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College Physics by Openstax Chapter 6 Problem 30

The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads

Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Solution:

Part A

The formula for an ideally banked curve is

\tan \theta = \frac{v^2}{rg}

Solving for $v$ in terms of all the other variables, we have

v= \sqrt{rg \tan \theta}

For this problem, we are given

radius, $r=100\ \text{m}$
acceleration due to gravity, $g=9.81\ \text{m/s}^2$
banking angle, $\theta = 15.0^\circ$

Substituting all these values in the 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)\tan 15.0^\circ }\\ v & = 16.2129\ \text{m/s} \\ v & = 16.2\ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Let us draw the free-body diagram of the car.

Summing forces in the vertical direction, we have

N \cos \theta + f \sin \theta -w= 0 \quad \quad \quad \color{Blue} \text{Equation 1}

Summing forces in the horizontal directions taking to the left as the positive since the centripetal force is directed this way, we have

N \sin \theta - f \cos \theta = F_c \quad \quad \quad \color{Blue} \text{Equation 2}

We are given the following quantities:

radius of curvature, $r=100\ \text{meters}$
banking angle, $\theta = 15.0^\circ$
velocity, $v=20\ \text{km/h} = 5.5556\ \text{m/s}$
We also know that the friction, $f=\mu_{s} N$ and weight, $w=mg$

We now use equation 1 to solve for N in terms of the other variables.

\begin{align*} N \cos \theta + f \sin \theta -w & = 0 \\ N \cos \theta +\mu_s N\sin \theta & = mg \\ N \left( \cos \theta + \mu_s \sin\theta \right) & =mg \\ N & = \frac{mg}{\cos \theta + \mu_s \sin\theta} \ \qquad \ \color{Blue} \left( \text{Equation 3} \right) \end{align*}

We also solve for N in equation 2.

\begin{align*} N \sin \theta - \mu_s N \cos \theta & = m \frac{v^2}{r} \\ N \left( \sin \theta-\mu_s \cos \theta \right) & = m \frac{v^2}{r} \\ N & = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \ \qquad \ \color{Blue} \left( \text{Equation 4} \right) \end{align*}

Now, we have two equations of $N$ . We now equate these two equations.

\frac{mg}{\cos \theta + \mu_s \sin\theta} = \frac{mv^2}{r \left( \sin \theta-\mu_s \cos \theta \right)}

We can now use this equation to solve for $\mu_s$ .

\begin{align*} \frac{\bcancel{m}g}{\cos \theta + \mu_s \sin\theta} & = \frac{\bcancel{m}v^2}{r \left( \sin \theta-\mu_s \cos \theta \right)} \\ rg\left( \sin \theta-\mu_s \cos \theta \right) & = v^2 \left( \cos \theta + \mu_s \sin\theta \right) \\ rg \sin \theta - \mu_s rg \cos \theta & = v^2 \cos \theta +\mu_s v^2 \sin \theta \\ \mu_s v^2 \sin \theta + \mu _s rg \cos \theta & = rg \sin \theta - v^2 \cos \theta \\ \mu _s \left( v^2 \sin \theta + rg \cos \theta \right) & = rg \sin \theta - v^2 \cos \theta \\ \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \end{align*}

Now that we have an equation for $\mu_s$ , we can substitute the given values.

\begin{align*} \mu _s & = \frac{rg \sin \theta - v^2 \cos \theta}{v^2 \sin \theta + rg \cos \theta} \\ \mu_s & = \frac{100\ \text{m}(9.81\ \text{m/s}^2) \sin 15.0^\circ -\left( 5.5556\ \text{m/s} \right)^2 \cos 15.0^\circ }{\left( 5.5556\ \text{m/s} \right)^2 \sin 15.0^\circ +100\ \text{m}\left( 9.81\ \text{m/s}^2 \right) \cos 15.0^\circ } \\ \mu_s & = 0.2345 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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College Physics by Openstax Chapter 6 Problem 29

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration $10g$ if the rider is 15.0 m from the center of rotation?

(b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure 6.34(b). At what angle $\theta$ below the horizontal will the cage hang when the centripetal acceleration is $10g$ ? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle $10g$ should be.)

**Figure 6.34** (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and reentries. (credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to always be along its axis.

Solution:

Part A

The centripetal acceleration, $a_c$ , is calculated using the formula $a_c = r \omega ^2$ . Solving for the angular velocity, $\omega$ , in terms of the other variables, we should come up with

\omega = \sqrt{\frac{a_c}{r}}

We are given the following values:

centripetal acceleration, $a_c = 10g = 10 \left( 9.81\ \text{m/s}^2 \right) = 98.1\ \text{m/s}^2$

radius of curvature, $r = 15.0\ \text{m}$

Substituting the given values into the equation,

\begin{align*} \omega & = \sqrt{\frac{a_c}{r}} \\ \\ \omega & = \sqrt{\frac{98.1\ \text{m/s}^2}{15.0\ \text{m}}} \\ \\ \omega & = 2.5573\ \text{rad/sec} \\ \\ \omega & = 2.56\ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The free-body diagram of the force is shown

The free-body diagram of the rider’s cage that hangs on a pivot at the end of the arm of a large centrifuge. College Physics Problem 6-29 — The free-body diagram of the rider’s cage hangs on a pivot at the end of the arm of a large centrifuge.

Summing forces in the vertical direction, we have

\begin{align*} \sum_{}^{} F_y & = 0 \\ \\ F_{arm} \sin \theta-w & = 0 \\ \\ F_{arm} & = \frac{w}{\sin \theta} \ \quad \quad \color{Blue} \text{Equation 1} \end{align*}

Now, summing forces in the horizontal direction, taking into account that $F_c$ is the centripetal force which is the net force. That is,

\begin{align*} F_c & = m a_c \end{align*}

We know that $F_c$ is equal to the horizontal component of the force $F_{arm}$ . That is $F_c = F_{arm} \cos \theta$ . Therefore,

\begin{align*} F_{arm} \cos \theta & = m a_c \\ \end{align*}

Now, we can substitute equation 1 into the equation, and the value of the centripetal acceleration given at $10g$ . Also, we note that the weight $w$ is equal to $mg$ . So, we have

\begin{align*} F_{arm} \cos \theta & = m a_c \\ \\ \frac{w}{\sin \theta} \cos \theta & = m (10g) \\ \\ \frac{mg \cos \theta}{\sin \theta} & = 10 mg \\ \\ \end{align*}

From here, we are going to use the trigonometric identity $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$ . We can also cancel $m$ , and g since they can be found on both sides of the equation.

\begin{align*} \frac{1}{\tan \theta} & = 10 \\ \\ \tan \theta & = \frac{1}{10} \\ \\ \theta & = \tan ^{-1} \left( \frac{1}{10} \right) \\ \\ \theta & = 5.71 ^\circ \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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

(b) Calculate $\theta$ 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 $N$ and $F_c$ are the vertical and horizontal components of the force $F$ .

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

\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

\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.

\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 $m$ from both sides, and we can apply the trigonometric identity $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$ . We should come up with

\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, $v = 12.0\ \text{m/s}$

radius of curvature, $r=30.0\ \text{m}$

acceleration due to gravity, $g = 9.81\ \text{m/s}^2$

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

\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 27

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

For ideally banked curved, the ideal banking angle is given by the formula $\displaystyle \tan \theta = \frac{v^2}{rg}$ . We can solve for $r$ in terms of all the other variables, and we should come up with

r = \frac{v^2}{g \tan \theta}

We are given the following values:

ideal banking angle, $\displaystyle \theta = 75.0\ ^\circ$

linear speed, $\displaystyle v=30.0\ \text{m/s}$

acceleration due to gravity, $\displaystyle g=9.81\ \text{m/s}^2$

If we substitute all the given values into our formula for $r$ , we have

\begin{align*} r & = \frac{v^2}{g \tan \theta} \\ \\ r & = \frac{\left( 30.0\ \text{m/s} \right)^2}{\left( 9.81\ \text{m/s}^2 \right)\left( \tan 75^\circ \right)} \\ \\ r & = 24.5825\ \text{m} \\ \\ r & = 24.6\ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The radius of the ideally banked curve is approximately $24.6\ \text{m}$ .

Part B

The centripetal acceleration $a_c$ can be solved using the formula

a_c = \frac{v^2}{r}

Substituting the given values, we have

\begin{align*} a_c & = \frac{v^2}{r} \\ \\ a_c & = \frac{\left( 30.0\ \text{m/s} \right)^2}{24.5825\ \text{m}} \\ \\ a_c & = 36.6114\ \text{m/s}^2 \\ \\ a_c & = 36.6 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The centripetal acceleration is about $36.6\ \text{m/s}^2$ .

Part C

To know how large is the computed centripetal acceleration, we can compare it with that of acceleration due to gravity.

\frac{a_c}{g} = \frac{36.6114\ \text{m/s}^2}{9.81\ \text{m/s}^2} = 3.73

The computed centripetal acceleration is 3.73 times the acceleration due to gravity. That is $a_c = 3.73g$ .

This does not seem too large, but it is clear that bobsledders feel a lot of force on
them going through sharply banked turns!

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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, $m=4\ \text{kg}$

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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The Ideal Speed and the Minimum Coefficient of Friction in Icy Mountain Roads

Problem:

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve (a real problem on icy mountain roads).

(a) Calculate the ideal speed to take a 100 m radius curve banked at 15.0º.

(b) What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

The centripetal acceleration of a large centrifuge as experienced in rocket launches and atmospheric reentries of astronauts

Problem:

A large centrifuge, like the one shown in Figure 6.34(a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric reentries.

(a) At what angular velocity is the centripetal acceleration 10g if the rider is 15.0 m from the center of rotation?

Solution:

Part A

Part B

Riding a Bicycle in an Ideally Banked Curve

Problem:

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

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

Solution:

Part A

Part B

The radius and centripetal acceleration of a bobsled turn on an ideally banked curve

Problem:

(a) What is the radius of a bobsled turn banked at 75.0° and taken at 30.0 m/s, assuming it is ideally banked?

(b) Calculate the centripetal acceleration.

(c) Does this acceleration seem large to you?

Solution:

Part A

Part B

Part C

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 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:

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:

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

Part B

Part C

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

Problem:

(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:

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:

Part A

Part B

Part C

Part D

(a) At what angular velocity is the centripetal acceleration $10g$ if the rider is 15.0 m from the center of rotation?

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

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

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