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College Physics by Openstax Chapter 6 Problem 28

Riding a Bicycle in an Ideally Banked Curve

Problem:

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure 6.33. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that θ\theta (as defined in the figure) is related to the speed vv and radius of curvature rr of the turn in the same way as for an ideally banked roadway—that is, θ=tan⁡−1(v2/rg)\theta = \tan ^{-1} \left( v^2/rg \right)

(b) Calculate θ\theta for a 12.0 m/s turn of radius 30.0 m (as in a race).

Figure 6.33 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system, while its horizontal component must supply the centripetal force. This process produces a relationship among the angle θ, the speed v, and the radius of curvature r of the turn similar to that for the ideal banking of roadways.

Solution:

Part A

Let us redraw the given forces in a free-body diagram with their corresponding components.

The force NN and FcF_c are the vertical and horizontal components of the force FF.

If we take the equilibrium of forces in the vertical direction (since there is no motion in the vertical direction) and solve for FF, we have

∑Fy=0Fcos⁡θ−mg=0Fcos⁡θ=mgF=mgcos⁡θEquation 1\begin{align*} \sum F_y & = 0 \\ \\ F \cos \theta - mg & = 0 \\ \\ F \cos \theta & = mg \\ \\ F & = \frac{mg}{\cos \theta} \quad \quad & \color{Blue} \small \text{Equation 1} \end{align*}

If we take the sum of forces in the horizontal direction and equate it to mass times the centripetal acceleration (since the centripetal acceleration is directed in this direction), we have

∑Fx=macFsin⁡θ=macFsin⁡θ=mv2rEquation 2\begin{align*} \sum F_x & = ma_c \\ \\ F \sin \theta & = m a_c \\ \\ F \sin \theta & = m \frac{v^2}{r} \quad \quad & \color{Blue} \small \text{Equation 2} \end{align*}

We substitute Equation 1 to Equation 2.

Fsin⁡θ=mv2rmgcos⁡θsin⁡θ=mv2rmgsin⁡θcos⁡θ=mv2r\begin{align*} F \sin \theta & = m \frac{v^2}{r} \\ \\ \frac{mg}{\cos \theta} \sin \theta & = m \frac{v^2}{r} \\ \\ mg \frac{\sin \theta}{\cos \theta} & =m \frac{v^2}{r} \\ \\ \end{align*}

We can cancel mm from both sides, and we can apply the trigonometric identity tan⁡θ=sin⁡θcos⁡θ\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}. We should come up with

gtan⁡θ=v2rtan⁡θ=v2rgθ=tan⁡−1(v2rg)  (Answer)\begin{align*} g \tan \theta & = \frac{v^2}{r} \\ \\ \tan \theta & = \frac{v^2}{rg} \\ \\ \theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are given the following values:

  • linear velocity, v=12.0 m/sv = 12.0\ \text{m/s}
  • radius of curvature, r=30.0 mr=30.0\ \text{m}
  • acceleration due to gravity, g=9.81 m/s2g = 9.81\ \text{m/s}^2

We substitute the given values to the formula of θ\theta we solve in Part A.

θ=tan⁡−1(v2rg)θ=tan⁡−1[(12.0 m/s)2(30.0 m)(9.81 m/s2)]θ=26.0723∘θ=26.1∘  (Answer)\begin{align*} \theta & = \tan ^ {-1} \left( \frac{v^2}{rg} \right) \\ \\ \theta & = \tan ^ {-1} \left[ \frac{\left( 12.0\ \text{m/s} \right)^2}{\left( 30.0\ \text{m} \right)\left( 9.81\ \text{m/s}^2 \right)} \right] \\ \\ \theta & = 26.0723 ^\circ \\ \\ \theta & = 26.1 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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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 mr=100\ \text{m}
  • angular velocity, ω=0.5 rev/sec×2π rad1 rev=3.1416 rad/sec\omega = 0.5\ \text{rev/sec}\times \frac{2\pi \ \text{rad}}{1\ \text{rev}} = 3.1416\ \text{rad/sec}
  • mass, m=4 kgm=4\ \text{kg}

Centripetal force FcF_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 vv and has magnitude Fc=macF_c = m a_c which can also be expressed as

Fc=mv2ror Fc=mrω2F_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 Fc=mrω2F_c = mr \omega^2. If we substitute the given values, we have

Fc=mrω2Fc=(4 kg)(100 m)(3.1416 rad/sec)2Fc=3947.8602 NFc=4×103 N  (Answer)\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 4×103 N4 \times 10^3\ \text{N}.

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Problem 6-19: The angular velocity of an “artificial gravity”

A rotating space station is said to create “artificial gravity”—a loosely-defined term used for an acceleration that would be crudely similar to gravity. The outer wall of the rotating space station would become a floor for the astronauts, and centripetal acceleration supplied by the floor would allow astronauts to exercise and maintain muscle and bone strength more naturally than in non-rotating space environments. If the space station is 200 m in diameter, what angular velocity would produce an “artificial gravity” of 9.80 m/s2 at the rim?

Solution:

We are given the following quantities:

radius=diameter2=200 m2=100 m\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
centripetal acceleration,ac=9.80 m/s2\text{centripetal acceleration}, a_c = 9.80 \ \text{m/s}^2

Centripetal acceleration is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. The formula for centripetal acceleration is

ac=rω2a_{c} = r \omega ^2

If we solve for the angular velocity in terms of the other quantities, we have

ω=acr\omega = \sqrt{\frac{a_c}{r}}

Substituting the given quantities,

ω=acrω=9.80 m/s2100 mω=0.313 rad/sec  (Answer)\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{9.80 \ \text{m/s}^2}{100\ \text{m}}} \\ \\ \omega & = 0.313 \ \text{rad/sec} \ \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 km/s0.50\ \text{km/s}. We are asked to verify this value if we are given a radius of r=0.100 mr=0.100\ \text{m} and angular velocity of ω=50000 rev/min \omega = 50000 \ \text{rev/min}. We are going to use the formula

v=rωv = r \omega

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

r=0.100 m×1 km1000 m=0.0001 kmr=0.100\ \text{m} \times \frac{1\ \text{km}}{1000\ \text{m}} = 0.0001\ \text{km}
ω=50000 rev/min×2π rad1 rev×1 min60 sec=5235.9878 rad/sec\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

v=rωv=(0.0001 km)(5235.9878 rad/sec)v=0.5236 km/s  (Answer)\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)
SunEarth1.496×1081.496 \times 10^{8} 13.35×10243.35 \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=1.496×108 kmr=1.496 \times10^{8} \ \text{km}

The angular velocity is

ω=1 revyear×2π rad1 rev×1 year365.25 days×1 day24 hours×1 hour3600 secω=1.9910×10−7 rad/sec\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

v=rωv=(1.496×108 km)(1.9910×10−7) rad/secv=29.7854 km/s  (Answer)\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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College Physics by Openstax Chapter 3 Problem 5

Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.54, then this problem finds their sum R=A+B.)

Figure 3.54

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

β=20∘+(90∘−40∘)=70∘\beta = 20^\circ +\left( 90^\circ -40^\circ \right) = 70^\circ

We can use cosine law to solve for R.

R2=A2+B2−2ABcos⁡βR2=(12.0 m)2+(20.0 m)2−2(12.0 m)(20.0 m)cos⁡70∘R=(12.0 m)2+(20.0 m)2−2(12.0 m)(20.0 m)cos⁡70∘R=19.4892 mR=19.5 m  (Answer)\begin{align*} R^2 & =A^2+B^2 -2AB \cos \beta \\ R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right) \cos 70^\circ \\ R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right) \cos 70^\circ} \\ R & =19.4892 \ \text{m} \\ R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

We can solve for α using sine law.

sin⁡αB=sin⁡βRsin⁡α20.0 m=sin⁡70∘19.4892 msin⁡α=20.0 sin⁡70∘19.4892α=sin⁡−1(20.0 sin⁡70∘19.4892)α=74.6488∘\begin{align*} \frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\ \frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\ \sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\ \alpha & = \sin ^{-1} \left( \frac{20.0 \ \sin 70^\circ }{19.4892} \right) \\ \alpha & = 74.6488 ^\circ \end{align*}

Then we solve for the value of θ by subtracting 70° from α.

θ=74.6488∘−70∘=4.65∘\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65∘,South of West  (Answer)4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}
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College Physics by Openstax Chapter 3 Problem 4

Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A⃗ \vec{A} and B⃗ \vec{B} , as in Figure 3.53, then this problem asks you to find their sum R⃗=A⃗+B⃗\vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors A⃗\vec{A} and B⃗ \vec{B} is labeled R⃗\vec{R}. This R⃗\vec{R} is directed θ∘\theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

R=A2+B2R=(18.0 m)2+(25.0 m)2R=30.8 m  (Answer)\begin{align*} R & = \sqrt{A^2 +B^2} \\ R & = \sqrt{\left(18.0 \ \text{m} \right)^2+\left( 25.0 \ \text{m} \right)^2} \\ R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Solve for the value of θ \theta .

θ=arctan⁡(BA)θ=arctan⁡(25.0 m18.0 m)θ=54.2∘\begin{align*} \theta & = \arctan \left( \frac{B}{A} \right) \\ \theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\ \theta & = 54.2^\circ \end{align*}

We need the complementary angle for the compass angle.

90∘−54.2∘=35.8∘\begin{align*} 90^\circ -54.2^\circ =35.8^\circ \end{align*}

Therefore, the compass angle reading is

35.8∘,W of N  (Answer)\begin{align*} 35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 3

Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

snorth=(5.00 km)sin⁡40∘snorth=3.21 km  (Answer)\begin{align*} \text{s}_{\text{north}} & = \left( 5.00 \ \text{km} \right)\sin 40^\circ \\ \text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Using the same right triangle, the east component is computed as follows.

seast=(5.00 km)cos⁡40∘seast=3.83 km  (Answer)\begin{align*} \text{s}_{\text{east}} & = \left( 5.00 \ \text{km} \right)\cos 40^\circ \\ \text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side

Solution:

Part A

The total distance traveled is 

d=(4×120m)+(3×120 m)+(3×120 m)d=1200 m  (Answer)\begin{align*} \text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\ \text{d} & = 1 200\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s=(sx)2+(sy)2s=(1×120 m)2+(3×120 m)2s=(120 m)2+(360 m)2s=379 m  (Answer)\begin{align*} \text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\ \text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\ \text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\ \text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=arctan⁡(sysx)θ=arctan⁡(360 m120 m)θ=71.6∘, N of E (Answer)\begin{align*} \theta & = \arctan \left( \frac{s_y}{s_x} \right) \\ \theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\ \theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 64

(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t=2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64

Solution:

Part A

To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t=0 st = 0 \ \text{s}, x=0 mx = 0 \ \text{m}, and when t=5 st = 5 \ \text{s}, x=17.5 mx = 17.5 \ \text{m}. The velocity at t=2.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=17.5 m−0 m5 s−0 sv=3.5 m/s  (Answer)\begin{align*} \text{velocity} & =\text{slope} \\ \text{v} & =\frac{\Delta x}{\Delta t} \\ \text{v} & = \frac{x_2-x_1}{t_2-t_1} \\ \text{v} & = \frac{17.5\ \text{m}-0\ \text{m}}{5 \ \text{s}-0\ \text{s}} \\ \text{v} & =3.5 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}

Part B

When t=10 st = 10 \ \text{s}, x=2.5 mx=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=2.5 m−17.5 m10 s−5 sv=−3.0 m/s  (Answer)\begin{align*} \text{velocity} & =\text{slope} \\ \text{v} & =\frac{\Delta x}{\Delta t} \\ \text{v} & = \frac{x_2-x_1}{t_2-t_1} \\ \text{v} & = \frac{2.5\ \text{m}-17.5\ \text{m}}{10 \ \text{s}-5\ \text{s}} \\ \text{v} & =-3.0 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\ \end{align*}
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College Physics by Openstax Chapter 2 Problem 63

Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Figure 2.18

Solution:

The figure with the corresponding examples are shown on this page: https://openstax.org/books/college-physics/pages/2-4-acceleration#import-auto-id2590556

The position-vs-time graph of the train’s motion is also graphed in the first figure here: https://openstax.org/apps/archive/20210713.205645/resources/a697c43432cdf2d09a02df47d2b746283b841fcd

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

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