Chap 15 Exercises

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Exercise 1 For each mathematical operation, identify the operation as valid or invalid according to the rules of dimensional arithmetic.

1. In this formula $$\frac{8\text{m}-2.5\text{km}}{2\text{min}-32\text{s}}$$ choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential

It is valid. No rules are violated.

question id: dim-1

2. In this formula $$\frac{3\text{g}\times 2\text{m}}{3{\text{km}}^2}$$ choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential

It is valid. No rules are violated.

question id: dim-2

3. For this formula $${10}^{\frac{4\text{hr}}{3\text{g}}}$$ choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential [correct hint: Nice. The exponent is 4 ft / 3 g, which has dimension L / M. Exponents must always have dimension

It is valid. No rules are violated.

question id: dim3

4. In this formula $$6^{\frac{2\text{hr}}{3\text{min}}}$$choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential

It is valid. No rules are violated.

question id: dim-4

5. In this formula $$5\text{g}\times 3\text{kg}-7\text{lbs}$$ choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential

It is valid. No rules are violated.

question id: dim-5

6. In this formula $$\sqrt[3]{8m^3+27{\text{ft}}^2}$$ choose which rule (if any) is violated.

Addition or Subtraction rule

Multiplication or Division rule

Exponential

It is valid. No rules are violated.

question id: dim-6

Exercise 2 Consider the road descent summarized by this sign …

The “grade” of road is defined as rise-over-run. Since both rise and run have dimension $L$, the ratio is dimensionless. The grade in percent is 100 times rise-over-run. A 100% grade corresponds to a ${45}^{\circ }$ angle.

1. Taking the “run” to be the 5 miles indicated on the sign, what’s the “rise”? (You can find unit conversion data on the internet.)

0.91 miles

4752 feet

1.5 km

292 rods

7.36 furlongs

22.64 $\sqrt{\text{acre}}$

question id: rooster-blue-1

Since a vehicle’s odometer measures distance along the road surface rather than along the horizontal “run,” it is likely that the sign-makers had in mind 5 miles being the length of the hypotenuse of the triangle rather than the horizontal leg.

2. Which of these expressions gives the horizontal run for a rise of $y$ miles?

$\sqrt{5-y^2}$
      $\sqrt{25-y^2}$       $\sqrt{y^2+25}$       $\sqrt{y^2+5}$      

question id: rooster-blue-2

3. Given that the grade is 18% and that the length of the road surface is 5 miles, what is the “rise”?

0.85 miles

1.40 km

283.5 rods

6.8 furlongs

22.01 $\sqrt{\text{acre}}$

question id: rooster-blue-3

Exercise 3 Newton’s law of universal gravitation—also known as the inverse square law—is generally written

$$F=G\frac{m_1{m}_2}{r^2}.$$ $m_1$ and $m_2$ are the masses of the two objects (say, Earth and Sun). $r$ is the distance between the two objects (about 150,000,000 km). $F$ is the gravitational force and $G$ is a fixed quantity called the “gravitational constant.”

Recall that the dimension of force is $ML{T}^{-2}$.

What is the dimension of the gravitational constant, $G$?

$L^3{M}^{-1}{T}^{-2}$

$L^2{M}^{-2}$

$L^2{M}^2{T}^{-2}$

$L^2{M}^{-2}{T}^{-2}$

question id: slc-7-1

The quantity $G$ is $6.674\times {10}^{-11}$ when $L$ is in meters, $M$ is in kilograms, and $T$ in seconds. What is the gravitational force between Earth (mass $6\times {10}^{24}$ kg) and Sun (mass $2\times {10}^{30}$ kg) ?

$3.6\times {10}^{28}$ Newtons

$3.6\times {10}^{31}$ meters

$3.6\times {10}^{28}$ meters per second-squared

$3.6\times {10}^{31}$ meter seconds per kg

question id: slc-7-2

Exercise 4 A simple model of the distance travelled by a tennis ball after launch from a slingshot is $$\text{hdist}(v_0,\theta )=2v_0^2\cos (\theta )\sin (\theta )/g$$ where $\theta$ is the launch angle, measured from the horizontal, $v_0$ is the initial velocity, and $g$ is the acceleration due to gravity.

1. What is the dimension of $v_0$?

L T       L / T       L / T${}^2$       L${}^2$ / T${}^2$      

question id: child-iron-laundry-1

2. What is the dimension of $g$?

M L       M L / T       L / T${}^2$       M L / T${}^2$      

question id: child-iron-laundry-2

3. What is the dimension of $\theta$?

L       L^2       T/L       It is dimensionless      

question id: child-iron-laundry-3

4. What is the dimension of $\sin (\theta$)?

L       L^2       T/L       It is dimensionless      

question id: child-iron-laundry-4

5. What is the dimension of $2v_0^2\cos (\theta )\sin (\theta )/g$?

L       L^2       L/T       L/T${}^2$      

question id: child-iron-laundry-5

6. Suppose the initial velocity of the ball is $v_0=10$ meters/second. Since we are on the surface of Earth, $g=9.8$ meters/second-squared. At the optimal launch angle $\theta$, how far does the model predict the ball will travel?

10 meters       25 meters       50 meters       75 meters      

question id: child-iron-laundry-6

Exercise 5 The “Energy-maneuverability Theory” (E-M) of aircraft performance was developed by renowned fighter pilot Col John Boyd and mathematician Thomas Christie in the 1960s. The theory posits that the available maneuverability of an aircraft is closely related to its specific energy $E_s$, that is, the kinetic plus potential energy of the aircraft divided by aircraft weight. To be highly maneuverable, an aircraft must be able to change it is specific energy rapidly in time. Let’s call this ability the specific power (that is, power divided by mass), $P_s$. An aircraft with large $P_s$ is more maneuverable than one with small $P_s$.

An important formula in E-M Theory is $$P_s=\frac{T-D}{W}V$$ where $T$ is aircraft thrust, $D$ is drag, $W$ is weight, and $V$ is velocity. $(T-D)$. Thrust minus drag, is the net forward force on the aircraft.

Recall these facts about the dimension of physical quantities:

• Velocity has dimension $L^1{T}^{-1}$ (e.g. meters per second)
• Acceleration has dimension $L^1{T}^{-2}=$ [Velocity] $\times T^{-1}$ (e.g. meters per second-squared)
• Force has dimension $M\times$ [Acceleration]
• Energy has dimension [Force] $\times L$
• Power has dimension [Energy] $\times T^{-1}$.

1. Which of the following is a correct dimensional formulation of power?

[Force][Velocity]

[Energy][Velocity]

[Force] / [Velocity]

[Energy] / [Velocity]

question id: boyd-1

2. What is the dimension of $P_s$ in E-M Theory?

[Power] $\times M^{-1}$

[Force] $\times$ [Acceleration]

[Force] $\times$ [Velocity]

[Power]

question id: boyd-2

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