General Physics 1 - Rotational Kinematics PDF

AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE FIRST SEMESTER, S.Y. 2020-2021 GENERAL PHYSICS 1 Quarte...

AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE FIRST SEMESTER, S.Y. 2020-2021 GENERAL PHYSICS 1 Quarter 2 - Week 1 TOPIC: ROTATIONAL KINEMATICS LESSON 1 At the end of this handout, you are expected to: Define kinematic rotational variables such as angular position, angular velocity, and angular acceleration Derive rotational kinematic equations, and Solve for the angular position, angular velocity, and angular acceleration of a rotating body. Determine whether a system is in static equilibrium or not Determine the conditions of a system under equilibrium and solve static equilibrium problems Define torque and learn how force should be applied in a body to attain maximum torque Determine angular momentum of different systems Apply the torque-angular momentum relation in solving problems Rotational Kinematics Kinematics is the description of motion. It is concerned with the description of motion without regard to force or mass. But what exactly is rotational kinematics? From the word, you can describe that it’s all about any object that can rotate or spin. It’s different from linear motion when object simply moves forward. The kinematics of rotational motion describes the relationships among rotation angle (θ), angular velocity (ω), angular acceleration ( α) , and time (t). You will find that translational kinematic quantities, such as displacement, velocity, and acceleration have direct analogs in rotational motion. Axis of Rotation Everything that you are all rotating about a line somewhere within the object called the axis of rotation. We are also going to assume that all these objects are rigid bodies, that is, they keep their shape and are not deformed in any way by their motion. Look at Figure 1 below. It shows the wheel and axle of a bike. Is the axle (axis of rotation) part of the wheel (rigid body)? The answer is NO. If you were to spin the wheel around its center, the axis of Figure 1.1 Wheel and Axle Source: rotation (axle) would be pointing perpendicular to the motion of the wheel. https://www.pinterest.ph/pin/764134261 749726038/ https://bit.ly/3mmkceW Angular Displacement The symbol generally used for angular displacement is θ pronounced "teta" or "theta." θ is the angle swept by the radius of a circle that points to a rotating object. Look at the circle below and assume its rotating about its middle so the axis of rotation is pointing out of the page. Start with a piece of the circle at point A. as the circle rotates counterclockwise, the piece of the circle reaches point B. The point traveled a distance of s along the circumference, and swept out an angle θ. We can also say that the angle θ “substends’ an arc length of s. Note that the points A and B are always at the same distance, r, from the axis of rotation. Figure 1.2 The Angle of Displacement Source: file:///C:/Users/USER/Downloads/rotati onal_motion_-_day_1___2%20(3).pdf Where : θ is the angle of rotation, S is the arc length, and r is the radius. 1  Angular displacement is unitless since it is the ratio of two distances but, we will say that the angular displacement is measured in radians. We know degrees, and we know that when a point on a circle rotates and comes back to the same point, it has performed one revolution; let us say from point A, and rotate until we come back to point A. Refer to Figure 2 again, what distance (s) was covered? How many degrees were swept by this full rotation? The point moved around the entire circumference, so it traveled 2πr while an angle of 3600 was swept through. Using the angular displacement definition: When an object makes one complete revolution, it sweeps out an angle of 360 0 or 2π radians. One radian is the angle at which the arc has the same length as the radius r. 1 radian = 57.30 The radian is frequently abbreviated as rad. Sample Problems 1. ) An object travels around a circle10.0 full turns in 2.5 seconds. Calculate (a) the angular displacement, θ in radians. Given: # of turns/complete rotations = 10 turns Time = 2.5 seconds Find: Angular displacement (θ) in radians *Note that 1 complete rotation = 3600 = 2π radians = 6.28 rd Solution: Θ = 10.0 turns ( 6.28 rd / turn ) = 62.8 radians. 2. ) A girl goes around a circular track that has a diameter of 12 m. If she runs around the entire track for a distance of 100 m, what is her angular displacement? Given: Diameter of the curved path = 12m ; *Note that diameter = 2r therefore, r= d/2 so, r= 12m/2= 6m Linear displacement, s = 100 m. Find: Angular displacement θ Solution: Θ = s/r → θ = 100m / 6 m = 16.67 radians Angular displacement can now be related to linear displacement. Working on Kinematics problems with linear displacement was tackled in your previous lessons. What other quantities played a key role in linear displacement? Angular Velocity In linear motion, velocity (v) is defined as the rate of change of the object's position with respect to a frame of reference and time that is, 𝑣 = ∆𝑥/∆𝑡 while acceleration (a) is the rate of change of velocity. In symbol, we have: 𝑎 = ∆𝑣/∆𝑡 ; 𝑎 = (𝑣2 − 𝑣1 )/∆𝑡 In rotational motion, angular velocity (ω) is defined as the change in angular displacement (θ) per unit of time (t). In symbol, 𝜔 = ∆𝜃/∆𝑡 The symbol ω is pronounced "omega" is used to denote angular velocity. We usually describe the angular velocity as revolution per second (rev/sec, rps), or radian per second. See Figure 1.3. You will often have to convert this number, since it is usually given as a frequency (revolutions per time frame). 2 From linear velocity conversion, we have: Figure 1.3 Angular Velocity Source: http://230nsc1.phy- ω = v/r, astr.gsu.edu/hbase/rotq.html where: ω is the angular velocity (rad/s), v is the tangential velocity (m/s), and r is the radius in circular path (meters). Starting from angular velocity, let’s substitute the linear displacement for the angular displacement we have: As you can see, the tangential velocity (v) is directly proportional to the product of the angular velocity and the radius of the moving object. This confirms your feeling when riding a merry-go-round. Thus, the farther you are from the center, the faster you feel you are moving. Sample Problems 1.) If an object travels around a circle with an angular displacement of 70.8 radian in 3.0 seconds, what is its average angular velocity ω in (rd/s)? Answer Given: Δθ = 70.8 rd ; Δt = 3 s Find: ω =? Solution: ω = Δθ / Δt = 70.8 rd / 3.0 s = 23.6 rd/s. 2.) A bicycle wheel with a radius of 0.28 m starts from rest and accelerates at a rate of 3.5 rad/s 2 for 8 s. What is its final angular velocity? Answer Given: r = 0.28 m; α = 3.5 rd/s2 t =8s Find: ω =? Solution: From the equation α = ω / t , we can have ω = αt = 3.5 rd/s2 ( 8s ) = 28 rd/s Angular Acceleration If the angular velocity of the rotating object increases or decreases with time, we say that the object experiences an angular acceleration, α. The angular acceleration of a rotating object is the rate at which the angular velocity changes with respect to time. It is the change in the angular velocity, divided by the change in time. The average angular acceleration is the change in the angular velocity, divided by the change in time. The angular acceleration is a vector that points in a direction along the rotation axis. The magnitude of the angular acceleration is given by the formula below. The unit of angular acceleration is radians/s2. Insymbol, 3 Where: α = angular acceleration, (radians/s2) Δω = change in angular velocity (radians/s) Δt = change in time (s) ω1 = initial angular velocity (radians/s) ω2= final angular velocity (radians/s) t1 = initial time (s) t2= final time (s). The symbol α is pronounced "Alpha". The unit of measure is radian per second squared (rd/s 2). All points in the object have the same angular acceleration. Every point on a rotating has, at any instant a linear velocity (v) and a linear acceleration (a). Look at the illustration in Figure 1.4 below, we can relate the linear quantities (v and a) to the angular quantities (ω and α). Linear velocity and angular velocity are related since v = rω Where; v is the linear velocity, Figure 1.4 Relation of Linear r is the radius of the object, and Velocity and Angular Velocity: ω is the angular acceleration. Points farther than the axis of rotation will move faster (linear velocity) but the angular velocity for all points is the same. Source: https://www.slideshare.net/ibidorigin/1 2-rotational-motion It is not only the point (we measure) move in that angular velocity. All points in the object rotate with the same angular velocity. Every position in the object move through the same time interval. Conventionally, object moving counterclockwise has a value of positive (+) angular acceleration, while the one moving the clockwise direction has negative (-) value. Angular acceleration occurs when the angular velocity changes over time. It acts in the direction of rotation in a circular motion (not the same as centripetal acceleration). In this case, we must also introduce tangential acceleration at, since the tangential velocity is changing. If there is angular acceleration, there will also be tangential acceleration 𝑎𝑡𝑎𝑛 = 𝑟𝛼 and 𝑎𝑟𝑎𝑑 = 𝜔2 𝑟 where: at is the tangential (linear) acceleration (m/s) r is the radius of circular path (meters) α is the angular acceleration; ar is the radial (linear) acceleration (m/s);and ω is the angular acceleration (rd/s2). Sample Problems 1. A disc in a DVD player starts from rest, and when the user presses “Play”, it begins spinning..The disc is spins at 160 radians/s after 4.0 s. What was the average angular acceleration of the disc? Given: T1 = 0 T2= 4.0 s ω1 = 0 ω2 =160 rd/s Find: Angular acceleration (α) =? Solution: Between the initial and final times, the average angular acceleration of the disc was 40.0 radians/s 2. 4 2.) A car tire is turning at a rate of 5.0 rd / sec as it travels along the road. The driver increases the car's speed, and as a result, each tire's angular speed increases to 8.0 rd /sec in 6.0 sec. Find the angular acceleration of the tire. Given: ω1 = 5.0 rd/s; ω2 = 8.o rd/s; Δt = 6.o s Find: α=? Solution: α = Δω / Δt = (ω2 - ω1) /Δt = ( 8.0 rd/s - 5.0 rd/s ) / 6.0s = 0.50 rd/s2. 2. As a car starts accelerating ( from rest ) along a straight road at a rate of 2.4 m/s2, each of its tires gains an angular acceleration of 6.86 rd/s2. Calculate (a) the radius of its tires, (b) the angular speed of every particle of the tires after 3.0s, and (c) the angle every particle of its tires travels during the 3.0-second period. Given: ω1 = 0 (from rest) at = 2.4 m/s2 α = 6.86 rd/s2 Δt = 3.0 s Find: a) r b) ω2 c) θ Solution: a. ) From the equation at = rα, we get r = at /α = [2.4 m/s2] / [6.86 rd/s2] = 0.35m (b) From the equation α = (ω2- ωi1)/Δt, we get α Δt = ω2- ω1 ω2= ω1+ α Δt ω2= 0 +(6.86rd/s2)(3.0s) = 21 rd/s. (c) θ = (1/2)α t2 + ωi t = (1/2)( 6.86 rd/s2)(3.0s)2 + (0) (3.0s) = 31 rd. Putting these definitions together, you observe a very strong parallel between translational kinematic quantities and rotational kinematic quantities See Table 1.1 below. Table 1.1a Symbols for Translational ang Table 1.2 Translational ang Angular Angular Kinematics Quantities Kinematics Quantities (if radius is given) Source:https://www.aplusphysics.com/courses/honors/rotation/ Source:https://www.aplusphysics.com/courses/honors/rotation/ honors_rot_kinematics.html honors_rot_kinematics.html It’s quite straightforward to translate between translational and angular variables as well when you know the radius (r) of the point of interest on a rotating object. The rotational kinematic equations (See table below) can be used the same way you used the translational kinematic equations to solve problems. Once you know three of the kinematic variables, you can always use the equations to solve for the other two. Table 1.3 Translational and Rotational Kinematics Equations Source:https://www.aplusphysics.com/courses/hon ors/rotation/honors_rot_kinematics.html 5 ENGAGEMENT Write your answer on a separate sheet of paper. Activity 1 Think Critically Solve the following in a clean sheet of paper. Show your solution and box your final answer. 1. Mark bought a pizza of a radius of 0.5 m. A fly lands on the pizza and walks around the edge for a distance of 80 cm. Calculate the angular displacement of the fly? 2. What is the angular velocity of an object traveling in a circle of radius 0.75 m with a linear speed of 3.5 m/s? 3. What is the angular acceleration of a ball that starts at rest and increases its angular velocity uniformly to 5 rad/s in 10 s? Activity 2 Correct Me If I’m Wrong Explain briefly. Write your answer in a separate paper. 1.) Angular acceleration does not change with radius, but tangential acceleration does. _____________________________________________________________________________________ 2.) Differentiate angular acceleration from tangential (or linear) acceleration. _____________________________________________________________________________________ 3.) On a rotating carousel or merry-go-round, one child sits on a horse near the outer edge and another child sits on a lion halfway out from the center. Which child has the greater linear velocity? Which child has the greater angular velocity? _____________________________________________________________________________________ TOPIC: ROTATIONAL DYNAMICS LESSON 2 TORQUE (𝝉) Have you ever wondered why doorknobs are situated at the opposite end of the hinges and not near it? And why is it easier to use long-handled wrenches than the short-handled one in removing bolts? How about doing an arm-wrestling with a longer-arm person? What do you think would be your chances of winning? This lesson will enlighten you on the simple physics behind these things. With the understanding of Torque, you will be able to answer these questions. To understand this, let us imagine a door. The door has hinges on one side. To successfully open the door, you need to apply a force. This force will cause the door to rotate about its hinges, or its axis of rotation. This rotation creates Torque. Torque, also called the Moment of Force, is the result of the force that can cause an object to rotate about an axis. It is a vector quantity. It is the cross product of the vector Force and the distance from the axis of rotation. Mathematically, 𝜏⃑ = 𝑟⃑ 𝑥 𝐹⃑ whose magnitude is equal to 𝜏 = 𝑟⊥ 𝐹 𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 where 𝑟⊥ = 𝑟𝑠𝑖𝑛𝜃 And 𝜃 is the angle between 𝑟 and 𝐹 S.I. Unit: Nm 6  The direction of the torque may either be counterclockwise (CCW) or clockwise (CW). By convention, we take the counterclockwise direction to be positive and clockwise as negative. From the equation, we see that the effect of the Force on the motion of the rotating body depends on three factors as follows: 1. Magnitude of the Force 2. Lever Arm (Moment Arm) – perpendicular distance of the line of action to the axis of rotation 3. The angle between the Force vector and the lever arm The torque increases as the force increases, and also as the distance increases. That is why doorknobs are located at the opposite end of the hinges. It is easier to open the door in this case since small force is needed to cause torque to the door. Sample Problems: 1. A Force of (4𝑖̂ − 3𝑗̂ + 5𝑘̂)𝑁 is applied at a point whose position vector is (7𝑖̂ + 4𝑗̂ − 2𝑘̂)𝑚. Find the Torque of force about the origin. Solution: 𝐹⃑ = (4𝑖̂ − 3𝑗̂ + 5𝑘̂)𝑁 𝑟⃑ = (7𝑖̂ + 4𝑗̂ − 2𝑘̂)𝑚 𝑖̂ 𝑗̂ 𝑘̂ 𝜏⃑ = 𝑟⃑ 𝑥 𝐹⃑ = | 7 4 −2| = (20 − 6)𝑖̂ − (35 + 8)𝑗̂ + (−21 − 16)𝑘̂ = (14𝑖̂ − 43𝑗̂ − 37𝑘̂)𝑁𝑚 4 −3 5 2. A crane has an arm length of 20 m inclined at 30º with the vertical. It carries a container of mass of 2 ton suspended from the top end of the arm. Find the torque produced by the gravitational force on the container about the point where the arm is fixed to the crane. [Given: 1 ton = 1000 kg; neglect the weight of the arm. g = 9.8 m/s2] Solution: 𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 where 𝐹 = 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 = 𝑟(𝑚𝑔)𝑠𝑖𝑛𝜃 = (20𝑚)(2000𝑘𝑔)(9.8𝑚/𝑠 2 )(𝑠𝑖𝑛30) 𝜏 = 1.96𝑥105 𝑁𝑚 Note: 𝜃 is the angle between 𝑟 and 𝐹 3. Consider the door shown in the figure, which is seen from an aerial view. The circle on the left is the hinge (pivot point). a. Find the Net Torque acting on the door. b. Which way will the door open, up or down? Solution: a. 𝜏𝑛𝑒𝑡 = 𝜏1 + 𝜏2 + 𝜏3 + 𝜏4 𝜏1 = 𝑟𝐹1 𝑠𝑖𝑛𝜃 = (0)(60𝑁)𝑠𝑖𝑛90 = 0 𝜏2 = 𝑟𝐹2 𝑠𝑖𝑛𝜃 = (0.20𝑚)(50𝑁)𝑠𝑖𝑛90 = 10𝑁𝑚 𝜏3 = 𝑟𝐹3 𝑠𝑖𝑛𝜃 = (0.2𝑚 + 0.6𝑚)(70𝑁)𝑠𝑖𝑛90 = 56𝑁𝑚 𝜏4 = 𝑟𝐹4 𝑠𝑖𝑛𝜃 = (0.2𝑚 + 0.6𝑚 + 0.2𝑚)(80𝑁)𝑠𝑖𝑛30 = 40𝑁𝑚 7 Before adding the torque, determine their corresponding direction according to the rotation of the door. 𝜏1 has no rotation since the torque is zero 𝜏2 and 𝜏4 : pulling the door upward would make it rotate in the CCW direction (+) 𝜏3 : pulling the door downward would make it rotate in the CW direction (-) 𝜏𝑛𝑒𝑡 = 0 + 10𝑁𝑚 + (−56𝑁𝑚) + 40𝑁𝑚 = −6𝑁𝑚 b. Since the result of the net torque is negative, this means that the door will rotate in the clockwise or downward direction. STATIC EQUILIBRIUM Static equilibrium occurs when an object is at rest – neither rotating nor translating. It is analogous to Newton’s 1st Law of motion for rotational system. An object which is not rotating remains not rotating unless acted on by an external torque. Similarly, an object rotating at constant angular velocity remains rotating unless acted on by an external torque. For an object to maintain in static equilibrium, the following conditions must be met: 1. The net force acting on the object must be zero: ∑ ⃑𝑭⃑⃑ = 𝟎 2. The net torque acting on the object must be zero: ∑ 𝝉 = 𝟎 Applications of Static Equilibrium is constantly seen and observed around us. A common example of balanced torques is two children on a see-saw. If the fulcrum is in the center of the see-saw, the two children must have equal mass for it to be balanced. If the fulcrum is not in the center, their masses must vary to create equal torques. This topic will also help you understand important applications in the field of engineering such as building bridges, the Physics behind crane towers and many more. Sample Problems: 1. A 0.15kg meterstick is supported at the 50cm mark. A mass of 0.5kg is attached at the 80cm mark. a. How much mass should be attached to the 40cm mark to keep the meterstick horizontal? b. Determine the supporting force from the fulcrum on the meterstick. Solution: a. From the 2nd condition of Equilibrium: ∑𝜏 = 0 → 𝜏1 + 𝜏2 = 0 Where 𝜏1 is the torque caused by the force exerted by mass m 𝜏2 is the torque caused by the force exerted by the 0.5kg mass Hanging mass m would cause the stick to rotate in the CCW direction, thus 𝜏1 is (+) Hanging the 0.5kg-mass would cause the stick to rotate in the CW direction, thus 𝜏2 is (-) ∑ 𝜏 = 𝜏1 − 𝜏2 = 𝑟1 𝐹1 𝑠𝑖𝑛𝜃 − 𝑟2 𝐹2 𝑠𝑖𝑛𝜃 = 0 where 𝐹 = 𝑤𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 [(0.10𝑚)(𝑚)(9.8𝑚/𝑠 )𝑠𝑖𝑛90] − [(0.30𝑚)(0.5𝑘𝑔)(9.8𝑚/𝑠 2 )𝑠𝑖𝑛90] = 0 2 (0.98𝑚2 /𝑠 2 )𝑚 − 1.47𝑁𝑚 = 0 → 𝑚 = 1.47𝑁𝑚/0.98𝑚2 /𝑠 2 ) = 1.5𝑘𝑔 𝑚 = 1.5𝑘𝑔 8  b. From the 1st condition of Equilibrium: ∑ 𝐹⃑ = 0 = ⃑⃑⃑⃑ ⃑⃑⃑⃑2 + ⃑⃑⃑⃑⃑ 𝐹1 + 𝐹 𝐹𝑚 + ⃑⃑⃑⃑ 𝐹𝑓 = 0 Where ⃑⃑⃑⃑ 𝐹1 = 𝑤 = 𝑚1 𝑔 is the downward force due to mass 𝑚1 = 1.5𝑘𝑔 ⃑⃑⃑⃑2 = 𝑤 = 𝑚2 𝑔 is the downward force due to mass 𝑚2 = 0.5𝑘𝑔 𝐹 ⃑⃑⃑⃑⃑ 𝐹𝑚 = 𝑤 = 𝑚𝑔 is the downward force due to the mass of the meterstick ⃑⃑⃑⃑ 𝐹𝑓 = is the upward force exerted by the fulcrum to support the weight of the meterstick and masses ⃑⃑⃑⃑1 − 𝐹 −𝐹 ⃑⃑⃑⃑2 − ⃑⃑⃑⃑⃑ 𝐹𝑚 + ⃑⃑⃑⃑ 𝐹𝑓 = 0 → ⃑⃑⃑⃑ 𝐹𝑓 = ⃑⃑⃑⃑ ⃑⃑⃑⃑2 + ⃑⃑⃑⃑⃑ 𝐹1 + 𝐹 𝐹𝑚 = 𝑚1 𝑔 + 𝑚2 𝑔 + 𝑚𝑔 ⃑⃑⃑⃑ 𝐹𝑓 = (1.5𝑘𝑔)(9.8𝑚/𝑠 2 ) + (0.5𝑘𝑔)(9.8𝑚/𝑠 2 ) + (0.15𝑘𝑔)(9.8𝑚/𝑠 2 ) ⃑⃑⃑⃑ 𝐹𝑓 = 21.07𝑁 2. A firefighter who weighs 800N climbs a uniform ladder and stops one-third of the way up the ladder. The ladder is 5m long and weighs 180N. It rests againsts a vertical wall making an angle 53º with the ground. Find the normal and the frictional forces on the ladder at its base. Solution: We first contruct the Free-Body Diagram to identity the forces present. Since the Normal Force is located at the y-axis, we get the net force along this axis. ∑ 𝐹𝑦 = 0 𝑁 − 𝑊𝑝 − 𝑊𝑙 = 0 𝑁 = 𝑊𝑝 + 𝑊𝑙 = 800𝑁 + 180𝑁 𝑁 = 980𝑁 To solve the friction Force, we use get the net force along the x-axis. ∑ 𝐹𝑥 = 0 𝐹−𝑓 =0 → 𝐹=𝑓 To solve for F, we use the 2nd condition of equilibrium. ∑𝜏 = 0 𝜏𝑁 − 𝜏𝑊𝑝 − 𝜏𝑊𝑙 + 𝜏𝐹 = 0 𝑟𝑁 𝑁𝑠𝑖𝑛𝜃 − 𝑟𝑤𝑝 𝑊𝑝 𝑠𝑖𝑛𝜃 − 𝑟𝑤𝑙 𝑊𝑙 𝑠𝑖𝑛𝜃 + 𝑟𝑓 𝐹𝑠𝑖𝑛𝜃 = 0 −𝑟𝑁 𝑁𝑠𝑖𝑛𝜃 + 𝑟𝑤𝑝 𝑊𝑝 𝑠𝑖𝑛𝜃 + 𝑟𝑤𝑙 𝑊𝑙 𝑠𝑖𝑛𝜃 𝐹= 𝑟𝑓 𝑠𝑖𝑛𝜃 1 (0)(980𝑁𝑠𝑖𝑛𝜃) + ( ) (5𝑚)(800𝑁)(𝑠𝑖𝑛37°) + (2.5𝑚)(180𝑁)(𝑠𝑖𝑛37°) 𝐹= 3 (5𝑚)𝑠𝑖𝑛53° 0 + 802.42𝑁𝑚 + 270.82𝑁𝑚 = 3.99𝑚 𝐹 = 268.98𝑁 = 𝑓 WORK DONE BY A TORQUE We have seen how Newton’s Laws of motion is similar to rotational motion. Newton’s Laws may be stated in terms of rotational motion. 1st Law of Rotational Motion: A body in motion at constant angular velocity will continue in motion at the same angular velocity, unless acted upon by some unbalanced external torque. 9  2nd Law of Rotational Motion: When an unbalanced external torque acts on a body with moment of inertia, I, it gives that body an angular acceleration α, which is directly proportional to the torque and inversely proportional to the moment of Inertia. 3rd Law of Rotational Motion: If body A and body B have the same axis of rotation, and if body A exerts a torque on body B, then body B exerts an equal but opposite torque on body A. We can derive the equation of Torque in terms of the angular acceleration , 𝛼,from Newton’s 2nd Law of Motion: 𝐹⃑ = 𝑚𝑎⃑ multiplying both sides with 𝑟 ⃑ 𝑟𝐹 = 𝑟𝑚𝑎⃑ where 𝑟𝐹⃑ = 𝜏 and 𝑎 = 𝑟𝛼 𝜏 = (𝑟𝑚)(𝑟𝛼 ) 𝜏 = 𝑚𝑟 2 𝛼 ; 𝐼 = 𝑚𝑟 2 𝜏 = 𝐼𝛼 Rotational Work To calcula the work done by the torque, we derive it from the translational equation of Work. 𝑊 = 𝐹𝑑 where 𝑑 = 𝑠 = 𝑟𝜃 (𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛) 𝑊 = 𝐹𝑟𝜃 ; 𝐹𝑟 = 𝜏 𝑊 = 𝜏𝜃 Rotational Kinetic Energy 1 𝐾𝐸 = 𝑚𝑣 2 𝑣 = 𝑟𝜔 (𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑚𝑜𝑡𝑖𝑜𝑛) 2 1 1 𝐾𝐸 = 𝑚(𝑟𝜔)2 = 𝑚𝑟 2 𝑣 2 ; 𝑚𝑟 2 = 𝐼 2 2 1 𝐾𝐸 = 2 𝐼𝜔2 For vehicles such as cars and bicycles, the tires exert rotational and translational kinetic energy. Thus, the total kinetic energy is equal to: 1 1 𝐾𝐸 = 𝑚𝑣 2 + 𝐼𝜔2 2 2 Angular Momentum Angular momentum is a quantity that tells us how hard it is to change the rotational motion of a particular spinning body. For a single particle with known momentum. The angular momentum can be calculated using the relationship: 𝐿=𝑟𝑥𝑝 Where 𝐿 is the angular momentum of the object; 𝑟 is the distance of the particle from the point of rotation and 𝑝 is the linear momentum 𝐿 = 𝑟 𝑥 (𝑚𝑣) 𝑤ℎ𝑒𝑟𝑒 𝑣 = 𝑟𝜔 𝐿 = 𝑟 𝑥 (𝑚)(𝑟𝜔) = 𝑚𝑟 2 𝜔 𝑤ℎ𝑒𝑟𝑒 𝑚𝑟 2 = 𝐼 Therefore: 𝐿 = 𝐼𝜔 The higher the angular momentum of the object, the harder it is to stop. Objects with higher angular momentum have greater orientational stability. That is why in riding a bicycle, if you are going faster, you will not fall ober easily as when you are going slower. Conservation of Momentum: “The momentum of a system will not change unless an external torque is applied.” 𝐿𝑓 = 𝐿𝑖 (𝐹𝑖𝑛𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚) 10 Sample Problems: 1. Janelle uses a 20cm long wrench to tighten a nut. The wrench handle is tilted 30º above the horizontal and Janelle pulls straight down on the end with a force of 100N. How much torque does Janelle exert on the nut? Solution: 𝜏 = 𝑟𝐹⊥ = 𝑟𝐹𝑠𝑖𝑛𝜃 𝜏 = (0.20𝑚)(100𝑁)(𝑠𝑖𝑛60°) = 17.3𝑁𝑚 2. A flywheel of mass 182kg has a radius of 0.62m (assume the flywheel is a hoop). a. What is the torque required to bring the flywheel from rest to a speed of 120rpm in an interval of 30 sec? b. How much work is done in this 30-sec period? Solution a. 𝜏 = 𝑟𝐹 = 𝑟(𝑚𝑎) = 𝑟𝑚(𝑟𝛼 ) 𝑤ℎ𝑒𝑟𝑒 𝛼 = (∆𝜔/∆𝑡) ∆𝜔 𝜔𝑓−𝜔𝑖 𝜔𝑓 𝜏 = 𝑚𝑟 2 ( ∆𝑡 ) = 𝑚𝑟 2 ( ∆𝑡 ) = 𝑚𝑟 2 ( ∆𝑡 ) 𝑤ℎ𝑒𝑟𝑒 𝜔𝑖 = 0 (𝑓𝑟𝑜𝑚 𝑟𝑒𝑠𝑡) 𝑟𝑎𝑑 12.57 𝑠 𝜏 = (182𝑘𝑔)(0.62𝑚) [ ] = 29.31𝑁𝑚 30𝑠 Rememeber to be consistent with the units. Conversion of the angular velocity 𝜔 in 𝑟𝑎𝑑/𝑠𝑒𝑐: 𝑟𝑒𝑣 2𝜋𝑟𝑎𝑑 1𝑚𝑖𝑛 𝜔𝑓 = (120 )( )( ) = 12.57 𝑟𝑎𝑑/𝑠𝑒𝑐 𝑚𝑖𝑛 1 𝑟𝑒𝑣 60𝑠𝑒𝑐 b. 𝑊 = 𝜏𝜃 𝑤ℎ𝑒𝑟𝑒 𝜃 = 𝜔𝑎𝑣𝑒 ∆𝑡 𝜔𝑖 + 𝜔𝑓 12.57𝑟𝑎𝑑/𝑠) 𝑊 = 𝜏( ) ∆𝑡 = (29.31𝑁𝑚) ( ) (30𝑠) = 5,526.4𝐽 2 2 3. A bowling ball that has an 11cm radius and a 7.2kg mass is rolling without slipping at 2.0m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill. Model the bowling ball as uniform sphere and calculate h. Solution: Since the problem involves the presence of kinetic 𝐾, and potential energy 𝑈, we use the conservation of mechanical energy to calculate h. ∆𝐸 = 0 → 𝐸𝑓 − 𝐸𝑖 = 0 → 𝐸𝑓 = 𝐸𝑖 where 𝐸𝑖 = 𝐾𝑖 + 𝑈𝑖 and 𝐸𝑓 = 𝐾𝑓 + 𝑈𝑓 𝐾𝑓 + 𝑈𝑓 = 𝐾𝑖 + 𝑈𝑖 1 1 1 𝑚𝑔ℎ𝑓 + 𝑚𝑣𝑓2 = 𝑚𝑔ℎ𝑖 + 𝑚𝑣𝑖2 + 𝐼𝜔2 ; 𝑣𝑓 = 0 𝑎𝑛𝑑 ℎ𝑖 = 0 2 2 2 1 1 2 𝑚𝑔ℎ𝑓 + 0 = 0 + 𝑚𝑣𝑖2 + 2 𝐼𝜔 2 ; 𝐼𝑠𝑝ℎ𝑒𝑟𝑒 = 5 𝑚𝑟 2 𝑎𝑛𝑑 𝜔 = 𝑣/𝑟 2 1 1 2 𝑣 1 1 1 2.0𝑚 2 1 2.0𝑚 2 (7.2𝑘𝑔) ( 𝑚𝑣𝑖2 + 2 (5 𝑚𝑟 2 )( 𝑟 )2 2 𝑚𝑣𝑖2 + 5 𝑚𝑣 2 2 𝑠 ) + 5 (7.2𝑘𝑔)( 𝑠 ) ℎ𝑓 = 2 = ℎ𝑓 = = 0.29𝑚 𝑚𝑔 𝑚𝑔 9.8𝑚 (7.2𝑘𝑔)( 2 ) 𝑠 ENGAGEMENT Write your answer on a separate sheet of paper. Activity 2 Geometric Center Solve the following problems in a separate paper. Show your solutions systematically and clearly 1. A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut. 11  2. A cylinder of mass m and radius R has a moment of inertia 1 of 2 𝑚𝑟 2. The cylinder is released from rest at a height ℎ on an inclined plane, and rolls down the plane without slipping. What is the velocity of the cylinder when it reaches the bottom of the incline? 3. A uniform solid cylinder, sphere, and hoop roll without slipping from rest at the top of an incline. Find out which object would reach the bottom first. Activity 35-36.2 Composite Areas Copy the figure in a separate paper and calculate the mass of each item. Show your solutions. 12  AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE FIRST SEMESTER, S.Y. 2020-2021 GENERAL PHYSICS 1 Quarter 2 - Week 2 TOPIC: NEWTONS LAW OF UNIVERSAL GRAVITATION LESSON 3 At the end of this handout, you are expected to: Use Newton’s Law of gravitation to infer gravitational force, weight, and acceleration due to gravity; Discuss the physical significance of gravitational; Apply the concept of gravitational potential energy in physics problems; Calculate quantities related to planetary or satellite motion, and; For circular orbits, relate Kepler’s third law of planetary motion to Newton’s law of gravitation and centripetal acceleration. Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. Translating this into an equation, we have where Fg is the magnitude of the gravitational force on either particle, m1 and m2 are their masses, r is the distance between them, and G is a fundamental physical constant called the gravitational constant. The numerical value of G depends on the system of units used. This equation tells us that the gravitational force between two particles decreases with increasing distance r: If the distance is doubled, the force is only one-fourth as great, and so on. Although many of the stars in the night sky are far more massive than the sun, they are so far away that their gravitational force on the earth is negligibly small. Determining the Value of G To determine the value of the gravitational constant G, we have to measure the gravitational force between two bodies of known masses m1 and m2 at a known distance r. The force is extremely small for bodies that are small enough to be brought into the laboratory, but it can be measured with an instrument called a torsion balance, which Sir Henry Cavendish used in 1798 to determine G. After calibrating the Cavendish balance, we can measure gravitational forces and thus determine G. The presently accepted value is G = 6.67428 (67) x 10-11 N ∙ m2 / kg2 To three significant figures, G = 6.67 x 10-11 N ∙ m2 / kg2. (Because 1 N = 1 kg ∙ m/s2, the units of G can also be expressed as m3 / kg ∙ s2.) Gravitational forces combine vectorially. If each of two masses exerts a force on a third, the total force on the third mass is the vector sum of the individual forces of the first two. Calculating Gravitational Force Example: The mass m1 of one of the small spheres of a Cavendish balance is 0.0100 kg, the mass m2 of the nearest large sphere is 0.500 kg, and the center-to-center distance between them is 0.0500 m. Find the gravitational force Fg on each sphere due to the other. 1  Solution: Because the spheres are spherically symmetric, we can calculate Fg by treating them as particles separated by 0.0500 m. Each sphere experiences the same magnitude of force from the other sphere. We use Newton’s law of gravitation to determine Fg: Acceleration due to Gravitational Attraction Example: Suppose the two spheres in the previous example are placed with their centers 0.0500 m apart at a point in space far removed from all other bodies. What is the magnitude of the acceleration of each, relative to an inertial system? Solution: Each sphere exerts on the other a gravitational force of the same magnitude Fg , which we found in the previous example. We can neglect any other forces. The acceleration magnitudes a1 and a2 are different because the masses are different. To determine these, we’ll use Newton’s second law: ENGAGEMENT Write your answer on a separate sheet of paper. Activity 3.1 Problem Solving 1. The moon has a mass of 7.34  10 22 kg and a radius of 1.74  106 meters. If you have a mass of 66 kg, how strong is the force between you and the moon? 2. A distance of 0.002 m separates two objects of equal mass. If the gravitational force between them is 0.0104 N, find the mass of each object. TOPIC: GRAVITATIONAL FIELD LESSON 4 Gravitation is the most important force on the scale of planets, stars, and galaxies. It is responsible for holding our earth together and for keeping the planets in orbit about the sun. The mutual gravitational attraction between different parts of the sun compresses material at the sun’s core to very high densities and temperatures, making it possible for nuclear reactions to take place there. These reactions generate the sun’s energy output, which makes it possible for life to exist on earth and for you to read these words. The gravitational force is so important on the cosmic scale because it acts at a distance, without any direct contact between bodies. Electric and magnetic forces have this same remarkable property, but they are less important on astronomical scales because large accumulations of matter are electrically neutral; that is, they contain equal amounts of positive and negative charge. As a result, the electric and magnetic forces between stars or planets are very small or zero. The strong and weak interactions that we discussed also act at a distance, but their influence is negligible at distances much greater than the diameter of an atomic nucleus (about 10 -14 m). Our solar system is part of a spiral galaxy like the figure below, which contains roughly 1011 stars as well as gas, dust, and other matter. The entire assemblage is held together by the mutual gravitational attraction of all the matter in the galaxy. 2 ENGAGEMENT Write your answer on a separate sheet of paper. Activity 4.1 Think Critically Answer as required. 1. Compare the gravitational attraction between objects on earth and interaction of celestial bodies in space. Which gravitational force is almost negligible? Why? ______________________________________________________________ 2. Discuss why the study of a gravitational field is important. ___________________________________________________________________ TOPIC: GRAVITATIONAL POTENTIAL ENERGY LESSON 5 When we first introduced gravitational potential energy in our previous lessons, we assumed that the gravitational force on a body is constant in magnitude and direction. This led to the expression U = mgh. But the earth’s gravitational force on a body of mass m at any point outside the earth is given more generally by Fg = (GmEm) / r2, where mE is the mass of the earth and r is the distance of the body from the earth’s center. Consider a block with mass, and is tied to the end of a rope and goes up over a pulley while the other end is being pulled by a man. If the man lets go of the rope, the rope will be pulled downward with a force equal to the force of gravity of the block. The work performed by the block depends on the weight and height, Δh. The work done will be: W = Fd Since F = mg, then: W = (mg)( Δh) where d = Δh The more work to perform and energy stored in the block, the higher the block is from the ground. Gravitational Potential Energy, Eg , is the energy stored of an object because of its distance above the surface of the Earth. The change in gravitational potential energy of an object is expressed as: Eg = mgΔh where: m – is the mass of the object in kilograms g – is the acceleration due to gravity at 9.8 m/s2 Δh – is the vertical displacement of the object in meters ΔEg - is the object’s change in gravitational potential energy in Joules Sample Problems: 1. How much gravitational potential energy does a 4.0 kg block has if it is lifted 25 m? Eg = mgΔh = (4.0 kg) (9.80 N/kg) (25 m) = 9.8 x 102 J 2. A 61.2 kg boy fell 0.500 m out of the bed. How much potential energy is lost? Eg = mgΔh = (61.2 kg) (9.80 N/kg) (-0.500 m) = -299.8 J ≈ 300 J ENGAGEMENT Write your answer on a separate sheet of paper. Activity 5.1 Think Critically Answer as required. 1. How much potential energy does a car gain if a crane lifts the car with a mass of 1,500 kg and 20 m straight up? 2. A basketball of mass 0.0400 kg is dropped from a height of 5.00 m to the ground and bounces back to a height of 3.00 m. a. On its way down, how much potential energy does the ball lose? b. On its way back, how much potential energy does the ball regain? 3 TOPIC: ORBITS LESSON 6 A circular orbit, is the simplest case. It is also an important case, since many artificial satellites have nearly circular orbits and the orbits of the planets around the sun are also fairly circular. The only force acting on a satellite in circular orbit around the earth is the earth’s gravitational attraction, which is directed toward the center of the earth and hence toward the center of the orbit. This means that the satellite is in uniform circular motion and its speed is constant. The satellite isn’t falling toward the earth; rather, it’s constantly falling around the earth. In a circular orbit the speed is just right to keep the distance from the satellite to the center of the earth constant. The radius of the orbit is r, measured from the center of the earth; the acceleration of the satellite has magnitude arad = v2 / r and is always directed toward the center of the circle. By the law of gravitation, the net force (gravitational force) on the satellite of mass m has magnitude F g = GmEm / r2 and is in the same direction as the acceleration. Newton’s second law (F = ma) then tells us that GmEm = mv2 r2 r Solving this to find the circular orbit, we find This relationship shows that we can’t choose the orbit radius r and the speed independently; for a given radius r, the speed for a circular orbit is determined. The satellite’s mass m doesn’t appear in the equation above, which shows that the motion of a satellite does not depend on its mass. If we could cut a satellite in half without changing its speed, each half would continue on with the original motion. For example, an astronaut on board a space shuttle is herself a satellite of the earth, held by the earth’s gravitational attraction in the same orbit as the shuttle. The astronaut has the same velocity and acceleration as the shuttle, so nothing is pushing her against the floor or walls of the shuttle. She is in a state of apparent weightlessness, as in a freely falling elevator. True weightlessness would occur only if the astronaut were infinitely far from any other masses, so that the gravitational force on her would be zero. Apparent weightlessness is not just a feature of circular orbits; it occurs whenever gravity is the only force acting on a spacecraft. Hence it occurs for orbits of any shape. We can derive a relationship between the radius r of a circular orbit and the period T, the time for one revolution. The speed is the distance traveled in one revolution, divided by the period: To get an expression for T, we use the equation above to solve for T and substitute: We have talked mostly about earth satellites, but we can apply the same analysis to the circular motion of any body under its gravitational attraction to a stationary body. Other examples include the earth’s moon and the moons of other planets. Sample Problem: You wish to put a 1000-kg satellite into a circular orbit 300 km above the earth’s surface. What speed, period, and radial acceleration will it have? Solution: The radius of the satellite’s orbit is r = 6380 km + 300 km = 6680 km = 6.68 x 10 6 m. The orbital speed is 4  = 7720 m/s To find the orbital period, we have: Finally, the radial acceleration is Activity 5.1 Think Critically Answer as required. 1. NASA is expected to send a 2600-kg satellite 450 km above the earth’s surface. (Hint: Earth’s mass is 5.97 x 1024 kg) a. What is its radius? b. What speed will it have? c. What is its orbital period? d. What is its radical acceleration? TOPIC: KEPLER’S LAWS OF PLANETARY MOTION LESSON 7 The following are the laws developed by Johannes Kepler: 1. LAW OF ORBITS The first law explains that all planets move in elliptical orbits with the sun at one focus. Kepler’s first law means that planets move around the Sun in elliptical orbits. An ellipse is a shape that resembles a flattened circle. How much the circle is flattened is expressed by its eccentricity. The eccentricity is a number between 0 and 1. It is zero for a perfect circle. The eccentricity of an ellipse measures how flattened a circle it is. For a perfect circle, a and b are the same such that the eccentricity is zero. Earth’s orbit has an eccentricity of 0.0167, so it is very nearly a perfect circle. 2. LAW OF AREAS The second law describes a line that connects a planet to the sun and sweeps out equal areas in equal times. When a planet is near the sun, it travels faster and sweeps through a longer path in a given time. 5  It follows from Kepler’s second law that Earth moves the fastest when it is closest to the Sun. This happens in early January, when Earth is about 147 million km (91 million miles) from the Sun. When Earth is closest to the Sun, it is traveling at a speed of 30.3 kilometers (18.8 miles) per second. 3. LAW OF PERIODS The third law mathematically expressed as the square of the period of any planet is proportional to the cube of the semi-major axis of its orbit. Knowledge of these laws, especially the second (the law of areas), proved crucial to Sir Isaac Newton in 1684– 1685, when he formulated his famous law of gravitation between Earth and the Moon and between the Sun and the planets, postulated by him to have validity for all objects anywhere in the universe. Newton showed that the motion of bodies subject to central gravitational force need not always follow the elliptical orbits specified by the first law of Kepler but can take paths defined by other, open conic curves; the motion can be in parabolic or hyperbolic orbits, depending on the total energy of the body. Thus, an object of sufficient energy—e.g., a comet—can enter the solar system and leave again without returning. From Kepler’s second law, it may be observed further that the angular momentum of any planet about an axis through the Sun and perpendicular to the orbital plane is also unchanging. 6  AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE GENERAL PHYSICS 1 SECOND QUARTER WEEK 3-4 LESSON 1: KEPLER’S THIRD LAW OF PLANETARY MOTION, NEWTON’S LAW OF GRAVITATION AND CENTRIPETAL ACCELERATION Most Essential Learning Competency: For circular orbits, relate Kepler’s third law of planetary motion to Newton’s law of gravitation and centripetal acceleration. Kepler's laws and Newton's laws taken together imply that the force that holds the planets in their orbits by continuously changing the planet's velocity so that it follows an elliptical path is (1) directed toward the Sun from the planet, (2) is proportional to the product of masses for the Sun and planet, and (3) is inversely proportional to the square of the planet-Sun separation. This is precisely the form of the gravitational force, with the universal gravitational constant G as the constant of proportionality. Thus, Newton's laws of motion, with a gravitational force used in the 2nd Law, imply Kepler's Laws, and the planets obey the same laws of motion as objects on the surface of the Earth! Newton's comparison of the acceleration of the moon to the acceleration of objects on earth allowed him to establish that the moon is held in a circular orbit by the force of gravity - a force that is inversely dependent upon the distance between the two objects' centers. Establishing gravity as the cause of the moon's orbit does not necessarily establish that gravity is the cause of the planet's orbits. How then did Newton provide credible evidence that the force of gravity is meets the centripetal force requirement for the elliptical motion of planets? Johannes Kepler proposed three laws of planetary motion. His Law of Harmonies suggested that the ratio of the period of orbit squared (T2) to the mean radius of orbit cubed (R3) is the same value k for all the planets that orbit the sun. Known data for the orbiting planets suggested the following average ratio: k = 2.97 x 10-19 s2/m3 = (T2)/(R3) Newton was able to combine the law of universal gravitation with circular motion principles to show that if the force of gravity provides the centripetal force for the planets' nearly circular orbits, then a value of 2.97 x 10-19 s2/m3 could be predicted for the T2/R3 ratio. Here is the reasoning employed by Newton: Consider a planet with mass Mplanet to orbit in nearly circular motion about the sun of mass MSun. The net centripetal force acting upon this orbiting planet is given by the relationship Fnet = (Mplanet * v2) / R This net centripetal force is the result of the gravitational force that attracts the planet towards the sun, and can be represented as Fgrav = (G* Mplanet * MSun ) / R2 Since Fgrav = Fnet, the above expressions for centripetal force and gravitational force are equal. Thus, (Mplanet * v2) / R = (G* Mplanet * MSun ) / R2 Since the velocity of an object in nearly circular orbit can be approximated as v = (2*pi*R) / T, v2 = (4 * pi2 * R2) / T2 Substitution of the expression for v2 into the equation above yields, (Mplanet * 4 * pi2 * R2) / (R T2) = (G* Mplanet * MSun ) / R2 By cross-multiplication and simplification, the equation can be transformed into T2 / R3 = (Mplanet * 4 * pi2) / (G* Mplanet * MSun ) The mass of the planet can then be canceled from the numerator and the denominator of the equation's right-side, yielding T2 / R3 = (4 * pi2) / (G * MSun ) The right side of the above equation will be the same value for every planet regardless of the planet's mass. Subsequently, it is reasonable that the T 2/R3 ratio would be the same value for all planets if the force that holds the planets in their orbits is the force of gravity. Newton's universal law of gravitation predicts results that were consistent with known planetary data and provided a theoretical explanation for Kepler's Law of Harmonies. LESSON 2: PERIODIC MOTION Most Essential Learning Competency: Relate the amplitude, frequency, angular frequency, period, displacement, velocity and acceleration of oscillating systems. Recognize the necessary conditions for an object to undergo simple harmonic motion.  Calculate the period and the frequency of spring mass, simple pendulum and physical pendulum. Periodic Motion - motion that is regular and repeating - for examples a rocking chair, a bouncing ball, a vibrating tuning fork, a swing in motion, the Earth in its orbit around the Sun, and water wave. Period – the interval of time for a repetition, or cycle, of the motion Frequency – number of periods per unit time ✓ The period of the Earth’s orbit is one year and its frequency is one orbit per year. Simple Harmonic Motion - the repetitive movement back and forth through an equilibrium, or central, position so that the maximum displacement on one side of this position is equal to the maximum displacement on one side of this position is equal to the maximum displacement on the other side. - the time interval of each complete vibration is the same, and the force responsible for the motion is always directed toward the equilibrium position and is directly proportional to the distance from it. - an object is undergoing simple harmonic motion (SHM) if the acceleration of the object is directly proportional to its displacement from its equilibrium position and the acceleration is always directed towards the equilibrium position. ✓ The frequency (f) of an oscillation is measured in hertz (Hz) it is the number of oscillations per second. The time for one oscillation is called the period (T) it is measured in seconds. 𝟏 f=𝑻 ✓ We can calculate the acceleration of the object t any point in its oscillation using the equation below. a = - (2πf)2 x a = acceleration in m/s2, f = frequency in Hz, x = displacement from the central position in m.  ✓ We can calculate the displacement of the object at any point in its oscillation using the equation below. x = Acos 2πft A = the amplitude (maximum displacement) in m, t = the time since the oscillation begin in s. ✓ We can calculate the velocity of the object at any point in its oscillation using the equation below. v = ± 2πf A2 – x2 ✓ The velocity equation simplifies to the equation below when we just want to know the maximum speed. Speedmax = 2πfA ✓ The acceleration equation simplifies to the equation below when we just want to know the maximum acceleration. Accelerationmax = (2πf)2 A ✓ Time period of mass-spring system. 𝒎 T = 2π  𝒌 ✓ Time period of pendulum. 𝒍 T = 2π  𝒈 SAMPLE PROBLEM: Ultrasound machines are used by medical professionals to make images for examining internal organs of the body. An ultrasound machine emits high-frequency sound waves, which reflect off the organs, and a computer receives the waves, using them to create a picture. Consider a medical imaging device that produces ultrasound by oscillating with a period of 0.400 s. What is the frequency of this oscillation? Sol. T = 0.400 s 1 1 f= = 0400 𝑥 10−6 𝑠 = 2.50 x 106 Hz 𝑇 PROBLEM SOLVING: Solve the given problem: Identify the given quantities, unknown quantities, and the equation to be used based on the given problem. BOX the final answer with proper unit. 1. A simple pendulum has a length of 0.25 m and the mass attached to it is 0.50 kg. What is the period of oscillation of the pendulum? How long will the pendulum reach its maximum velocity if it is initially displaced 3.50? What is the maximum velocity of the pendulum mass? 2. A mass on a spring system was initially displaced at 10 cm and was released. The resulting oscillation completed 10 cycles in 3 seconds. What is the mass of the system if the spring stiffness is 2.40 N/m? What should be the mass to increase the oscillation frequency by two-fold, given the same spring? LESSON 3: DAMPED AND DRIVEN OSCILLATION Most Essential Learning Competency: Differentiate underdamped, overdamped and critically damped motion. ✓ If a frictional force (damping) proportional to the velocity is also present, the harmonic oscillator is described as a damped oscillator. Depending on the friction coefficient, the system can: Oscillate with a frequency lower than in the non-damped case, and an amplitude decreasing with time (underdamped oscillator). Decay to the equilibrium position, without oscillations (overdamped oscillator). ✓ Under Damped – The condition in which damping of an oscillator causes it to return to equilibrium with the amplitude gradually decreasing to zero; system returns to equilibrium faster but overshoots and crosses the equilibrium position one or more times. ✓ Critically Damped – The condition in which the damping of an oscillator causes it to return as quickly as possible to its equilibrium position without oscillating back and forth about this position. ✓ Over Damped – The condition in which damping on an oscillator causes it to return to equilibrium without oscillating; oscillator moves more slowly toward equilibrium than in the critically damped system. LESSON 4: MECHANICAL WAVES Most Essential Learning Competency: Define mechanical wave, longitudinal wave, transverse wave, periodic wave, and sinusoidal wave. From a given sinusoidal wave function infer the speed, wavelength, frequency, period, direction, and wave number. Mechanical Waves – is a wave that is not capable of transmitting its energy through a vacuum. It requires a medium in order to transport their energy from one location to another. Sound wave is an example of mechanical wave and are incapable of traveling through a vacuum. Slinky waves, water waves, stadium waves and jump rope waves are other examples of mechanical waves; each requires some medium in order to exist. Types of Mechanical Waves Transverse Wave – a wave in which particles of the medium move in a direction perpendicular to the direction that the wave moves. Longitudinal Wave – a wave in which particles of the medium move in a direction parallel to the direction of the wave moves Periodic Waves – regularly repeating waves such as electromagnetic wave in optical fiber, sound from a guitar string. It can characterize either by the length scale, wavelength or the time scale, period, at which they repeat. ✓ The speed of the wave is the distance the wave travels per unit time. A unit time is one period, as that is the time it takes the oscillation to return to tits  original point. The distance traveled in one period is one wavelength, as that is the distance between two maxima. The speed therefore simply the ratio, which can also be expressed in terms of the wave number and frequency.   vw = 𝑻 = 𝒌 LESSON 5: SOUND Most Essential Learning Competency: Apply the inverse-square relation between the intensity of waves and the distance from the source. Sound - A wave that is created by vibrating objects and propagated through a medium from one location to another. - A wave can be described a s a disturbance that travels through a medium, transporting energy from one location to another location. - The medium is simply the material through which the disturbance is moving; it can be thought of as a series of interacting particles. Wave Intensity - The intensity of any wave is the time average power (P) it transfers per area (A) through some region of space. - the SI unit is Watt/m2. I= (𝑷) 𝑨 ✓ As a sound wave carries its energy through a two-dimensional or three- dimensional medium, the intensity of the sound wave decreases with increasing distance from the source. ✓ The mathematical relationship between intensity and distance is sometimes referred to as an inverse square relationship. ✓ The intensity varies inversely with the square of the distance from the source. Prepared by: LOIDA A. ARCE Subject Teacher  AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE GENERAL PHYSICS 1 SECOND QUARTER WEEK 5-6 Lesson 1 Sound and Waves Most Essential Learning Competencies: Describe qualitatively and quantitatively the superposition of waves. Apply the condition for standing waves on a string. Relate the frequency and wavelength of sound with the motion of the source and the listener. Superposition of Waves ✓ It may be applied to waves whenever two or more waves traveling through the same medium at the same time. ✓ The waves pass through each other without being disturbed. ✓ The net displacement of the medium at any point in space or time is simply the sum of the individual wave displacements. ✓ This is true of waves which are finite in length (wave pulses) or which are continuous sine waves. The Principle of Superposition “When two waves interfere, the resulting displacement of the medium at any location is the algebraic sum of the displacements of the individual waves at the same location”. Example: Displacement of Pulse 1 Displacement of Pulse 2 Resulting Displacement +1 +1 +2 -1 -1 -2 +1 -1 0 +1 -2 -1 Pitch and Frequency ❖ The frequency of a wave refers on how often the particles of the medium vibrate when a wave passes through the medium. ❖ The frequency of a wave is measured as the number of complete back and forth vibrations of a particle of the medium per unit time. ❖ The unit of frequency is Hertz (Hz) = vibration / second Example: If the particle of air undergoes 1000 longitudinal vibrations in 2 seconds. What is the frequency? 1 Given: vibrations = 1000 time = 2 seconds frequency =? Solution: f = vibrations / time f = 1000 / 2 s = 500 vibrations/second or Hz Interference of Waves It occurs when two waves meet while traveling along the same medium and causes the medium to take on a shape that results from the net effect of the two individual waves upon the particles of the medium. Constructive Interference A type of interference that occurs any location along the medium where the two interfering waves have a displacement in the same direction. Destructive Interference A type of interference that occurs any location along the medium where the two interacting waves have a displacement in the opposite direction Standing Waves ✓ A vibrational pattern created within a medium when the vibrational frequency of the source causes reflected waves from one end of the medium to interfere with incident waves from the source. ✓ It occurs because of the specific point along the medium appears to be standing still and are only created within the medium at specific frequencies of vibration and these are cold harmonic frequencies of harmonics. Doppler Effect ✓ The Doppler effect (or the Doppler shift) is the change in frequency or wavelength of a in relation to observer who is moving relative to the wave 2  source. It is named after the Austrian physicist Christian Doppler, who described the phenomenon in 1842. ✓ A common example of Doppler shift is the change of pitch heard when a vehicle sounding a horn approach and recedes from an observer. Compared to the emitted frequency, the received frequency is higher during the approach, identical at the instant of passing by, and lower during the recession. ✓ As a wave source approaches, an observer encounters wave with higher frequency. And as the wave source move away, an observer encounters wave with lower frequency. ✓ The Doppler effect is observed because the distance between the source of the sound and the observer is changing. If the source and the observer are approaching, then the distance is decreasing and if the source and the observer are receding, then the distance is increasing. The source of sound always emits the same frequency. ASSESSMENT (PRE-TEST AND POST-TEST) 1. What is Periodic motion? a. any movement of an object that is repeated in a given length of time. b. any circular motion c. any movement of an object that has consistent pattern d. any movement that is predictable and continuous forever 2. Which of the following represents the time period of a pendulum hang in a satellite? a. 2𝜋 b. Zero c. T (period of time on earth) d. infinite 3. Which of the following describes the equilibrium point of an oscillating system? a. the velocity is zero b. the velocity is minimum c. the force is zero d. the velocity is maximum 3 4. Which of the following describes the periodic motion? a. objects move in circular b. objects move in constant velocity c. objects move in constant acceleration d. objects move and return to its original position 5. In simple harmonic motion, what will be the velocity if the position is in equilibrium? a. constant position b. minimum position c. maximum position d. no position 6. In simple harmonic motion what will be the maximum speed will be proportional to the following a. acceleration b. frequency c. velocity d. time 7. The tachometer in of a jeepney measures the revolutions per minute of the tires. The jeepney is travelling at a constant speed, and the tachometer reads 2100 revolutions per minute. What is the frequency of the tires spinning, measured in revolution per sec? a. 34 revolution/sec c. 35 revolution/sec b. 38 revolution/sec d. 40 revolution/sec 8. A pendulum takes 4.00 seconds to complete one back-and-forth cycle. What is the frequency of the pendulum’s motion? a. 0.20 cycles/sec b. 0.25 cycles/sec c. 0.26 cycles/sec d. 1 cycles/sec 9. Which of the following makes the oscillations damped? a. normal force makes the oscillations damped b. frictional force makes the oscillation damped c. tangential force makes the oscillation damped d. parallel force makes the oscillation damped 10. Which of the following is the unit of loudness? a. hertz b. Luminous c. Decibel d. watts 11. Which of the following describes the sound waves? a. surface waves b. latitudinal mechanical waves c. transverse waves d. longitudinal mechanical waves 4  12. Which of the following applies ultrasonic waves? a. measure the coastal areas of the sea b. sterilizing surgical instruments c. mother takes ultrasound for her baby d. drinks sterilized milk 13. Which of the following best describes the variation of a quantity such as voltage or current shown in a graph with the x-axis as time? a. waveform b. wavelength c. circular form d. line form 14. Which of the following is the value of a given waveform at any instant time? a. waveform b. instantaneous value c. cycle value d. time value 15. Which of the following is the repetition of a variable quantity, recurring at equal intervals? a. waveform b. instantaneous value c. cycle d. time value 16. An ambulance is approaching a stationary observer. For the observer, the apparent frequency of the siren compared to the actual frequency of the siren. a. increases or decreases depending on the speed of the ambulance b. resonates c. decreases d. stays the same 17. The speed of any mechanical wave as it propagates through a medium is dependent mainly on the a. frequency of the wave source b. wavelength c. period of the wave d. type of medium through which the wave travels e. amplitude 18. The shortest time travel in which a wave motion completely repeats itself (makes one complete vibration or oscillation) is called the a. amplitude b. period c. wavelength d. frequency e. speed 19. In an ideal wave, which of the following wave properties does not change as it travels through the same medium? a. amplitude b. wavelength c. frequency d. period e. all of them 20. Consider the following wave properties: I. speed II. Frequency III. Wavelength IV. Period Which of the following does not change as a wave changes medium? a. I only b. II only c. I and III d. I and IV e. II and IV 5 Lesson 2 Fluid Mechanics Most Essential Learning Competencies: Relate density, specific gravity, mass and volume to each other. Relate pressure to area and force. Relate pressure to fluid density and depth. Apply Pascal’s principle in analyzing fluids in various systems. Apply the concept of buoyancy and Archimedes Principle. Density and Specific Gravity ✓ Density is mass per unit volume ✓ ρ=m/v m = kg and v=m3 Example: A golf ball has a diameter of 42 mm and a mass of 45 g. Given: d = 42 mm = 0.042 m m = 45 g = 0.045 kg Sol. V = 4/3 πr3 V = 4/3 (3.14) (0.021)3 V = 3.84 x 10-5 m3 d = m / V = 0.045 kg / 3.84 x 10-5 m3 = 1172 kg / m3 Specific Gravity ✓ is a dimensionless unit defined as the ratio of the density of a substance to the density of water at a specified temperature sg = ρ substance / ρ water ρsubstance – density of the fluid or substance kg/m3 ρwater – density of water kg/m3 Example: The density of iron is 7850 kg/m3. The specific gravity of iron related to water with density 1000 kg/m3 is Given: ρiron = 7850 kg/m3 ρwater = 1000 kg/m3 Solution: sg = 7850 kg/m3 / 1000 kg/m3 = 7.85 6 Pressure ✓ Force per unit area. P = F / A ✓ Units are Pascal and N/m2. Static Fluid Pressure ✓ The pressure exerted by a static fluid depends on only upon the depth of the fluid, the density of the fluid, and the acceleration due to gravity. ✓ P = ρgh ✓ ρ – fluid density, g – acceleration due to gravity and h – depth of fluid Example: The density of water at 40C is 1000 kg/m3. The pressure acting in water is 1 m can be calculated as. Given: ρ = 1000 kg/m3 h = 1m g = 9.8 m/s2 Sol. P = (1000 kg/m3) (9.8 m/s2) (1m) P = 9800 Pa Pascal’s Principle ❖ It states that the pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid such that the pressure variations remain the same. ❖ A change in pressure at any point in an enclosed fluid at rest is transmitted undiminished to all points in the fluid. ❖ This principle states that in a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the fluid and to the walls of the container. ❖ The principle was first enunciated by the French scientist Blaise Pascal. ❖ The pressure inside an enclosed system is the same. The pressure at each point is F/A. According to Pascal’s principle, the pressure is the same everywhere inside the fluid. ❖ F = PA ❖ F1 / A1 = F2 / A2 7 Archimedes Principle A fluid exerts an upward buoyant force on a submerged object equal in magnitude to the weight of the volume of fluid displaced by the object. FB = pgdA = pgV the weight of an object is GREATER than the buoyant force, the object will SINK. If the weight of an object is LESS than the buoyant force, the object will FLOAT. If the weight of an object and the buoyant force is EQUAL, the object will FLOAT UNDERWATER. ASSESSMENT (Pre-test and Post-test): Read and analyze the following given questions. 1. How is mass differ from weight? a. Mass and weight are the same. b. Mass is measured of matter and Weight is measured of matter plus gravity c. Mass is the measurement due to gravity and weight is the measurement of matter d. Mass is the amount of matter and weight is the amount of matter plus energy 2. You are asked to bring stone in the class. Your teacher asks you to determine if the stone your brig in the class is more or less dense than water. You placed the stone in the glass with water and it sinks. Which statement best fit to describe the situation? a. The stone is denser than water. b. The stone is less dense than water. c. Water is denser than the stone. d. They have both the same mass. 3. Which statement will best fit in finding density? a. The volume will be divided by the mass. b. The mass will be multiplied by the volume. c. The mass will be divided by the volume. d. The volume will be multiplied by the mass. 8 4. Which of the following describe the difference of measurement when you were on the moon versus on earth? a. Mass b. Weight c. Mass and weight d. none of them 5. Which of the following method determining mass measures how much a body resist acceleration when responding to any types of force? a. Atomic mass b. Gravitational mass c. Inertial mass d. Newton mass 6. What kind of tool is used to determine the measurement of mass? a. graduated cylinder b. meter stick c. scale d. triple beam balance 7. Which of the following determines pressure? a. Force times area. b. Force per area. c. Density times volume. d. Mass per volume. 8. When force acting in smaller area, what pressure you need to exert? a. less pressure b. more pressure c. no pressure d. minimal pressure 9. Which of the following is the best suited example of atmospheric pressure? a. pouring of water in a container b. drinking of soft drinks in glass c. drinking of juice with a straw d. a stirrer dip in a cup 10. Which of the following pressure is the exerted by air molecules to earth’s surface due to weight? a. absolute pressure b. atmospheric pressure c. differential pressure d. gauge pressure 11. What is the reciprocal of the specific volume of a liquid? a. density b. mass c. specific weight d. specific volume 9  12. Which of the following situation best describes the application of Pascal’s Principle? a. car brakes b. push a car c. pull the door d. an electric fan 13. Which of the following word best describes fluid systems that apply Pascal’s Principle? a. hydraulics b. pneumatics c. pushing d. air pressure 14. Which of the following best describes the Pascal’s Law? a. The liquid at motion applies pressure at a point is the same in all directions. b. The liquid at rest applies pressure at different point in different directions. c. The liquid at rest does not apply pressure at a point in different directions. d. The liquid at rest applies pressure at a point is the same in all directions. 15. Which of the following the buoyant force of a floating object in a surface depend? a. viscosity of the liquid b. volume of the object c. density of the liquid d. mass of the liquid Prepared by: LOIDA A. ARCE General Physics Teacher 10  AMADEO NATIONAL HIGH SCHOOL BY PASS ROAD, BRGY. 1, AMADEO, CAVITE GENERAL PHYSICS 1 SECOND QUARTER WEEK 7-8 LESSON 1: ZEROTH LAW OF THERMODYNAMICS AND TEMPERATURE MEASUREMENT Most Essential Learning Competencies: Apply Bernoulli’s principle and continuity equation, whenever appropriate, to infer relations involving pressure, elevation, speed and flux. Explain the connection between the Zeroth Law of Thermodynamics, temperature, thermal equilibrium and temperature scales. Convert temperatures and temperature differences in the following scales: Fahrenheit, Celsius and Kelvin. Define coefficient of thermal expansion and coefficient of volume expansion. Calculate volume or length changes of solids due to changes in temperature. Solve problems involving temperature, thermal expansion, heat capacity heat transfer, and thermal equilibrium in contexts such as, but not limited to, the design of bridges and train rails using steel, relative severity of steam burns and water burns, thermal insulation, sizes of stars, and surface temperatures of planets. The Zeroth Law of Thermodynamics There are a few ways to state the Zeroth Law of Thermodynamics, but the simplest is as follows: systems that are in thermal equilibrium exist at the same temperature. Systems are in thermal equilibrium if they do not transfer heat, even though they are in a position to do so, based on other factors. For example, food that’s been in the refrigerator overnight is in thermal equilibrium with the air in the refrigerator: heat no longer flows from one source (the food) to the other source (the air) or back. What the Zeroth Law of Thermodynamics means is that temperature is something worth measuring, because it indicates whether heat will move between objects. This will be true regardless of how the objects interact. Even if two objects don’t touch, heat may still flow between them, such as by radiation (as from a heat lamp). However, according to the Zeroth Law of Thermodynamics, if the systems are in thermal equilibrium, no heat flow will take place. 1 There are more formal ways to state the Zeroth Law of Thermodynamics, which is commonly stated in the following manner: Let A, B, and C be three systems. If A and C are in thermal equilibrium, and A and B are in thermal equilibrium, then B and C are in thermal equilibrium. This statement is represented symbolically in. Temperature is not mentioned explicitly, but it’s implied that temperature exists. Temperature is the quantity that is always the same for all systems in thermal equilibrium with one another. Zeroth Law of Thermodynamics: The double arrow represents thermal equilibrium between systems. If systems A and C are in equilibrium, and systems A and B are in equilibrium, then systems B and C are in equilibrium. The systems A, B, and C are at the same temperature. The Celsius, Kelvin, and Fahrenheit temperature scales are shown in relation to the phase change temperatures of water. The Kelvin scale is called absolute temperature and the Kelvin is the SI unit for temperature. 2 The Fahrenheit Temperature Scale The Fahrenheit temperature scale sets the freezing point of water at 32 degrees Fahrenheit. Water boils at 212 degrees in this scale. Hence a Fahrenheit degree is 1/180 the temperature difference between the boiling and freezing points of water. The Centigrade or Celsius Temperature Scale In the Celsius temperature scale the freezing point of water is defined as 0 degrees Celsius. Water boils at 100 degrees Celsius. Hence a Celsius degree is 1/100 the difference between the boiling and freezing point of water. A Centigrade degree is 100/180 or 5/9 times the size of a Fahrenheit degree. That is why the factors 9/5 or 5/9 appear in the formulas to convert Celsius to Fahrenheit or Fahrenheit to Celsius. The Kelvin Temperature Scale A Kelvin degree is the same size as a Centigrade degree. This temperature scale however uses absolute zero, rather than the freezing point of water, as the zero point. In this temperature scale water freezes at 273.15 Kelvins and boils at 373.15 Kelvins. Absolute Zero When an object's temperature increases its internal heat energy causes the random motions of individual atoms and molecules to increase. Random molecular motions are faster at higher temperatures. As the temperature decreases, the random atomic and molecular motions decrease. The lowest possible temperature is when these random atomic and molecular motions are at the minimum possible energy. This temperature is absolute zero. It is zero in the Kelvin temperature scale. Absolute zero is -273.15 degrees Celsius and -459.67 degrees Fahrenheit. And so, to convert: from Celsius to Fahrenheit: first multiply by 180100, then add 32 from Fahrenheit to Celsius: first subtract 32, then multiply by 100180 180100 can be simplified to 95, and 100180 can be simplified to 59, so this is the easiest way: °C to °F: Divide by 5, then multiply by 9, then add 32 3 °F to °C: Subtract 32, then multiply by 5, then divide by 9 Example: Convert 25° Celsius (a nice warm day) to Fahrenheit First: 25° / 5 = 5 Then: 5 × 9 = 45 Then: 45 + 32 = 77° F Example: Convert 98.6° Fahrenheit (normal body temperature) to Celsius First: 98.6° − 32 = 66.6 Then: 66.6 × 5 = 333 Then: 333 / 9 = 37° C Thermal expansion refers to a fractional change in size of a material in response to a change in temperature. This includes… o changes in length compared to original length (∆ℓ/ℓ0) called linear expansion o changes in area compared to original area (∆A/A0) called areal expansion or superficial expansion o changes in volume compared to original volume (∆V/V0) called volumetric expansion or cubical expansion For most materials, over small temperature ranges, these fractional changes… o are directly proportional to temperature change (∆T) and o have the same sign (i.e., materials usually expand when heated and contract when cooled) o are larger for liquids than solids A coefficient of thermal expansion… o is the ratio of the fractional change in size of a material to its change in temperature o is represented by the symbol α (alpha) for solids and β (beta) for liquids o uses the SI unit inverse kelvin (K−1 or 1/K) or the equivalent acceptable non SI unit inverse degree Celsius (°C−1 or 1/°C). 4  Solids… o tend to retain their shape when not constrained and so are best described by a linear coefficient of thermal expansion, α (alpha). o have a real expansion that is very nearly twice their linear expansion, 2α (since two perpendicular linear measurements describe an area) o have a volumetric expansion that is very nearly three times their linear expansion, 3α (since three perpendicular linear measurements describe a volume) Liquids… o tend to take on the shape of their container and so are best described by a volumetric coefficient of thermal expansion, β (beta). Gases… o have a thermal expansion that is best described using the ideal gas law equation solids ∆ℓ = ℓ0 α∆T linear expansion ∆A = A0 2α∆T areal (or superficial) expansion ∆V = V0 3α∆T volumetric (or cubical) expansion equation liquids ∆V = V0 β∆T volumetric (or cubical) expansion equation gases PV = nRT ideal gas law Equations of thermal expansion EXAMPLE 1. CALCULATING LINEAR THERMAL EXPANSION: THE GOLDEN GATE BRIDGE The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from –15ºC to 40ºC. What is its change in length between these temperatures? Assume that the bridge is made entirely of steel. 5 Strategy Use the equation for linear thermal expansion ΔL = αLΔT to calculate the change in length , ΔL. Use the coefficient of linear expansion, α, for steel from Table 1, and note that the change in temperature, ΔT, is 55ºC. Solution Plug all of the known values into the equation to solve for ΔL. ΔL=αLΔL=(12×10−6∘C)(1275 m)(55∘C)=0.84 mΔL=αLΔL=(12×10−6∘C)(1275 m)(55∘C)=0.84 m Discussion Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small. EXAMPLE 2. CALCULATING THERMAL EXPANSION: GAS VS. GAS TANK Suppose your 60.0-L (15.9-gal) steel gasoline tank is full of gas, so both the tank and the gasoline have a temperature of 15.0ºC. How much gasoline has spilled by the time they warm to 35.0ºC? Strategy The tank and gasoline increase in volume, but the gasoline increases more, so the amount spilled is the difference in their volume changes. (The gasoline tank can be treated as solid steel.) We can use the equation for volume expansion to calculate the change in volume of the gasoline and of the tank. Solution 1. Use the equation for volume expansion to calculate the increase in volume of the steel tank: ΔVs = βsVsΔT. 2. The increase in volume of the gasoline is given by this equation: ΔVgas = βgasVgasΔT. 3. Find the difference in volume to determine the amount spilled as Vspill=ΔVgas − ΔVs. Alternatively, we can combine these three equations into a single equation. (Note that the original volumes are equal.) 6 Vspill=(βgas−βs)VΔT=[(950−35)×10−6/∘C](60.0L)(20.0∘C)=1.10LVspill=(βgas−βs)VΔT=[(950−35) ×10−6/∘C](60.0L)(20.0∘C)=1.10L Discussion This amount is significant, particularly for a 60.0-L tank. The effect is so striking because the gasoline and steel expand quickly. The rate of change in thermal properties is discussed in the chapter Heat and Heat Transfer Methods. If you try to cap the tank tightly to prevent overflow, you will find that it leaks anyway, either around the cap or by bursting the tank. Tightly constricting the expanding gas is equivalent to compressing it, and both liquids and solids resist being compressed with extremely large forces. To avoid rupturing rigid containers, these containers have air gaps, which allow them to expand and contract without stressing them. Activity: Solve the given problems. Identify the given quantities and unknown quantity. BOX the final answer. 1. A vertical steel antenna tower is 200 m high. Calculate the change in height of the tower that takes place when the temperature changes from -200C on a winter day to 300C on a summer day. 2. A metal rod is 200 cm long at 00C and 200.18 cm long at 600C. What is its coefficient of linear expansion? 3. A round hole of 6 cm in diameter at 00C is cut in a sheet  = 0.000018 /C. Find the new diameter of the hole at 500C. Assessment: Choose the letter of the best answer. Write the CAPITAL LETTER of your choice on your answer sheet. (2 points each) 1. What is the name of the following statement: “When two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other”? A. First Law of Thermodynamics B. Second Law of Thermodynamics C. Mechanical equivalent of heat D. Zeroth Law of Thermodynamics 7 2. Dry ice has a temperature of -110.20F. What would its temperature be on Celsius scale? A. -7110C B. -790C C. -142.20C D. -198.30C 3. The constant of proportionality is called of a substance and it plays a role in thermal expansion similar to that of the elastic modulus in tensile stress. A. coefficient of linear expansion C. coefficient of volume expansion B. coefficient of thermal expansion D. coefficient of friction 4. A metal box is heated until each of its sides has expanded bu 0.1%. By what percent has the volume of the box changed? A. -0.3% B. -0.2% C. 0.1% D. 0.2% 5. Room temperature is often identified as 680F. What temperature is this on the Kelvin scale? A. 20 K B. 100 K C. 293 K D. 341 K LESSON 2: IDEAL GAS LAW, THERMODYNAMIC PROCESSES, HEAT ENGINES, ENTROPY AND SECOND LAW OF THERMODYNAMICS Most Essential Learning Competencies Enumerate the properties of an ideal gas. Solve problems involving ideal gas equations in contexts such as, but not limited to, the design of metal containers for compressed gases. Interpret PV diagrams of a thermodynamic process. Compute the work done by a gas using dW=PdV. State the relationship between changes internal energy, work done, and thermal energy supplied through the First Law of Thermodynamics. Differentiate the following thermodynamic processes and show them in PV diagram: isochoric, isobaric, isothermal, adiabatic and cyclic. Calculate the efficiency of heat engine. Describe reversible and irrever