Tag Archives: uniform circular motion

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

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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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 6-14: The centripetal acceleration and a linear speed of a point on an edge of an ordinary workshop grindstone

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.

(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.

(b) What is the linear speed of a point on its edge?

Solution:

We are given the following values: $r=7.50\ \text{cm}$ , and $\omega = 6500\ \text{rev/min}$ . We need to convert these values into appropriate units so that we can come up with sensical units when we solve for the centripetal acceleration.

r = 7.50 \ \text{cm} = 0.075 \ \text{m}

\omega = 6500 \ \text{rev/min} \times\frac{2\pi \ \text{rad}}{1\ \text{rev}} \times \frac{1 \ \text{min}}{60\ \text{sec}} = 680.6784 \ \text{rad/sec}

Part A

We are asked to solve for the centripetal acceleration $a_{c}$ . Basing on the given data, we are going to use the formula

a_{c} = r \omega ^{2}

Substituting the given values, we have

\begin{align*} a_{c} & = r \omega ^2 \\ \\ a_{c} & = \left( 0.075 \ \text{m} \right) \left( 680.6784 \ \text{rad/sec} \right)^2 \\ \\ a_{c} & = 34749.2313 \ \text{m/s}^2 \\ \\ a_{c} & = 3.47 \times 10^{4} \ \text{m/s} ^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Now, we can convert the centripetal acceleration in multiples of g.

\begin{align*} a_{c} & = 34749.2313 \ \text{m/s}^2 \times \frac{g}{9.81 \ \text{m/s}^2}\\ \\ a_{c} & =3542.2254g \\ \\ a_{c} & = 3.54\times 10^3 g \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

We are then asked for the linear speed, $v$ of the point on the edge. So, we can use the given values to find the linear speed. We are going to use the formula

v=r\omega

If we substitute the given values, we have

\begin{align*} v & = r \omega \\ \\ v & = \left( 0.075 \ \text{m} \right)\left( 680.6784\ \text{rad/sec} \right) \ \ \\ \\ v & = 51.0509 \ \text{m/s} \\ \\ v & = 51.1 \ \text{m/s} \ \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 6-10: 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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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 10g10g10g 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 θ\thetaθ (as defined in the figure) is related to the speed vvv and radius of curvature rrr 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)θ=tan−1(v2/rg)

(b) Calculate θ\thetaθ 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:

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

An ordinary workshop grindstone has a radius of 7.50 cm and rotates at 6500 rev/min.

(a) Calculate the magnitude of the centripetal acceleration at its edge in meters per second squared and convert it to multiples of g.

(b) What is the linear speed of a point on its edge?

Solution:

Part A

Part B

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:

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:

(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?

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).