General Physics 1 - Rotational Motion PDF

ROTATIONAL MOTION for GENERAL PHYSICS 1/ Grade 12 Quarter 2/ Week 1 1  FOREWORD This Self-learning Kit is designed to holistically developed STEM Students. The activities provided are MELC-aligned and is simplified for better understand...

ROTATIONAL MOTION for GENERAL PHYSICS 1/ Grade 12 Quarter 2/ Week 1 1  FOREWORD This Self-learning Kit is designed to holistically developed STEM Students. The activities provided are MELC-aligned and is simplified for better understanding. This kit deals with calculations on the moment of inertia and magnitude & direction of torque. Students will also learn to describe rotational quantities. Determining a system to be in static equilibrium or not is also given emphasis. Moreover, concepts presented will lead students to apply rotational kinematic relations for systems with constant angular accelerations. They will also apply the torque-angular momentum relation. Students’ skills in solving static equilibrium problems will also be enhanced. Thus, this kit may become an instrument in the development of students who are critical problem solvers, lifetime learners, responsible stewards of Mother Earth, truth seekers, impartial decision makers, responsive and ingenious citizens, and effective communicators. 2  OBJECTIVES At the end of this Self-Learning Kit, you should be able to: K: describe rotational quantities using vectors; S: calculate the moment of inertia about a given axis; : calculate magnitude and direction of torque; : solve static equilibrium problems; and A: display appreciation on the application of rotational motion to people’s daily lives. LEARNING COMPETENCIES Calculate the moment of inertia about a given axis of single-object and multiple-object systems (STEM_GP12RED-IIa-1). Calculate magnitude and direction of torque using the definition of torque as a cross product (STEM_GP12REDIIa-3). Describe rotational quantities using vectors (STEM_GP12REDIIa-4). Determine whether a system is in static equilibrium or not (STEM_GP12REDIIa-5). Apply the rotational kinematic relations for systems with constant angular accelerations (STEM_GP12REDIIa-6). Determine angular momentum of different systems (STEM_GP12REDIIa-9). Apply the torque-angular momentum relation (STEM_GP12REDIIa- 10). Solve static equilibrium problems in contexts but not limited to see-saws, cable-hinge-strut system, leaning ladders, and weighing a heavy suitcase using a small bathroom scale (STEM_GP12REDIIa-8). 3 I. WHAT HAPPENED PRE-TEST: I. CONCEPTS IN A BOX: Complete the chart by supplying the right words/phrases related to rotational motion. The words/phrases are provided in the box below. Write your answers on your notebook/Activity Sheet. Kepler’s laws of planetary motion linear motion kinematics motion about an axis law of periods escape velocity law of equal areas law of ellipses angular quantities planetary motion 4  II. MODIFIED TRUE OR FALSE: Read each statement carefully. Write TORQUE if the statement is true and FORCE if it is false. Write your answer on your notebook/answer sheet. 1. Torques are associated with rotation. 2. Moment arm of a force is the perpendicular distance from the force’s line of action to the axis of rotation. 3. If you want to make an object move, apply force. 4. The other name for moment of arm is leverage. 5. Torque is the quantity that measures the ability of a force to rotate n object around some axis. 6. Positive torque is when turned counterclockwise and negative torque is when turned clockwise. 7. The center of mass is the perpendicular distance between the fulcrum and the point of application of the force. 8. Inertia is the measurement of a rotation caused by a force. 9. Fulcrum or pivot point refers to the point of rotation. 10. Newton-meter is the unit of torque. II. WHAT I NEED TO KNOW DISCUSSION: INERTIA Inertia is defined as the tendency of an object at rest to remain at rest and an object in motion to stay moving in a straight line at a constant velocity. A similar principle applies to objects moving in rotational motion. Moment Of Inertia Moment of inertia also known as rotational inertia, is defined as the property of a rotating body to resist change in its state of rotation. The SI unit for the moment of inertia is kg· m 2. An object rotating about an axis tends to continue rotating about that axis unless an unbalanced external force (torque) tries to stop it. This is because objects tend to resist any change in their state of motion. This resistance is physically embodied in the mass of the object. Moment of inertia depends on the distribution of the mass. A mass which is at greater distance from the axis of rotation has a Figure 1. Dumbbell B is easier to greater moment of inertia compared to the rotate because its masses are same mass which is near the axis of rotation. near its axis of rotation; hence Thus, dumbbell A is more difficult to rotate than the dumbbells’ moment of dumbbell B. inertia is smaller. The opposite can be said of dumbbell a. (Padua and Crisostomo 2003) 5  Figure 2. The moment of inertia in A comparatively small because the limbs are drawn near the spin axis. The moment of inertia in B increases because of the bent knees. The moment of inertia reaches its maximum in C because of the extended arms and legs which are perpendicular to the axis. (Padua and Crisostomo, 2003) The moment of inertia gives a measurement of the resistance of the body to a change in its rotational motion. The larger the moment of inertia of a body, the more difficult it is to put that body into rotational motion or the larger the moment of inertia of body, the more difficult it is to stop its rotational motion. The moment of inertia of a particle about an axis is obtained by multiplying the mass by the square of its distance from the axis. where is the moment of inertia; is mass; and is distance from the axis of rotation For a system made up of several particles, the moment of inertia of the system is the sum of the individual moments of inertia. Radius of gyration is the distance from an axis of rotation where the mass of a body may be assumed to be concentrated without altering the moment of inertia of the body about that axis. Radius of gyration is analogous to the center of mass. 6  It is important to remember that when moment of inertia is asked for, it is a must to determine first about what axis the rotation will takes place. Because, is different from each axis and, since differs as , is also different from each axis. Calculus is usually used to solve for the moment of inertia. However, for simplicity, figure 3 below shows how values of the moments of inertia for some symmetrical bodies about different axis can be determined. Figure 3. Moment of inertia of uniform and regular shaped bodies (Silverio, 2017) 7 Example: 8 POINTS TO PONDER  The moment of inertia also known as rotational inertia is the measure of the resistance of a body to a change in its rotational motion.  The larger the moment of inertia of a body, the more difficult it is to change its rotational motion.  Radius of gyration (k) is the distance from an axis of rotation where the mass of a body may be assumed to be concentrated without altering the moment of inertia of the body about that axis.  Before solving for the moment of inertia, consider first the shape of the object and its specified axis of rotation. 9  TORQUE To understand the concept of torque, we begin by investigating the rotation of a door hinged at one edge. Try pulling the door with a constant force applied at different points (Figure 4a) and at varying angles (Figure 4b). This simple activity shows that it is easiest to rotate the door when the force is applied farthest from the hinge. That is the reason why doorknobs are placed at the other edge of the door farthest from the hinge. Furthermore, rotation is greatest when force is applied perpendicularly to the door. Retrieved from The effectiveness of a force in rotating https://images.app.goo.gl/rrnFmnxJvq xH9jPAs6 a body on which it acts is called torque or moment of force. The Greek letter tau 𝜏 is Figure 4. The orientation of the usually used to represent torque. Torque can applied force affects the rotation of the door. be determined by multiplying the force applied F by the perpendicular distance from the axis of rotation to the line along which the force acts. This perpendicular distance is called moment arm or lever arm, represented by r. The magnitude of the torque can be calculated using, Maximum torque can be obtained if θ is 90°. The value for sin 90° is 1. If the angle is perpendicular or is in 90°, you can use this formula, 𝜏 This formula is consistent with the definition of torque as a cross product of force and displacement/distance. An equivalent way of finding torque associated with a given force into components parallel and perpendicular to the axis of rotation. The parallel component F1 exerts no torque. F2 also exerts no torque since the force applied 10 is directed to the axis of rotation. The perpendicular component of the force F3 produces torque. Retrieved from https://images.app.goo.gl/tBN7qxhuNfpjPM3M9 Figure 5. Determining torque using the orientation of the force being applied The SI unit of torque is meter· newton (m· N). Torque is a vector quantity. Torque may also be positive or negative depending on the sense of rotation. By convention, torque is positive if it tends to produce counterclockwise rotation. It is negative if it tends to produce clockwise rotation. The greater torque applied to an object, the greater its tendency to rotate. Torque is also applied when you turn on a water faucet or tighten a nut with a wrench. Example: 1. A particle located at 0.56 m is acted upon by a constant force of 4.7 N. The force makes the particles to rotate in clockwise direction. Calculate the torque about the origin that the particle experiences as a result of this force. Given: r = 0.56 m Find: 𝜏 F = 4.7 N Assuming that the angle is 90°, 𝜏 = F r sin ϴ = (4.7 N) (0.56 m) (sin 90) = 2.632 N·m = 2.6 N·m 11  2. If the force applied is perpendicular to the handle of the wrench as shown in the diagram, find: a. torque exerted by the force about the center of the nut b. type of rotation Given: F = 2.5 N r = 15 cm = 0.15 m ϴ = 90˚ (since the arm is perpendicular to the handle of the wrench) a. Find 𝜏 𝜏 = F r sin ϴ = (2.5 N)(0.15 m) (sin 90) = 0.375 N · m = 0.38 N · m b. Type of rotation The rotation is counterclockwise since the force is pointed upwards (right-left) based from the picture..·. The torque is positive since the rotation is in counterclockwise direction. POINTS TO PONDER  Torque is the cross product between displacement and force.  Moment of force or lever arm is the perpendicular distance from the axis of rotation to the direction or line of action of the force.  Torque may also be positive or negative depending on the sense of rotation.  By convention, torque is positive if it tends to produce counterclockwise rotation. It is negative if it tends to produce clockwise rotation.  The greater torque applied to an object, the greater its tendency to rotate.  The SI unit of torque is meter· newton (m· N). 12  ROTATIONAL QUANTITIES Rotation refers to the motion of a body turning about an axis, where each particle of the body moves along a circular path. Examples of rotation include the movement of the hands of a clock, the motion of the blades of an electric fan, and Earth rotating about its axis. Physicists use angular quantities to describe rotation. The equations of kinematics and dynamics of rectilinear motion can be written in terms of angular quantities. Each of the rectilinear motion quantities has its own rotational analog. For the purpose of distinction, Greek letters are used to represent angular quantities. Kinematics of Rotation Consider a rotating disk with a particle on its edge. The angular position is represented by ϴ. When the disk rotates from an initial position ϴ0 to a final position ϴ, its angular displacement represented by Δϴ is (Equation 1) Figure 6. Rotational Quantities (Silverio, 2017) The units for angular displacement may be degrees, radians, or revolutions. Recall from trigonometry that 1 revolution = 360° = 6.28 radians = 2 π radians The linear distance s along an arc subtended by an angle ϴ in radians in a circle of radius r is given by Retrieved From htttps://images.app.goo.gl/X2t2 cYLbpV5Ekd6d6 s=rϴ (Equation 2) Figure 7. The angular speeds Angular velocity is defined in analogy with of the hour hand and minute linear velocity. Instead of liner displacement, hand of a clock are 30°/h angular displacement is used. and 6°/min, respectively. Angular velocity ω is the rate at which angular displacement changes with time. In equation form, (Equation 3) Angular velocity may be expressed in deg/s, rad/s, or rev/s. The linear velocity of the bug on the disk has a magnitude of s/t. Thus, the relationship between angular velocity and linear velocity may be obtained using Eq. 2 and 13 Eq. 3. Dividing both sides of Eq. 2 by t gives (Equation 4) As in the case of linear motion, rotational motion may also be accelerated. Angular acceleration 𝛼 is the time rate of change of angular velocity. (Equation 5) Angular acceleration may be expressed in deg/s 2, rad/s2, or rev/s2. Linear acceleration a and angular acceleration 𝛼 are related as follows: a = r𝛼 (Equation 6) The equations of kinematics developed for constant acceleration for linear motion can be rewritten in angular quantities and may be used to solve problems on rotational motion. Table 1 shows the kinematic equations for translation (linear) motion with their rotational motion counterparts. Example: A rider notices that the wheels of his bicycle make 12 rev in 15 s. a. What is the average angular speed of the wheel in radians? b. What distance in meters does the wheel travel if its radius is 33 cm? Given: = 12 rev t = 15s r = 33 cm = 0.33 m 14 a. Find average angular speed 𝜔. Convert first revolutions to radians. 2 2 ( ) The average angular speed may be computed by using Eq. 3 𝜔 2 b. Find linear distance. The linear distance traveled by the wheel may be computed using Eq. 2 s=rϴ = (0.33 m)(75.36 rad) = 24.8688 m = 25 m POINTS TO PONDER  Rotation is the motion of the body turning about an axis, where each particle of the body moves along a circular path.  Kinematics of rotation are similar to those of linear motion. These equations may be obtained by using the following substitutions:  Angle for distance d  Angular velocity 𝜔 for velocity v  Angular acceleration 𝛼 for acceleration a  Torque 𝜏 for force F  Moment of inertia I for mass m 15  STATIC EQUILIBRIUM Statics is concerned with the calculation of forces acting on and within structures that are in equilibrium. The basic principles were discovered early. The fundamentals of levers, inclined planes and other principles were used by early civilizations in the construction of huge structures like the pyramids. Archimedes introduced the concept of lever equilibrium as early as 287-212 BC. Leonardo da Vinci conceptualized moments in relation to equilibrium in the latter part of the fifteenth century. Static equilibrium is defined as a body at rest having zero acceleration and zero net force. Center of Gravity The center of mass or center of gravity is a useful concept when dealing with equilibrium problems. The center of gravity of a body is the point where its entire weight may be assumed to be concentrated. The body’s center of gravity and center of mass are identical on the assumption that the body is small enough for g not to vary over its extents. The center of gravity may be inside or outside the body. For a circular plate, the center of gravity is at the center and is inside it. For the traditional doughnut, the center of gravity is also at the center but outside the doughnut. Retrieved from https://images.app.goo.gl/q7nZj For a regular-shaped object, the center of wk6s5VcqADT8 gravity is its geometric center. If the object is Figure 8. The simplest way to irregular in shape, the center of gravity may be find the center of gravity of determined experimentally either by balancing or an object is to balance it on using the plumb line method. a knife edge. The center of gravity is just above the point of support. Whether a body is stable or unstable depends upon the location of its center of gravity. In general, an object will be stable if a plumb line drawn from its center of gravity falls within its base. This is called the condition of stability. Some objects are easier to knock over than others even though they will not tip over themselves. In this sense, the greater the weight, the broader the base and the lower the center of gravity, the greater is the stability of the object. 16  Retrieved from Retrieved from https://images.app.goo.gl/pAxBAv4ceMRATzNH8 https://images.app.goo.gl/NK6ym22S eNbvS7 Figure 9. The Marina Sands Hotel in Singapore is a Figure 10. The Leaning Tower of perfect example where principle of statics, structural Pisa in Italy satisfies the condition loads and rigid body equilibrium are considered. for stability. Conditions for Equilibrium Newton’s first law of motion applied to translation and rotation gives the so-called conditions for equilibrium. The first condition for equilibrium states that an object that has no net force acting on it is said to be in translational equilibrium. In equation form, ∑F Force, being a vector quantity may be resolved into its horizontal and vertical components. Thus, the first condition for equilibrium may be equivalently expressed in the form, ∑F x = 0 ; ∑F y ; and ∑F z A necessary condition for a body to be in rotational equilibrium is that the sum of the torques about any point must be zero. ∑𝜏=0 This condition is known as the second condition for equilibrium. 17 Equilibrant If the resultant of the forces acting on a body is not zero, the addition of another force called the equilibrant is needed to balance the resultant to produce translational equilibrium. This force is equal on magnitude to the resultant force but oppositely directed. If the resultant of the forces acting on a body is zero, then any of the forces is the equilibrant of the other forces. Example: Eight boys are equally spaced about a pushball and each exerts a force upon it toward the center, as shown. Each boy in positions A, C and E exerts a force of 70 N and the rest of the boys exert 50 N each. What force is needed to keep the ball at rest? Given: FA = 70 N south FB = 50 N southwest FC = 70 N west FD = 50 N northwest FE = 70 N north FF = 50 N northeast FG = 50 N east FH = 50 N southeast Solution: From the figure, we can see that the forces exerted at the following positions cancel out: F A and FE, FB and FF, and FD and FH. FA = FE FB = FF FD = FH 70 N = 70 N 50 N = 50 N 50 N = 50 N The net force that will accelerate the ball is the resultant of the forces exerted at positions G and C, which is 20 N, west. FG ≠ FC Take note, a negative N≠ N and positive forces are used to indicate that the forces are in x = FG + FC opposing directions. = 50 N + (-70 N) = 50 N – 70 N | 2 N| = 20 N To keep the ball at rest, a force of 20 N east must be applied. 18  POINTS TO PONDER  Static equilibrium is defined as a body at rest having zero acceleration and zero net force.  The center of gravity of a body is the point where its entire weight may be assumed to be concentrated.  The center of gravity of regularly shaped (symmetrical) objects is located at their geometric centers. However, some objects have their centers of gravity located toward their thicker (massive) end and some objects located outside them.  There are two conditions for a body to be in rotational equilibrium:  The vector sum of all forces acting on it must be zero. (Translational)  The sum of all torques about any point must be zero. (Rotational) ROTATIONAL KINEMATICS At this point, we covered a lot of the basics about how things move. But we’ve mostly been focusing on only one type of motion…translational motion. If you can recall, translational motion is that which something moves through space but doesn’t rotate. For example, a soccer player kicks a soccer ball. In translational motion, we are perhaps interested in calculating how fast the ball is moving and the distance it would cover. However, in the game of soccer, it also important to study how the ball is rotating around its axis as it also does affect its movement through space. In rotational motion, similar to translational motion, we will still be discussing familiar quantities like position or displacement, velocity and acceleration. Many of the equations we used in our study of rotational motion will surely be familiar to you already although with some important differences. For example, instead of positions, there are angles in rotational motion and instead of following points along a line; we shall look at points along an arc. Just by using our intuition, we can begin to see how rotational quantities like θ (theta), ω (omega) and (alpha) are related to one another. For example, if a motorcycle wheel has a large angular acceleration for a fairly long time; it ends up spinning rapidly and rotates through many revolutions. In more technical terms, if the wheel’s angular acceleration, , is large for a long 19 period of time t, then the final angular velocity, ω, and angle of rotation θ are large. The wheel’s rotational motion is exactly analogous to the fact that the motorcycle’s large translational acceleration produces a large final velocity, and the distance traveled will also be large. Kinematics is the description of motion. The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. Let us start by finding an equation relating ω, and t. To determine this equation, we recall a familiar kinematic equation for translational, or straight- line, motion: v = vo + at ; a is constant Note that in rotational motion a = at, and we shall use the symbol a for tangential or linear acceleration from now on. As in linear kinematics, we assume a is constant, which means that the angular acceleration (alpha) is also constant, because a = r. Now, let us substitute v = rω and a = r into the linear equation above: rω = rωo + r t The radius r cancels in the equation, yielding ω = ωo + t, where ωo is the initial angular velocity. This last equation is a kinematic relationship among ω, and t --- that is, it describes their relationship without reference to forces or masses that may affect rotation. It is also precisely analogous in form to its translational counterpart. Kinematics for rotational motion is completely analogous to translational kinematics. Kinematics is concerned with the description of motion without regard to force or mass. We will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. 20  In these equations, the subscript 0 (zero) denotes initial values, and the average angular velocity 𝜔 ̅, and average velocity ̅ are defined as follows: 𝜔 𝜔 𝜔 ̅ ̅ 2 2 The equations in the table above can be used to solve any rotational or translational kinematics problem in which both linear and angular accelerations are constant. Before we dive into some practical examples, let discuss about some important conventions and concepts involving uniform circular motion. ROTATION ANGLE When objects rotate about its axis, for example, when a CD (compact disc) in the figure below rotates about its center, each point in the object follows a circular path, an arc. Consider a line from the center of the CD to its edge. Each pit used to record sounds along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle Δθ to be the ratio of the arc length Δs, to the radius of curvature: All points on a CD travel in circular arcs. The pits along a line from the center to the edge all move through the same angle Δθ in a time Δt. The arc length Δs is the distance traveled along a circular path as shown in the figure below. Note that r is the radius of curvature of the circular path. We know that for one complete revolution, the arc length is the circumference of a circle of radius r. The circumference of a circle is 2πr. Thus 21 for one complete revolution, the rotation angle is 2 2 This result is the basis for defining the units used to measure rotation angles, Δθ to be radians (rad), defined so that 2π rad = 1 revolution. A comparison of some useful angles expressed in both degrees and radians is shown in the table below. If Δθ = 2π rad, the CD in our example has made one complete revolution, and every point on the CD is back to its original position. Because there are 360 degrees in a circle or one revolution, the relationship between radians and degrees is thus, 2π rad = 360 degrees. So that. ANGULAR VELOCITY How fast is an object rotating? We define angular speed, ω, as the rate of change of an angle. In symbols, this is 𝜔 where an angular rotation Δθ takes place in a time Δt. The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s). 22  Angular velocity ω is analogous to linear velocity v. To get the precise relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length Δs in a time Δt, and so it has a linear velocity From , we see that. Putting this into the expression for v gives 𝜔 We write this relationship in two different ways and gain two different insights: 𝜔 𝜔 The first relationship in v = rω or ω = v/r states that the linear velocity Is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest radius, r)as you might expect. We can also call this linear speed of a point on the rim as the tangential speed. The second relationship in v = rω or ω = v/r can be illustrated by considering the tire of a moving car. Note that the speed of a point on the rim of the tire is the same as the speed v, of the car. Look at the illustration below. The faster the car moves, the faster the tire spins --- larger speed, v, means a large ω, because v = rω. Similarly, a larger-radius tire rotating at the same angular velocity ω, will produce a greater linear speed for the car. To sum up, angular displacement (θ), is the angle swept out by any line passing through a rotating body that intersects the axis of rotation. This value is positive if going counter-clockwise, negative if going clockwise and its SI unit is radians (sometimes its rpm). 23  Angular velocity (ω), is derivative of the change of angular displacement over a change of time. Counterclockwise rotation is positive angular velocity and negative if clockwise. Angular acceleration (⍺), is the change of angular velocity per unit time and measured in radians per second squared (rad/s2). 24 Discussion points for Example 1: This example illustrates that relationships among rotational quantities are highly analogous to those among linear quantities. We also see in this example how linear and rotational quantities are connected. The answers to the questions are realistic. After unwinding for 2 seconds, the reel is found to spin at 220 rad/s, which is 2,100 rpm. The amount of fishing line played out is 9.00 meters, about right for when the big fish bites. 25  ANGULAR MOMENTUM Have you seen a spinning ice skater on the rink? Notice that the skater spins faster or slower depending on whether she stretches one of her arms or puts both of them close to her chest. This can be explained using the concept of angular momentum. You have previously learned a moving body has momentum. The linear or translational momentum also has a rotational analogue for rotational motion. Recall that linear momentum is ⃗ ⃗. The rotational analogue to this is called angular momentum and is represented by the symbol L. Just like in linear momentum, the magnitude of angular momentum is defined as the product of an inertial quantity and speed. Here is the general equation for L. The SI unit for L is kg∙m 2/s. 𝜔 Recall that the second law of motion can be expresses in terms of the time rate of change in momentum. ⃗ ∑⃗ Similarly, you can express the rotational analogue of Newton’s second law in terms of angular momentum as follows: 𝜔 𝜔 𝜔 𝜔 𝜔 ∑𝜏 𝛼 This can be written in a simpler form as follows: ∑ You can manipulate this further by multiplying both sides of the equation by the change in time. ∑𝜏 This equation is the rotational analogue of the impulse-momentum theorem. When the net external torque acting on the system is zero, you see that ∑𝜏 26  This means that the change in the angular momentum of the system is zero; that is, angular momentum is conserved. 𝜔 𝜔 This is the law of the conservation of angular momentum and can be stated as follows: “The total angular momentum of a rotating object is constant when the net external torque acting on it is zero”. Observe the conservation of angular momentum during the spin of a skater on ice shown in the figure on the right side. The skater has a slower speed (small ω) when her arms are stretched away from the body (large I). When she brings her arms close to her body (small I), she spins faster (large ω). Example: A figure skater spins on an icy surface. Her moment of inertia is 3. kg∙m 2 when her arms are stretched away from her and drops to 2.2 kg∙m 2 when her arms are close to her chest. If she makes 3 revolutions per Retrieved from DIWA Learning Systems, second (rps) when her arms are stretched, Inc. how many revolutions per second will she Figure 11. A skater spins faster when make when she puts her arms to her chest? she brings her arms close to her body. Solution: Her angular speed when her arms are stretched away from her is 2 ( ) When she puts her arms close to her chest, her angular momentum is conserved. In this scenario, net torque is zero. 𝜔 𝜔 Solving for final angular speed gives 𝜔 𝜔 2 22 27 Converting this to the number of revolutions per second gives 2 2 ( ) 2 Therefore, she makes 4 revolutions per second when she brings her arms close to her chest. Performance Task: Directions: Read and understand the given situation below. After which, do what is ask. Write your report in short bond paper. You are a member of an acrobatic team who will present a performance in the new theme park to be called Physics Land. Your new performance will showcase your balancing act involving a long stick with varying masses at the stick’s ends; this will be done while you are walking on a rope. To determine the correct point where you should hold the Retrieved from DIWA Learning Systems, Inc. stick, your trainer asked you to conduct an activity that will measure the location of this point where you will balance the hanging masses (whose values will be given by the trainer depending on the availability of materials) as shown in the figure here. Your task is to submit a report sheet explaining the methods you have used and the complete analysis of your results. Your trainer will rate your report sheet based on the accuracy of results. III. WHAT I HAVE LEARNED EVALUATION/POST TEST: I. MULTIPLE CHOICE: Encircle the letter of the correct answer. Show your solution for items that involve problem solving. Write your answers on your notebook/Answer Sheet. 1. A uniform rod of mass 2 kg and length of 2.0 m is pivoted about an axis perpendicular to the rod and 50 cm from its left end. Find the rotational inertia about this axis. a. 2.17 kg·m 2 b. 2.57 kg·m 2 c. 1.17 kg·m 2 d. 0.67 kg·m 2 2. Which has a greater moment of inertia? a. long fat leg b. long thin leg c. short fat leg d. short thin leg 28 3. In considering the moment of inertia of a body, it is convenient to assume as if all its mass were concentrated at a single point from the center of rotation. This distance is called __________. a. center of mass c. lever arm b. center of gravity d. radius of gyration 4. What is the rotational analog of mass in linear motion? a. angular speed c. moment of inertia b. angular momentum d. torque 5. What happens when a female gymnast moves from an extended position to a tucked position? a. She increases her rotational inertia. b. She decreases her rotational inertia. c. Neither A nor B. d. Cannot be determined. 6. Which of the forces pictured as acting upon the rod will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod? a. F1 b. F2 c. Both d. Neither 7. The two forces in the diagram have the same magnitude. Which force will produce the greater torque on the wheel? a. F1 b. F2 c. Both d. Neither 8. A 50-N force is applied at the end of the wrench handle that is 24-cm long. The force is applied in a direction perpendicular to the handle as shown. What is the torque applied to the nut by the wrench? a. 6 N· m c. 26 N· m b. b. 12 N· m d. 120 N· m 9. Refer to item #8, What would the torque be if the force were applied half-way up the handle instead of at the end? a. 6 N· m c. 26 N· m b. 12 N· m d. 120 N· m 10. The unit of torque is ________. a. N b. Pa c. N/m d. N· m 11. Which of the following is NOT a condition for an object to be in static equilibrium? a. The object is not moving. b. It is in translational equilibrium. c. It is in rotational equilibrium. d. The object is at constant velocity. 29 12. In which of the following items is the center of gravity located at a point where there is no mass? a. tennis racket c. boomerang b. billiard ball d. discus 13. When you carry a heavy load with one arm, why do you tend to hold your free arm away from your body? a. to change the mass of your body b. to change the weight of your body c. to feel good and look good d. to shift your body’s center of gravity 14. A resultant force of 4.5 N directed 30˚ north of east is acting on a particle. What force is needed to bring the particle into a state of equilibrium? a. 4.5 N, 30˚ north of east b. 4.5 N, 30˚ north of west c. 4.5 N, 30˚ south of east d. 4.5 N, 30˚ south of west 15. What is the angular velocity of a car whose tire radius is 0.300 meter when the car travels at a linear speed of 15 m/s? a. 50.4 rad/s b. 50.0 rad/s c. 51 rad/s d. 49.3 rad/s II. IDENTIFICATION: Determine whether a system is in static equilibrium or not. Write STATICS if it’s in static equilibrium and write X if not. 30  REFERENCES Arevalo, Ryan L. and Mulig, Charity I. General Physics 1. Makati City, Philippines: DIWA Learning Systems Inc., 2017. General Physics 1st Edition 2018, Department of Education Authors: Jose Perico H Esguerra PhD, Rommel G. Bacabac PhD, Jo-Ann M. Cordovilla, Ranziville Marianne L. Roxas-Villanueva PhD and John Keith V. Magali, Chapter 7, pp 135-158 Kinematics of Rotational Motion, Retrieved Sep 29, 2020 from https://course.lumenlearning.com/physics/chapter/10-2-kinematics-of- rotational-motion Earthquake Hazards, Retrieved Sep 29, 2020 from https://course.lumenlearning.com/physics/chapter/10--1-angular- acceleration Padua, Alicia L., and Ricardo M. Crisostomo. 2003. Practical and Explorational Physics Modular Approach. Quezon City: Vibal Publishing House, Inc. Silverio, Angelina A. 2017. General Physics 1. Quezon City: Phoenix Publishing House, Inc. University Physics 9th Edition 1996, Addison-Wesley Publishing Company Inc. Authors: Hugh D. Young and Roger A. Freedman, Chapter 9, pp 268-276 31  DEPARTMENT OF EDUCATION Division of Negros Oriental SENEN PRISCILLO P. PAULIN, CESO V Schools Division Superintendent FAY C. LUAREZ, TM, Ed.D., Ph.D. OIC - Assistant Schools Division Superintendent Acting CID Chief NILITA L. RAGAY, Ed.D. Assistant Schools Division Superintendent ROSELA R. ABIERA Education Program Supervisor – (LRMS) ARNOLD R. JUNGCO PSDS – Division Science Coordinator MARICEL S. RASID Librarian II (LRMDS) ELMAR L. CABRERA PDO II (LRMDS) MARY JOYCEN A. ALAM-ALAM ANGELO JANRY EMMANUEL D. ISO Writers/Illustrators/Lay-out Artists _________________________________ QUALITY ASSURANCE TEAM ARNOLD D. ACADEMIA ZENAIDA A. ACADEMIA LIEZEL A. AGOR MARY JOYCEN A. ALAM-ALAM EUFRATES G. ANSOK JR. JOAN Y. BUBULI LIELIN A. DE LA ZERNA ADELINE FE D. DIMAANO RANJEL D. ESTIMAR VICENTE B. MONGCOPA FLORENTINA P. PASAJINGUE THOMAS JOGIE U. TOLEDO DISCLAIMER The information, activities and assessments used in this material are designed to provide accessible learning modality to the teachers and learners of the Division of Negros Oriental. The contents of this module are carefully researched, chosen, and evaluated to comply with the set learning competencies. The writers and evaluator were clearly instructed to give credits to information and illustrations used to substantiate this material. All content is subject to copyright and may not be reproduced in any form without expressed written consent from the division. 32 SYNOPSIS AND ABOUT THE AUTHORS ANSWER KEY There are many analogies between linear motion and rotational motion. In linear motion, you can describe the motion of an object using kinematics and dynamics quantities such as velocity, acceleration, force, and energy. In rotational motion, you can describe the rotation of rigid bodies using the rotational analogues of these kinematics and dynamics quantities such as angular velocity, angular acceleration, torque, and rotational kinetic energy and angular momentum. The conservation laws are also applicable in rotational motion. Likewise, the total mechanical energy, including the linear and rotational components of energy, is conserved. A body is said to be in the state of equilibtium when two requirements are met. First, the vector sum of the forces acting on the body must be zero. Second, the net torque at any point in the rigid body must also be zero. MARY JOYCEN A. ALAM-ALAM, is a Senior High School teacher in Dauin Science High School. She earned her Bachelor of Secondary Education major in Biological Science in Negros Oriental State University in 2016. She was also a DOST-SEI scholar under R.A. No. 10612. ANGELO JANRY EMMANUEL D. ISO is a licensed professional teacher stationed in Benedicto P. Tirambulo Memorial National High School, SHS department. He is also a registered Electronics and Communications Engineer graduated from University of San Carlos, Cebu City. 33