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

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General

Physics 1

Quarter 2 – Module 1

Rotational Equilibrium and

Rotational Dynamics

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: [email protected] 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 For the learner: Welcome to the General Physics 1 Self – Learning Module (SLM) on Rotational Equilibrium and Rotational Dynamics. The hand is one of the most symbolized part of the human body. It is often used to depict skill, action and purpose. Through our hands we may learn, create and accomplish. Hence, the hand in this learning resource signifies that you as a learner is capable and empowered to successfully achieve the relevant competencies and skills at your own pace and time. Your academic success lies in your own hands! This module was designed to provide you with fun and meaningful opportunities for guided and independent learning at your own pace and time. You will be enabled to process the contents of the learning resource while being an active learner. This module has the following parts and corresponding icons: What I Need to Know This^ will give you an idea^ of the skills^ or competencies you are expected to learn in the module. What I Know This part^ includes an^ activity^ that aims^ to check what you already know about the lesson to take. If you get all the answers correct (100%), you may decide to skip this module. What’s In This is a brief drill or review to help you link the current lesson with the previous one. What’s New In^ this^ portion,^ the^ new^ lesson^ will^ be introduced to you in various ways such as a story, a song, a poem, a problem opener, an activity or a situation. What is It This section provides a^ brief discussion of the lesson. This aims to help you discover and understand new concepts and skills. What’s More This^ comprises^ activities^ for^ independent practice to solidify your understanding and skills of the topic. You may check the answers to the exercises using the Answer Key at the end of the module. What I Have Learned This^ includes^ questions^ or^ blank sentence/paragraph to be filled in to process what you learned from the lesson.

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.

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 kgm^2 /s b. 24 kgm^2 /s c. 6 kgm^2 /s d. 3 kgm^2 /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

10. A bowling ball has a mass of 5.5 kg and a radius of 12.0 cm. It is released so that it rolls down the alley at a rate of 12 revs/s. What is the magnitude of its angular momentum? a. 0.76 kgm^2 /s b. 1.8 kgm^2 /s c. 4.1 kgm^2 /s d. 5.1 kgm^2 /s
11. What is maximum torque on a 800 turn circular coil of radius 0.75 cm that carries a current of 1.9 mA and resides in a uniform magnetic field of 0.25T? a. 6.41x10-5 N.m b. 6.51x10-5 N.m c. 6.61x10-5 N.m d. 6.71x10-5 N.m
12. A child spins a top with a radius of 2 cm with a force of 0.5 N. How much torque if generated at the edge of the top? a. 0.05 N.m b. 0.01 N.m c. 0.75 N.m d. 0.25 N.m
13. An object moves at a constant speed of 9.0 m/s in a circular path of radius of 1.5 m. What is the angular acceleration of an object? a. 14 rad/s^2 b. 36 rad/s^2 c. 81 rad/s^2 d. 54 rad/s^2
14. Which quantity below refers to any system where the sum of the force and torque on every particle of the system happens to be zero? a. rotational kinematics b. Moment of inertia c. static equilibrium d. torque
15. Which quantity below refers to moment of force, in physics, the tendency of a force to rotate the body to which it is applied? a. rotational kinematics b. moment of inertia c. static equilibrium d. torque

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 m^4 or kgm^2. 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 = mr^2. 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:

A collection of two masses has a moment of inertia due to each separate mass. I = mr^2 + mr^2 = 2mr^2 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 = mr^2 = (01 kg)(0.3 m)^2 I = (0.1 kg)(0.09 m^2 ) I = 0.009 kgm^2

Lesson 2 Torque Learning Objectives:

1. Define and describe torque; and

2. Calculate the magnitude and direction of torque using the definition

of torque as a cross product.

When studying how objects rotate, it quickly becomes necessary to figure out how a given force results in a change in the rotational motion. The tendency of a force to cause or change rotational motion is called torque , and it's one of the most important concepts to understand in resolving rotational motion situations. Torque (also called moment — mostly by engineers) is calculated by multiplying force and distance. The SI units of torque are newton-meters, or N*m (even though these units are the same as Joules, torque isn't work or energy, so should just be newton-meters). Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration. Torque is a vector quantity. The direction of the torque vector depends on the direction of the force on the axis. What’s In How was your performance in the test? As you go through this module, you will be able to deepen your understanding in our topic and do better in the next test. In lesson 1, you were able to calculate the moment of inertia about a given axis of single-object and multiple-object systems. Here in lesson 2, you will be introduced to the concept on how to calculate magnitude and direction of torque using the definition of torque as a cross product.

Let us have a sample problem.

Example Number 1: A force of 5.0 N is applied at the end of a lever that has a length of 2.0 meters. If the force is applied directly perpendicular to the lever, as shown in the diagram above, what will be the magnitude of the torque acting on the lever?

Solution: This sample is a simple matter of plugging the values into the equation: Torque = F * 1 Torque = 5 .0N * 2.0m Torque = 10N*m Activity Number 1. Problem on Torque

1. If the same force as in example 1 is applied at an angle of 30

degrees at the end of the 2.0 meters lever, what will be the

magnitude of the torque?

2. What force is necessary to generate a 20.0 N*m torque at an angle

of 50 degrees from along a 3.00 m rod?

What’ New In a previous lecture, we discussed one type of product of two vectors known as dot product, which gives us a scalar value. A very common example of such a product is work. Work is a scalar quantity that has only magnitude and no direction and is given by taking the product of the force and displacement vectors. Cross product (or vector product) is the product of two vectors that produces a third vector. This means that the result has both magnitude as well as direction. We can use different methods to calculate the cross product. One way to calculate it is to take the product of the magnitudes of the two vectors and multiply by the sine of the angle between them. Note that the direction of the vector is given by the right hand rule and is always perpendicular to the two vectors. If we are given two three-dimensional vectors and we are asked to calculate the cross product vector, we have to use the method as described in the lecture. What is It Torque as a vector quantity The vector or cross product is another way to combine two vectors; it creates a vector perpendicular to both the originals. In vector form, torque is the cross product of the radius vector (from axis of rotation to point of application of force) and the force vector.

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^ = √( 10 𝑗^)^2 + (− 60 k^)^2 = √ 102 + 602 = √ 100 + 3600 = √ 3700 = 60. 83 N.m ( Magnitude of the Torque ) What’s More Activity Number 2. FIND MY A ⃗ ×B ⃗! 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.

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 A ⃗ ×B ⃗ and C ⃗ A⃗ = 4 i^ - 2 j^ - 6 k^ B⃗ = 5 i^ +4 j^ + 4 k^ Solve for:

1. A⃗ ×B⃗
2. C⃗

What’s New Using Physics, you can calculate the angular acceleration of an object in circular motion. Problem 1 : When you switch your room fan from medium to high speed, the blades accelerate at 1.2 radians per second squared for 1.5 seconds. If the initial angular speed of the fan blades is 3.0 radians per second, what is the final angular speed of the fan blades in radians per second? Angular is defined by α = ∆ŵ ∆t Where: ∆ŵ= final angular speed minus initial angular speed ∆t = time over which the angular speed changes, so you can write ∆ŵ=ŵf – ŵi where ŵi = 3.0 radians per second Solve the equation for acceleration for the final angular speed and plug in the known quantities to get the answer. The result is α= ∆ŵ ∆t = ŵf – ŵi ∆t α∆t = ŵf – ŵi ŵf = α∆t + ŵi = 1.2 rad/s^2 x 1.5 s + 3.0 rad/s = 4.8 rad/s What is It In a previous lecture, you calculated the magnitude and direction of torque using the definition of torque as a cross product. Torque can be described as the rotational equivalent of straight-line force. Angular acceleration can be thought of as the rotational equivalent of straight-line acceleration. There is also a rotational equivalent of mass: rotational inertia, represented by the upper case italic letter I, in some text it is called moment of inertia. Thus, following Newton’s second law of motion, torque is computed as t= Iα in N-m. If the rotation about an axis is to be changed, torque must be applied. The rotational inertia of a particle about the chosen axis is the product of the mass of the particle and the square of the radius, I = mr^2

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.