Category Archives: Engineering Mathematics Blog

Problem 6-9: A construct your own problem on College Physics involving angular motion


Construct Your Own Problem

Consider an amusement park ride in which participants are rotated about a vertical axis in a cylinder with vertical walls. Once the angular velocity reaches its full value, the floor drops away and friction between the walls and the riders prevents them from sliding down. Construct a problem in which you calculate the necessary angular velocity that assures the riders will not slide down the wall. Include a free body diagram of a single rider. Among the variables to consider are the radius of the cylinder and the coefficients of friction between the riders’ clothing and the wall.


Problem 6-8: An integrated problem involving circular motion, momentum, and projectile motion


Integrated Concepts

When kicking a football, the kicker rotates his leg about the hip joint.

(a) If the velocity of the tip of the kicker’s shoe is 35.0 m/s and the hip joint is 1.05 m from the tip of the shoe, what is the shoe tip’s angular velocity?

(b) The shoe is in contact with the initially stationary 0.500 kg football for 20.0 ms. What average force is exerted on the football to give it a velocity of 20.0 m/s?

(c) Find the maximum range of the football, neglecting air resistance.


Solution:

Part A

From the given problem, we are given the following values: v=35.0 m/sv=35.0\ \text{m/s} and r=1.05 mr=1.05\ \text{m}. We are required to solve for the angular velocity ω\omega.

The linear velocity, v v and the angular velocity, ω \omega are related by the equation

v=rω or ω=vrv=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into the formula, we can directly solve for the value of the angular velocity. That is,

ω=vrω=35.0 m/s1.05 mω=33.3333 rad/secω=33.3 rad/s  (Answer)\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{35.0\ \text{m/s}}{1.05\ \text{m}} \\ \\ \omega & = 33.3333\ \text{rad/sec} \\ \\ \omega & = 33.3 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

For this part of the problem, we are going to use Newton’s second law of motion in term of linear momentum which states that the net external force equals the change in momentum of a system divided by the time over which it changes. That is

Fnet=ΔpΔt=m(vfvi)tF_{net} = \frac{\Delta p}{\Delta t} = \frac{m\left( v_f - v_i \right)}{t}

For this problem, we are given the following values: m=0.500 kgm=0.500\ \text{kg}, t=20.0×103 st=20.0\times 10^{-3} \ \text{s}, vf=20.0 m/sv_{f}=20.0\ \text{m/s}, and vi=0v_{i}=0. Substituting all these values into the equation, we can solve directly for the value of the net external force.

Fnet=(0.500 kg)(20.0 m/s0 m/s)20.0×103 sFnet=500 N  (Answer)\begin{align*} F_{net} & = \frac{\left( 0.500\ \text{kg} \right)\left( 20.0\ \text{m/s}-0\ \text{m/s} \right)}{20.0\times 10^{-3}\ \text{s}} \\ \\ F_{net} & = 500\ \text{N} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

This is a problem on projectile motion. In this particular case, we are solving for the range of the projectile. The formula for the range of a projectile is

R=v02sin2θgR=\frac{v_{0}^2 \sin 2\theta}{g}

We are asked to solve for the maximum range, and we know that the maximum range happens when the angle θ\theta is 4545^\circ .

R=(20.0 m/s)2sin(2(45))9.81 m/s2R=40.7747 mR=40.8 m  (Answer)\begin{align*} R & = \frac{\left( 20.0\ \text{m/s} \right)^{2} \sin \left( 2\left( 45^\circ \right) \right)}{9.81 \ \text{m/s}^2} \\ \\ R & = 40.7747\ \text{m} \\ \\ R & = 40.8 \ \text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-7: Calculating the angular velocity of a truck’s rotating tires


A truck with 0.420-m-radius tires travels at 32.0 m/s. What is the angular velocity of the rotating tires in radians per second? What is this in rev/min?


Solution:

The linear velocity, vv and the angular velocity ω\omega are related by the equation

v=rω or ω=vrv=r\omega \ \text{or} \ \omega=\frac{v}{r}

From the given problem, we are given the following values: r=0.420 mr=0.420 \ \text{m} and v=32.0 m/sv=32.0 \ \text{m/s}. Substituting these values into the formula, we can directly solve for the angular velocity.

ω=vrω=32.0 m/s0.420 mω=76.1905 rad/sω=76.2 rad/s  (Answer)\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{32.0 \ \text{m/s}}{0.420 \ \text{m}} \\ \\ \omega & = 76.1905 \ \text{rad/s} \\ \\ \omega & = 76.2 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Then, we can convert this into units of revolutions per minute:

ω=76.1905 radsec×1 rev2π rad×60 sec1 minω=727.5657 rev/minω=728 rev/min  (Answer)\begin{align*} \omega & = 76.1905 \ \frac{\bcancel{\text{rad}}}{\bcancel{\text{sec}}}\times \frac{1 \ \text{rev}}{2\pi\ \bcancel{\text{rad}}}\times \frac{60\ \bcancel{\text{sec}}}{1\ \text{min}} \\ \\ \omega & = 727.5657\ \text{rev/min} \\ \\ \omega & = 728\ \text{rev/min} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-6: Calculating the linear velocity of the lacrosse ball with the given angular velocity


In lacrosse, a ball is thrown from a net on the end of a stick by rotating the stick and forearm about the elbow. If the angular velocity of the ball about the elbow joint is 30.0 rad/s and the ball is 1.30 m from the elbow joint, what is the velocity of the ball?


Solution:

The linear velocity, vv and the angular velocity, ω\omega of a rotating object are related by the equation

v=rωv=r\omega

From the given problem, we have the following values: ω=30.0 rad/s\omega=30.0 \ \text{rad/s} and r=1.30 mr=1.30 \ \text{m} . Substituting these values in the formula, we can directly solve for the linear velocity.

v=rωv=(1.30 m)(30.0 rad/s)v=39.0 m/s  (Answer)\begin{align*} v & =r\omega \\ \\ v & = \left( 1.30 \ \text{m} \right)\left( 30.0 \ \text{rad/s} \right) \\ \\ v & = 39.0 \ \text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-5: Calculating the angular velocity of a baseball pitcher’s forearm during a pitch


A baseball pitcher brings his arm forward during a pitch, rotating the forearm about the elbow. If the velocity of the ball in the pitcher’s hand is 35.0 m/s and the ball is 0.300 m from the elbow joint, what is the angular velocity of the forearm?


Solution:

We are given the linear velocity of the ball in the pitcher’s hand, v=35.0 m/sv=35.0\ \text{m/s}, and the radius of the curvature, r=0.300 mr=0.300 \ \text{m}. Linear velocity vv and angular velocity ω\omega are related by

v=rω or ω=vrv=r\omega \ \text{or} \ \omega=\frac{v}{r}

If we substitute the given values into our formula, we can solve for the angular velocity directly. That is,

ω=vrω=35.0 m/s0.300 mω=116.6667 rad/sω=117 rad/s  (Answer)\begin{align*} \omega & = \frac{v}{r} \\ \\ \omega & = \frac{35.0 \ \text{m/s}}{0.300 \ \text{m}} \\ \\ \omega & = 116.6667 \ \text{rad/s} \\ \\ \omega & = 117 \ \text{rad/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The angular velocity of the forearm is about 117 radians per second.


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Problem 6-4: Period, angular velocity, and linear velocity of the Earth


(a) What is the period of rotation of Earth in seconds? (b) What is the angular velocity of Earth? (c) Given that Earth has a radius of 6.4×106 m at its equator, what is the linear velocity at Earth’s surface?


Solution:

Part A

The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is

Period=24 hoursPeriod=24 hours×3600 seconds1 hourPeriod=86400 seconds  (Answer)\begin{align*} \text{Period} & = 24 \ \text{hours} \\ \\ \text{Period} & = 24 \ \text{hours} \times \frac{3600 \ \text{seconds}}{1 \ \text{hour}} \\ \\ \text{Period} & = 86400 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The angular velocity ω\omega is the rate of change of an angle,

ω=ΔθΔt,\omega = \frac{\Delta \theta}{\Delta t},

where a rotation Δθ\Delta \theta takes place in a time Δt\Delta t.

From the given problem, we are given the following: Δθ=2πradian=1 revolution\Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution}, and Δt=24 hours=1440 minutes=86400 seconds\Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds}. Therefore, the angular velocity is

ω=ΔθΔtω=1 revolution1440 minutesω=6.94×104 rpm  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{1 \ \text{revolution}}{1440 \ \text{minutes}}\\ \\ \omega & = 6.94 \times 10^{-4}\ \text{rpm}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

We can also express the angular velocity in units of radians per second. That is

ω=ΔθΔtω=2π radian86400 secondsω=7.27×105 radians/second  (Answer)\begin{align*} \omega & = \frac{\Delta\theta}{\Delta t} \\ \\ \omega & = \frac{2\pi \ \text{radian}}{86400 \ \text{seconds}}\\ \\ \omega & = 7.27 \times 10^{-5}\ \text{radians/second}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The linear velocity vv, and the angular velocity ω\omega are related by the formula

v=rωv = r \omega

From the given problem, we are given the following values: r=6.4×106 metersr=6.4 \times 10^{6} \ \text{meters}, and ω=7.27×105 radians/second\omega = 7.27 \times 10^{-5}\ \text{radians/second}. Therefore, the linear velocity at the surface of the earth is

v=rωv=(6.4×106 meters)(7.27×105 radians/second)v=465.28 m/s  (Answer)\begin{align*} v & =r \omega \\ \\ v & = \left( 6.4 \times 10^{6} \ \text{meters} \right)\left( 7.27 \times 10^{-5}\ \text{radians/second} \right) \\ \\ v & = 465.28 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-3: Calculating the number of revolutions given the tires radius and distance traveled


An automobile with 0.260 m radius tires travels 80,000 km before wearing them out. How many revolutions do the tires make, neglecting any backing up and any change in radius due to wear?


Solution:

The rotation angle Δθ\Delta \theta is defined as the ratio of the arc length to the radius of curvature:

Δθ=Δsr\Delta \theta = \frac{\Delta s}{r}

where arc length Δs\Delta s is distance traveled along a circular path and rr is the radius of curvature of the circular path.

From the given problem, we are given the following quantities: r=0.260 mr=0.260 \ \text{m}, and Δs=80000 km\Delta s = 80000 \ \text{km}.

Δθ=ΔsrΔθ=80000 km×1000 m1 km0.260 mΔθ=307.6923077×106 radians×1 rev2π radiansΔθ=48970751.72 revolutionsΔθ=4.90×107 revolutions  (Answer)\begin{align*} \Delta \theta & = \frac{\Delta s}{r} \\ \\ \Delta \theta & = \frac{80000 \ \text{km} \times \frac{1000 \ \text{m}}{1 \ \text{km}}}{0.260 \ \text{m}} \\ \\ \Delta \theta & = 307.6923077 \times 10^{6} \ \text{radians} \times\frac{1 \ \text{rev}}{2\pi \ \text{radians}} \\ \\ \Delta \theta & = 48970751.72 \ \text{revolutions} \\ \\ \Delta \theta & = 4.90 \times 10^{7} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Problem 6-2: Conversion of units from rpm to revolutions per second and radians per second


Microwave ovens rotate at a rate of about 6 rev/min. What is this in revolutions per second? What is the angular velocity in radians per second?


Solution:

This is a problem on conversion of units. We are given a rotation in revolutions per minute and asked to convert this to revolutions per second and radians per second.

For the first part, we are asked to convert 6 rev/min to revolutions per second.

6 revminute=6 revminute×1 minute60 seconds=0.1 rev/second  (Answer)\begin{align*} \frac{6 \ \text{rev}}{\text{minute}} & = \frac{6 \ \text{rev}}{\bcancel{\text{minute}}} \times \frac{1 \ \bcancel{\text{minute}}}{60 \ \text{seconds}} \\ \\ & = 0.1 \ \text{rev/second} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the next part, we are going to convert 6 rev/min to radians per second.

6 revminute=6 revminute×2π radians1 rev×1 minute60 seconds=0.6283 rad/sec  (Answer)\begin{align*} \frac{6 \ \text{rev}}{\text{minute}} & = \frac{6 \ \bcancel{\text{rev}}}{\bcancel{\text{minute}}} \times \frac{2\pi \ \text{radians}}{1 \ \bcancel{\text{rev}}} \times \frac{1 \ \bcancel{\text{minute}}}{60 \ \text{seconds}} \\ \\ & = 0.6283 \ \text{rad/sec} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

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Cover photo of Chapter 8: Partial Differentiation of the book Differential and Integral Calculus by Feliciano and Uy

Chapter 8: Partial Differentiation


8.1 Partial Derivative

8.2 Geometric Interpretation of Partial Derivative

Exercise 8.1

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

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8.3 Partial Derivatives of Higher Order

Exercise 8.2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

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8.4 Total Differentiation

Exercise 8.3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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8.5 Total Derivative

Exercise 8.4

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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8.6 Partial Differentiation of Implicit Functions

Exercise 8.5

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

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8.7 Tangent Plane and Normal Line

Exercise 8.6

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

8.8 Maxima and Minima

Exercise 8.7

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

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Cover photo of Chapter 7: Derivatives from Parametric Equations, Radius and Center of Curvature of the book Differential and Integral Calculus by Feliciano and Uy

Chapter 7: Derivatives from Parametric Equations, Radius and Center of Curvature


7.1 Derivatives in Parametric Form

Exercise 7.1

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

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7.2 Differential of Arc Length

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7.3 Radius of Curvature

Exercise 7.2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

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7.4 Center of Curvature

Exercise 7.2

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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