Understanding Torque: Definition, Calculation, and Applications - Prof. Taylor, Study notes of General Surgery

Download Understanding Torque: Definition, Calculation, and Applications - Prof. Taylor and more Study notes General Surgery in PDF only on Docsity! General Physics 1 Quarter 2 – Module 1 Rotational Equilibrium and Rotational Dynamics 12 i Subject Area – Grade Level: General Physics 1 – Grade 12 Self-Learning Module (SLM) Quarter 2 – Module 1: Rotational Equilibrium and Rotational Dynamics First Edition, 2020 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this module are owned by their respective copyright holders. Every effort has been exerted to locate and seek permission to use these materials from their respective copyright owners. The publisher and authors do not represent nor claim ownership over them. Printed in the Philippines by Department of Education – SOCCSKSARGEN Region Office Address: Regional Center, Brgy. Carpenter Hill, City of Koronadal Telefax: (083) 2288825/ (083) 2281893 E-mail Address: region12@deped.gov.ph Development Team of the Module Writers: Emma T. Suritta Judie E. Dela Cruz Jerold F. Villalobos John F. Villalobos Anna Liza C. Suello Jessa J. Jakil Ligaya D. Maneja Catherine S. Panaligan Editors: Retchie Joy B. Pisaña Reviewers: Blessy Mae M. Cabayao, Murdy F. Bautista, Jay Sheen A. Molina Illustrator: Merbin M. Sulit Layout Artists: Solomon P. Lebeco Jr. Cover Art Designer: Ian Caesar E. Frondoza Management Team: Allan G. Farnazo, CESO IV – Regional Director Atty. Fiel Y. Almendra, CESO V – Assistant Regional Director Ruth L. Estacio PhD, CESO VI – OIC-Schools Division Superintendent Jasmin P. Isla - Assistant Schools Division Superintendent Gilbert B. Barrera – Chief, CLMD Arturo D. Tingson Jr. – REPS, LRMS Peter Van C. Ang-ug – REPS, ADM and Science Lalaine SJ. Manuntag PhD - CID Chief Nelida A. Castillo – EPS, LRMS Marichu Jean R. Dela Cruz -EPS – Science, ADM iv What I Can Do This section provides an activity which will help you transfer your new knowledge or skill into real life situations or concerns. Assessment This is a task which aims to evaluate your level of mastery in achieving the learning competency. Additional Activities In this portion, another activity will be given to you to enrich your knowledge or skill of the lesson learned. This also tends retention of learned concepts. Answer Key This contains answers to all activities in the module. At the end of this module you will also find: The following are some reminders in using this module: 1. Use the module with care. Do not put unnecessary mark/s on any part of the module. Use a separate sheet of paper in answering the exercises. 2. Don’t forget to answer What I Know before moving on to the other activities included in the module. 3. Read the instruction carefully before doing each task. 4. Observe honesty and integrity in doing the tasks and checking your answers. 5. Finish the task at hand before proceeding to the next. 6. Return this module to your teacher/facilitator once you are through with it. If you encounter any difficulty in answering the tasks in this module, do not hesitate to consult your teacher or facilitator. Always bear in mind that you are not alone. We hope that through this material, you will experience meaningful learning and gain deep understanding of the relevant competencies. You can References This is a list of all sources used in developing this module. 1 What I Need to Know This module contains three (3) lessons which are: Lesson 1: Moment of Inertia Lesson 2: Torque Lesson 3: Static Equilibrium, Rotational Kinematics, and Work Done by a Torque Learning Objectives: 1. Define and describe operationally moment of inertia; 2. Calculate the magnitude of moment inertia in a given system; 3. Define and describe torque; 4. Calculate the magnitude and direction of torque using the definition of torque as a cross product; 5. Describe rotational quantities using vectors; 6. Determine whether a system is in static equilibrium or not; and 7. Apply the rotational kinematic relations for systems with constant angular acceleration. 2 What I Know Pre-Assessment Multiple Choice. Read the questions carefully. Write the letter of your answer in your notebook or answer sheet provided. 1. The moment of inertia of an object is greater when most of the mass is located in what area? a. off-center b. near the axis c. on the rotational axis d. away from the rotational axis 2. What quantity describes the ability of a force to rotate an object? a. mass b. moment of inertia c. torque d. velocity 3. Which of these statements best describes the torque produced when two children of different weights balance on a seesaw? a. equal torques b. unequal torques c. equal torques in the same direction d. equal torques in opposite direction 4. A force of 250 N is exerted on a cable wrapped around a drum that has a diameter of 120 mm. What is the torque produced about the center of the drum? a. 15 N 30 N c. 45 N d. 60 N 5. A 4.0 kg block travels around a 0.50 m radius circle with an angular velocity of 12 rad/s. What is its angular momentum at the center of the circle? a. 12 kgm2/s b. 24 kgm2/s c. 6 kgm2/s d. 3 kgm2/s 6. Which of the following tells if an object is in equilibrium? a. the object is in motion b. the forces acting are not equal c. the object moving at a constant speed d. the net force is equal to zero 7. What type of forces will cause an object to rotate? a. concurrent forces b. equilibrium forces c. normal forces d. non-concurrent forces 8. How many radians are there in 60 revolutions? a. 0.49 rad b. 0.98 rad c. 2.5 rad d. 18.8 rad 9. How long will the helicopter blade turn through 200 radians if its angular velocity is 60rad/s? a. 2 seconds b. 3.3 seconds c. 4 seconds d. 6 seconds 5 What Is It Moment of Inertia is defined as a measure of an object’s resistance to changes to its rotation. It also tells the capacity of a cross-section to resist bending. It must be specified with respect to a chosen axis of rotation. It is usually quantified in m4 or kgm2. It is also a name given to rotational inertia, the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation. For a point mass, the moment of inertia is just the mass times the square of perpendicular distance to the rotation axis, I = mr2. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses. The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared. The total moment of inertia is due to the sum of masses at a distance from the axis of rotation. It shows below: 6 A collection of two masses has a moment of inertia due to each separate mass. I = mr2 + mr2 = 2mr2 Since the moment of inertia of an ordinary object involves a continuous distribution of mass at a continually varying distance from any rotation axis, the calculation of moments of inertia generally involves calculus, the discipline of mathematics which can handle such continuous variables. Since the moment of inertia of a point mass is defined by Examples: 1. A 100 gram ball connected to one end of a cord with a length of 30 cm. What is the moment of inertia of a ball about the axis of rotation AB? Ignore cord’s mass. A Illustration: m B Given: The axis of rotation at AB Mass ball (m) = 100 gram/1000 =0.1 kg The distance between the ball and the axis rotation (r) = 30 cm = 0.3 m Unknown: Moment of inertia of a ball (I) Solution: I = mr2 = (01 kg)(0.3 m)2 I = (0.1 kg)(0.09 m2) I = 0.009 kgm2 7 What’s More Activity 3: Calculate Me! 1. A 100 gram ball m1, and a 200 gram ball, m2, connected by a rod with a length of 60 cm. The mass of the rod is ignored. The axis of rotation located at the center of the rod. What is the moment of inertia of the balls about the axis of rotation? Illustration: A m1 m2 B What I Have Learned Activity 4. Complete Me! Direction: Fill out the necessary word/s in the blank/s in every sentence below. Write your answer on your answer sheet. 1. The moment of inertia is defined with respect to a specific rotation ____________. 2. The moment of inertia of a ____________ with respect to an axis is defined as the product of the mass times the distance from the axis squared. 3. Moment of inertia is the name given to _______________________. 4. The point mass relationships become the __________ for all the other moments inertia. 5. The unit of moment of inertia of an area is ________________ . What I Can Do Draw or sketch a diagram that will describe the moment of inertia in a multi-system. 10 Torque is the twisting force that tends to cause rotation. The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters. Recall that torque is a vector and is obtained by taking the product of two vectors. Cross-product A⃗ ×B⃗ = C⃗ {/ C⃗/= / A⃗/./ B⃗/. sinθ} magnitude of the vector C⃗ Direction of C: Orient your fingers in direction of A⃗ and curl in direction of B⃗, then your thumb will point in direction of C⃗. Note that the direction of C⃗ is always perpendicular to A⃗ and perpendicular to B⃗. If A = Ax i^+Ay j^ +Az k^ and B = Bx i^+By j^ +Bz k^ For the first element of the first row, the i^i^, take the product down and to the right, ( this yields i^AyBzi^AyBz ) minus the product down and to the left ( the product down-and-to-the-left is i^AzByi^AzBy ) For the first element in the first row, we thus have: i^AyBz−i^AzByi^AyBz−i^Az By which can be written as: (AyBz−AzBy)i^(AyBz−AzBy)i^ . Repeating the process for the second and third elements in the first row (the j^j^ and the k^k^ ) we get (AzBx−AxBz)j^(AzBx−AxBz)j^ and (AxBy−AyBx)k^(AxBy−AyBx)k^ respectively. Adding the three results, to form the determinant of the matrix results in: A⃗ ×B⃗ =(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^ A→×B→=(AyBz−AzBy)i^+(AzBx−AxBz)j^+(AxBy−AyBx)k^ 11 Example Number 2: Suppose the lever arm vector r⃗ is given by the equation r⃗ = 5 i^+6 j^+1 k^. Calculate the torque if the force is F = 10 i^N Solution: Torque = (5 i^+6 j^+1 k^)(10 i^+0 j^+0 k^) = (6X0 – 1X0) i^ + (1X10 – 5X0) j^ + (5X0 – 6X10) k^ = (0 – 0) i^ + (10 – 0) j^ + (0 – 60) k^ = 0 i^ + 10 j^ + -60 k^ = √(10𝑗^)2 + (−60k^)2 = √102 + 602 = √100 + 3600 = √3700 = 60. 83 N.m ( Magnitude of the Torque ) What’s More Activity Number 2. FIND MY A⃗ ×B⃗! A⃗ = 3 i^ + 2 j^ -5 k^ B⃗ = 2 i^ -6 j^ + 4 k^ Solve for A⃗ ×B⃗ ? What I Have Learned From the very start of our lesson, you have learned a lot about torque as a vector. In the succeeding activities, you will make generalizations of your learnings. 12 Now after we have done varied activities let us check how far you understood our lesson. Answer the following questions as brief as you can. 1. Define operationally torque as a vector quantity. 2. Where does the direction of the torque vector depend? 3. What should always be the direction of C⃗? What I Can Do Activity Number 3. FIND MY A⃗ ×B⃗ and C⃗ A⃗ = 4 i^ -2 j^ -6 k^ B⃗ = 5 i^ +4 j^ + 4 k^ Solve for: 1. A⃗ ×B⃗ 2. C⃗ 15 With units of kilogram-meter squared (kg-m2) In the simplest kind of rotation, points on a rigid object move on circular paths around an axis of rotation. The term equilibrium has been used before to describe how an object experiences zero net force, and thus, its motion is unchanged. (If it rest, it will remain at rest; if in uniform motion, it will remain in uniform motion.) The Law of inertia actually describes the first condition of equilibrium, the case of linear equilibrium or zero linear acceleration. The second condition of equilibrium is rotational equilibrium. As discussed in the previous lesson, when a number of forces are acting on an object, the total torque is the sum of the individual torques produced by each force acting on the object. In the event that the sum of these individual torques turns out to be zero, this means we have attained rotational equilibrium. The second condition for equilibrium may be written as ∑t = 0 Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton. (Recall that the net force is "the vector sum of all the forces" or the resultant of adding all the individual forces head-to-tail.) Thus, an accurately drawn vector addition diagram can be constructed to determine the resultant. Sample data for such a lab are shown below. Force A Force B Force C Magnitude 3.4 N 9.2 N 9.8 N Direction 161 deg. 70 deg. 270 deg Another way of determining the net force (vector sum of all the forces) involves using the trigonometric functions to resolve each force into its horizontal and vertical components. Once the components are known, they can be compared to see if the vertical forces are balanced and if the horizontal forces are balanced. The diagram below shows vectors A, B, and C and their respective components. For vectors A and B, the vertical components can be determined using the sine of the angle and the horizontal components can be analyzed using the cosine of the angle. The magnitude and direction of each component for the sample data are shown in the table below the diagram. 16 The data in the table above show that the forces nearly balance. An analysis of the horizontal components shows that the leftward component of A nearly balances the rightward component of B. An analysis of the vertical components show that the sum of the upward components of A + B nearly balance the downward component of C. The vector sum of all the forces is (nearly) equal to 0 Newton. But what about the 0.1 N difference between rightward and leftward forces and the 0.2 N difference between the upward and downward forces? Why do the components of force only nearly balance? The sample data used in this analysis are the result of measured data from an actual experimental setup. The difference between the actual results and the expected results is due to the error incurred when measuring force A and force B. We would have to conclude that this low margin of experimental error reflects an experiment with excellent results. We could say it's "close enough for government work." In conclusion, equilibrium is the state of an object in which all the forces acting upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric functions can be utilized to determine the horizontal and vertical components of each force. If at equilibrium, then all the vertical components must balance and all the horizontal components must balance. What’s More Activity Number 1. Let’s Solve! 1. The radius of a car tire is about 0.35 meters. If the car accelerates in straight line from rest at 2.8 meters per second squared, what is the angular acceleration, both magnitude and direction, of the front passenger-side tire? 17 What I Have Learned From the very start of our lesson, you have learned a lot about describing rotational quantities using vectors as well as determining whether a system is in static equilibrium or not and applying the rotational kinematic relation for a system with constant angular acceleration. In the succeeding activities, you will make generalizations of your learnings. Now after we have done varied activities let us check how far you understood our lesson. 1. Suppose that a student pulls with two large forces (F1 and F2) in order to lift a 1-kg book by two cables. If the cables make a 1-degree angle with the horizontal, then what is the tension in the cable? What I Can Do A Jet Revving Its Engines As seen from the front of the engine, the fan blades are rotating with an angular speed of -110 rad/s. As the plane takes off, the angular velocity of the blades reaches -330 rad/s in a time of 14 s. Find the angular acceleration, assuming it to be constant. 20 Answer Key What's More ( Lesson 1) Solution: I = m1 + r2 + m2r2 I = (0.1kg)(0.3m)2+(0.2kg)(0.3m)2 I=(0.1kg)(0.09m2)+(0.2kg)(0.09m2) I=0.009kgm2 + 0.018kgm2 I=0.027 kgm2 What's In ( Lesson 1) 1.Center of m(0.2kg)ass 2.Momentum 3.impluse What I Know 1.B 2.C 3.D 4.A 5.A 6.C 7.D 8.D 9.B 10.A 11.D 12.B 13.B 14.C 15.D 21 What I Can Do Lesson 3 ǡ= -16 rad/s 2 Assessment POST TEST 1.B 2.B 3.D 4.B 5.C 6.D 7.B 8.A 9.C 10.C 11.D 12.A 13.A 14.C 15.d What I Have Learned Lesson 1: 1. Axis 2. Point mass 3. Rotational inertia 4. Basis 5. M4 For lesson 3 The tension 281 Newtons! Since the mass is 1 kg, the weight is 9.8 N. Each cable must pull upwards with 4.9 N of force. Thus, sine (1 degree) = (4.9 N) / (Ftens). Proper use of algebra leads to the equation Ftens = (4.9 N) / [ sine (1 degree) ] = 281 N. 22 References Aquino, M et al. Science Links, Physics. Rex Book Store Dadinirt, T. “Kepler's law of planetary motion.” https://www.slideshare.net/ Accessed November 3, 2020. Erickson, K. “What is Gravity.”. https://spaceplace.nasa.gov Accessed October 12, 2020 Faller, J. “Gravity”. https://www.britanicca.com Accessed October 12, 2020 Hayath, Zishaan. “State the Newton Law of Gravitation.” https://www.toppr.com Accessed October 15, 2020. Henderson, Tom. “Newton’s Law of Universal Gravitation.” https: https://www.physicsclassroom.com/class/circles/Lesson-3/Newton-s-Law-of- Universal-Gravitation Accessed October 10, 2020. Libretext Project. “Newton's Law of Universal Gravitation.” https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Phy sics_(OpenStax)/Newton's_Law_of_Universal_Gravitation Accessed October 12, 2020. Mann. A. “What is Gravity?”. https://www.livescience.com/ Accessed October 10, 2020 National Aeronautics and Space Administration. “More on Newton’s Law of Universal Gravitation.”. https://imagine.gsfc.nasa.gov/ Accessed October 12, 2020 National Aeronautics and Space Administration. “Kepler's Laws of Planetary Motion.”. https//solarsystem.nasa.gov Accessed November 3, 2020 Weber State University, “Universal Law of Gravitation,” http://physics.weber.edu/amiri/physics1010online/WSUonline12w/OnLineCourseMovi es/CircularMotion&Gravity/reviewofgravity/ReviewofGravity.html October 10, 2020. Wiley, David. “Newton’s Law of Universal Gravitation.” https://courses.lumenlearning.com/boundless-physics/chapter/newtons-law-of- universal-gravitation/ Accessed October15, 2020.