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MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 7: ROTATION (PART 2) TORQUE CLASS NOTES TORQUE Torque measures the force that causes a body to rotate about a specified axis. TORQUE EQUATIONS The torque equations develop from Newton’s second law: = I F┴r TORQUE AND NEWTON’S 2ND LAW Notice the following comparisons between = I and F = ma: (torque) ~ F┴r (force x radius) I (moment of inertia) ~ m (mass) (angular acceleration) ~ a (acceleration) Example 1. A pulley that rotates about a fixed axis O has string wrapped around it and attached to a mass m that hangs vertically as shown. T is the tension in the string. Example 1. Diagram 1a. According to the diagram, what is the net torque on the pulley about O? A. (1) = F┴r (2) = TR 1b. Choose the rotational equation that applies: (a) TR = I (b) T = I (c) T/R = I (d) TR = I CHEMISTRY PAGE 1 MR. SURRETTE VAN NUYS HIGH SCHOOL A. (1) = F┴r (2) = I (3) F┴r = I (4) TR = I(from 1a. ) 1c. According to the diagram, what is the net downward force (Fy) on mass m? A. (1) FNET = ma (2) T – mg = - ma (3) -1 (T – mg) = -1 (- ma) (4) mg – T = ma Fy: mg – T 1d. Express Fy in terms of torque. A. (1) Fy: mg – T (from part 1c.) (2) = TR (from part 1a.) (3) T=/R (4) Fy: mg – ( / R) MOMENT OF INERTIA The moment of inertia of a linear body depends on the distribution of mass relative to the axis of rotation: I = mr2) Example 2. A majorette takes two batons and fastens them together in the middle at right angles to make an “x” shape. Each baton is 0.8 m long and each ball on the end is 0.30 kg. (Ignore the mass of the rods.) What is the moment of inertia if the arrangement is spun around an axis formed by one of the batons? 2A. (1) I = mr2) Since there are two batons: (2) I = 2 [(0.30 kg)(0.4 m)2] (3) I = 0.096 kg.m2 CHEMISTRY PAGE 2 MR. SURRETTE VAN NUYS HIGH SCHOOL MOMENT OF INERTIA The moments of inertia for non-linear bodies are determined by experimental data. Some common moments of inertia are: CYLINDER: I = ½ mr2 HOOP: I = mr2 SPHERE: I = 2/5 mr2 ANGULAR MOMENTUM Angular momentum develops from the concept of linear momentum: L = I L = mvr TYPES OF MOMENTUM Notice the following comparisons between L = I and p = mv: I (moment of inertia) ~ m (mass) (angular velocity) ~ v (velocity) Example 3. Two particles move in opposite directions about their center of mass, a point halfway between them. The CM (center of mass) is 4 m from each particle. Example 3. Diagram 3a. Compute the angular momentum of one 2 kg particle about the axis through the CM in kg.m2/s units. Take counterclockwise as positive. A. (1) L = I (2) Find the value of I: I = mr2 (3) I = (2 kg)(4 m)2 (4) I = 32 kg.m2 (5) Find the value of : v = R (6) = v / R CHEMISTRY PAGE 3 MR. SURRETTE VAN NUYS HIGH SCHOOL 3a. (continued…) (7) = (- 15 m/s) / (4 m) (8) = - 3.75 rad /s (9) L = (32 kg.m2)(- 3.75 rad /s) (10) L = - 120 kg.m2/s 3b. Compute the angular momentum of the other 2 kg particle about the axis through the CM in kg.m2/s units. A. L = - 120 kg.m2/s (same steps as 3a.) 3c. Compute the total angular momentum of both particles about the axis through the CM in kg.m2/s units. A. (1) LT = LM1 + LM2 (2) LT = - 120 kg.m2/s – 120 kg.m2/s (3) LT = - 240 kg.m2/s Example 4. A girl of mass 60 kg runs to the left at 5 m/s and jumps on the rim of a merry-go-round that is rotating clockwise at 4 rad/s. The radius R is 3 m and the moment of inertia about the center is 400 kg.m2. Example 4. Diagram 4a. Let counterclockwise momentum be positive. What is the girl’s angular momentum about O in units of kg.m2/s? A. (1) L = mvR (2) L = (60 kg)(5 m/s) (3 m) (3) L = - 900 kg.m2/s (the girl adds clockwise momentum) 4b. What is the merry-go-round’s angular momentum about O before the girl jumps on it (in units of kg.m2/s)? A. (1) L = I (2) L = (400 kg.m2)(- 4 rad/s) (3) L = - 1600 kg.m2/s CHEMISTRY PAGE 4 MR. SURRETTE VAN NUYS HIGH SCHOOL 4c. What is the total angular momentum of the system of girl plus merry-go-round in units of kg.m2/s? A. (1) LT = LM + LG (2) L = - 1600 kg.m2/s – 900 kg.m2/s (3) L = - 2500 kg.m2/s 4d. Compute the moment of inertia of the whole system with the girl on the rim in kg.m2 units. A. (1) IT = IM + IG (2) IG = mR2 (3) IG = (60 kg)(3 m)2 (4) IG = 540 kg.m2 (5) IT = (400 kg.m2) + (540 kg.m2) (6) IT = 940 kg.m2 CHEMISTRY PAGE 5 MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 7: ROTATION (PART 2) ROTATIONAL DYNAMICS CLASS NOTES CONSERVATION OF ANGULAR MOMENTUM When the net external torque acting on a system is zero, the angular momentum of the system is conserved: LI = LF III = IFF CONSERVATION OF ANGULAR MOMENTUM The conservation of angular momentum can also be expressed in terms of one object: I11 = I1’1’ Example 1. A phonograph turntable has a moment of inertia of 2.50 x 10-2 kg.m2 and spins freely on a frictionless bearing at 33.3 rev/min. A 0.25 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.20 m from the center. What is the new rate of rotation (rev/min) of the 0.20 m system? 1A. (1) III = IFF (2) I = (33.3 rev / min)(1 min / 60 sec)(2 rad / 1 rev) (3) I = 3.49 rad/s (4) IF = ITT + IP (5) IF = ITT + mr2 (6) F = (II)(I) / (IF) (7) F = [(2.50 x 10-2 kg.m2)(3.49 rad/s)] / [(2.50 x 10-2 kg.m2) + (0.25 kg)(0.20 m)2] (8) f = 2.49 rad/s (9) f = (2.49 rad/s) (60 s / 1 minute)(1 rev / 2) (10) f = 23.8 rev/min ROTATIONAL KINETIC ENERGY The concept of energy can be used to analyze conceptual motion: KR = 1/2 I2 [Compare to translational (linear) kinetic energy: K = ½ mv2] Example 2. A solid cylinder with mass, M, and radius, R, rolls along a level surface without slipping with a linear velocity, v. What is the ratio of its rotational kinetic energy to its linear kinetic energy? (For a solid cylinder, I = ½ mr2.) 2A. (1) KR = ½ I2 (2) KR = ½ (1/2 MR2)(v/R)2 (3) KR = 1/4 Mv2 (rotational kinetic) (4) K = ½ Mv2 (translational kinetic) (5) KR ~ K = 1/4 ~ 1/2 (6) 0.25 ~ 0.5 (7) KR ~ K = 1 : 2 CHEMISTRY PAGE 6 MR. SURRETTE VAN NUYS HIGH SCHOOL ROTATIONAL POWER The power concept is also useful in simplifying the analysis of rotational motion: P = KR / t Example 3. A bus is designed to draw its power from a rotating flywheel that is brought up to its maximum speed (3000 rpm) by an electric motor. The flywheel is a solid cylinder of mass 1000 kg and diameter 1 m (ICYLINDER = ½ mr2). If the bus requires an average power of 10 kilowatts, how long will the flywheel rotate? 3A. (1) KR = ½ I2 (2) KR = ½ (1/2 mr2)2 (3) KR = ¼ mr22 (4) Solve for : (3000 rev/min)(2 rad / rev)(1 min / 60 s) (5) = 314 rad / s (6) KR = ¼ (1000 kg)(0.5 m)2(314 rad/s)2 (7) KR = 6.162 x 106 J (8) P = KR / t (9) tP = KR (10) t = KR / P (11) t = 6.162 x 106 J / 1.0 x 104 watts (12) t = 616 s CONSERVATION OF MECHANICAL ENERGY The conservation of mechanical energy can be applied to rotational systems: E = K + KR STATIC EQUILIBRIUM (re-visited) In the Dynamics chapter we learned that non-moving objects are in static equilibrium. In other words, the sum of the forces acting on a stationary object must equal 0: F = 0. STATIC EQUILIBRIUM (re-visited) It turns out that this definition is incomplete. To ensure that objects are in static equilibrium, they must not rotate either. The sum of the torques acting on a stationary object must also equal 0: = 0. In other words, F x r (clockwise) must equal F x r (counter-clockwise): CW = CCW. FREE-BODY DIAGRAMS Free-body diagrams can be used to analyze both conditions of static equilibrium. Examples 4 - 5. A uniform 10 m long plank weighing 150 N is supported by two blocks, A and B, as shown: CHEMISTRY PAGE 7 MR. SURRETTE VAN NUYS HIGH SCHOOL 4. What is the upward force supplied by block A? 4A. Solve in two steps. Step 1: (1) F = 0 (2) FUP = FDOWN (3) FBLOCK A + FBLOCK B = 150 N Step 2: (4) = 0 (5) CW = CCW Arbitrarily choose block B as a pivot: (6) (FBLOCK A)(8 m) = (FBOARD)(5 m) (7) (FBLOCK A)(8 m) = (150 N)(5 m) (8) FBLOCK A = (150 N)(5 m) / (8 m) (9) FBLOCK A = 94 N 5. A 10 N can of paint is placed on the plank directly over block B. What is the upward force supplied by block B? 5A. Before the paint can was added, the upward force supplied by block B was previously FBLOCK B = 150 N – 94 N = 56 N. Adding the can over block B supplies no additional torque if we still use B as our pivot. Therefore, now FBLOCK B = 56 N + 10 N = 66 N. CHEMISTRY PAGE 8 MR. SURRETTE VAN NUYS HIGH SCHOOL CHAPTER 7: ROTATION (PART 2) GRAVITY CLASS NOTES GRAVITY There are three major features to Newton’s Universal Law of Gravitation. 1. Gravity is an action-at-a-distance force that always exists between two particles regardless of the medium that separates them. 2. The force varies as the inverse square of the distance between the particles. 3. The force is proportional to the product of their masses. UNIVERSAL LAW OF GRAVITATION FG = - (Gm1m2) /r2 (G = constant = 6.67 x 10-11 m3/kg.s2) (The negative sign indicates that gravity is always an attractive force) Example 1. Two lead balls of mass 5 kg and 10 kg are 1.41 m apart. Calculate the magnitude of the gravitational force they exert on one another. 1A. (1) FG = Gm1m2/r2 (2) (6.67 x 10-11 m3/kg.s2)(5 kg)(10 kg) / (1.41 m)2 (3) FG = 1.67 x 10-9 N GRAVITY AS CENTRIPETAL FORCE Example 2. An artificial satellite of mass m travels at a constant speed in a circular orbit of radius r around the Earth mass M. What is the speed of the satellite? 2A. (1) FG = Gm1m2/r2 (2) FG = GmM/r2 (3) Fc = mv2/r (4) Set Fc to FG: Gm1m2/r2 = mv2/r (5) (r)GmM/r2 = (r)mv2/r (6) GmM/r = mv2 (7) GM/r = v2 (8) v2 = GM/r (9) v = (GM/r)1/2 GRAVITATIONAL POTENTIAL ENERGY (re-visited) Earlier we learned that gravitational potential energy Ug can be calculated: Ug = mgh. This is true for objects near or at the surface of the earth. This equation does not hold true for satellites high above the surface of the earth. CHEMISTRY PAGE 9 MR. SURRETTE VAN NUYS HIGH SCHOOL GRAVITATIONAL POTENTIAL ENERGY Satellites that orbit the earth have both kinetic and gravitational potential energy. Their kinetic energy K is calculated: K = ½ mv2. Their gravitational potential energy is calculated using a variation of Newton’s law of gravitation. GRAVITATIONAL POTENTIAL ENERGY UG = - (Gm1m2) / r (r = height above the earth) Example 3. A satellite of mass m is in a circular orbit of radius R around the earth (mass M). What is its total mechanical energy? 3A. (1) FC = mv2 / r (2) FG = - GmM / r2 (3) FC = FG (4) mv2 / r = - GmM /r2 (5) r [ mv2 / r = - GmM / r2] (6) mv2 = - GmM / r (7) K = ½ mv2 (8) ½ [mv2 = - GmM / r] (9) ½ mv2 = - GmM / 2r (10) K = | - GmM / r | (kinetic energy is scalar and positive) (11) K = GmM / 2r (12) E = K + U (13) E = GmM /2r – GmM / r (14) E = - GmM / 2r CHEMISTRY PAGE 10