Category Archives: Sciences

Includes Engineering Sciences such as Statics, Dynamics, Strength of Materials, Physics, Fluid Mechanics, Chemistry

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*}

Advertisements

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

Parent	Satellite	Average orbital radius r(km)	Period T(y)	r³ / T² (km³ / y²)
Sun	Earth	$1.496 \times 10^{8}$	1	$3.35 \times 10^{24}$

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.

Advertisements

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%.

Advertisements

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.

Advertisements

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.

Advertisements

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*}

Advertisements

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*}

Advertisements

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*}

Advertisements

Problem 6-11: Calculating the centripetal acceleration of a runner in a circular track

A runner taking part in the 200 m dash must run around the end of a track that has a circular arc with a radius of curvature of 30 m. If the runner completes the 200 m dash in 23.2 s and runs at constant speed throughout the race, what is the magnitude of their centripetal acceleration as they run the curved portion of the track?

Solution:

Centripetal acceleration $a_{c}$ is the acceleration experienced while in uniform circular motion. It always points toward the center of rotation. It is perpendicular to the linear velocity $v$ and has the magnitude

a_{c}=\frac{v^{2}}{r}

We can solve for the constant speed of the runner using the formula

v=\frac{\Delta x}{\Delta t}

We are given the distance $\Delta x = 200 \ \text{m}$ , and the total time $\Delta t = 23.2\ \text{s}$ . Therefore, the velocity is

\begin{align*} v & =\frac{\Delta x}{\Delta t} \\ \\ v & = \frac{200\ \text{m}}{23.2\ \text{s}} \\ \\ v & = 8.6207\ \text{m/s} \end{align*}

From the given problem, we are given the following values: $r=30\ \text{m}$ . We now have the details to solve for the centripetal acceleration.

\begin{align*} a_{c} & = \frac{v^{2}}{r} \\ \\ a_{c} & = \frac{\left( 8.6207\ \text{m/s} \right)^2}{30\ \text{m}} \\ \\ a_{c} & = 2.4772\ \text{m/s}^{2} \\ \\ a_{c} & = 2.5\ \text{m/s}^{2} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Advertisements

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*}

Advertisements

Solution:

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

Part B

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:

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:

Part A

Part B

Part C

Part D

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

Part B

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

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

Part B

Part C

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:

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:

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).