 31.4159 \ \text{rad/s} .

\begin{align*}
a_{c} & = r \omega ^2 \\ \\ 
a_{c} & = \left( 0.120 \ \text{m} \right) \left( 31.4159 \ \text{rad/s} \right)^2 \\ \\
a_{c} & = 118.4350 \ \text{m/s}^2 \\ \\
a_{c} & = 118 \ \text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part C

For this part, we are going to directly solve the centripetal acceleration.

\begin{align*}
a_{c} & = r \omega ^2 \\ \\ 
a_{c} & = \left( 0.120 \ \text{m} \right)\left( \frac{9\ \text{rev}}{\text{s}} \times \frac{2\pi \ \text{rad}}{1\ \text{rev}}\right)^2 \\ \\
a_{c} & = 383.7302 \ \text{m/s}^2 \\ \\
a_{c} & = 384 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part D

The centripetal acceleration felt by Olympic skaters is 12 times larger than the acceleration due to gravity. That is quite a lot of acceleration in itself. The centripetal acceleration felt by Button’s nose was 39.2 times larger than the acceleration due to gravity! It is no wonder that he ruptured small blood vessels in his spins.

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Problem 6-15: The centripetal acceleration at the tip of a helicopter blade

Helicopter blades withstand tremendous stresses. In addition to supporting the weight of a helicopter, they are spun at rapid rates and experience large centripetal accelerations, especially at the tip.

(a) Calculate the magnitude of the centripetal acceleration at the tip of a 4.00 m long helicopter blade that rotates at 300 rev/min.

(b) Compare the linear speed of the tip with the speed of sound (taken to be 340 m/s).

Solution:

Part A

We are given the following values: r=4.00\ \text{m}, and \omega = 300 \ \text{rev/min}.

Let us convert the angular velocity to unit of radians per second.

\omega = 300 \  \frac{\text{rev}}{\text{min}} \times \frac{2\pi \ \text{rad}}{1 \ \text{rev}}\times \frac{1\ \text{min}}{60 \ \text{sec}} = 31.4159 \ \text{rad/sec}

The centripetal acceleration at the tip of the helicopter blade can be computed using the formula

a_{c} = r \omega ^2

If we substitute the given values into the formula, we have

\begin{align*}
a_{c} & = r \omega^2 \\ \\
a_{c} & = \left( 4.00\ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right)^2 \\ \\
a_{c} & = 3947.8351 \ \text{m/s}^2 \\ \\
a_{c} & = 3.95 \times10^3 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

We are asked to solve for the linear velocity of the blade’s tip. We are going to use the formula

v=r \omega

We just needed to substitute the given values into the formula.

\begin{align*}
v & = r \omega \\ \\
v & = \left( 4.00 \ \text{m} \right)\left( 31.4159 \ \text{rad/sec} \right) \\ \\
v & = 125.6636 \ \text{m/s} \\ \\
v & = 126 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Let us compare this with the speed of light which is 340 m/s. 

\frac{125.6636 \ \text{m/s}}{340\ \text{m/s}} \times 100 \%= 36.9599 \% =37.0\%

The linear velocity of the blades tip is 37.0% of the speed of light.

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