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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 6-19: 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:

\text{radius} = \frac{\text{diameter}}{2} = \frac{200\ \text{m}}{2} = 100 \ \text{m}
\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

a_{c} = r \omega ^2

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

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

Substituting the given quantities,

\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\ \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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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 6-13: 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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