General Physics I (aka PHYS 2013)

Transcription

1 General Physics I (aka PHYS 2013) PROF. VANCHURIN (AKA VITALY) University of Minnesota, Duluth (aka UMD)

2 OUTLINE

3 CHAPTER 9: ROTATION OF RIGID BODIES Section 9.1: Angular velocity and acceleration Section 9.2: Rotation with constant angular acceleration Section 9.3: Relating linear and angular kinematics Section 9.4: Energy in rotational motion Section 9.5: Parallel-Axis Theorem Section 9.6: Moment-of-Inertia calculations

4 SECTION 9.1: ANGULAR VELOCITY / ACCELERATION Angular position is describe in terms of radians defined by or 2π rad 360 = 1 rev (1) 1 rad 360 2π = 1 rev. (2) 2π One can convert from one angular unit to another using, 1 = 1 = 1 = ( ) 2π rad 360 ( ) 1rev 2π rad ( ) 360 1rev (3)

5 SECTION 9.1: ANGULAR VELOCITY / ACCELERATION Average linear (velocity and acceleration) v avg x t a avg v t (4) Instantaneous linear (velocity and acceleration) x v lim t 0 t = dx dt v a lim t 0 t = dv dt = d2 x dt 2 (5) Average angular (velocity and acceleration) ω avg θ t α avg ω t (6) Instantaneous angular (velocity and acceleration) θ ω lim t 0 t = dθ dt α lim t 0 ω t = dω dt = d2 θ dt 2. (7)

6 SECTION 9.1: ANGULAR VELOCITY / ACCELERATION Example The angular position θ of a 0.36 m-diameter flywheel is given by ( θ(t) = 2.0 rad/s 3) t 3. (8) a) Find θ, in radians and in degrees, at t 1 = 2.0 s and t 2 = 5.0 s. b) Find the total distance (not displacement) that a particle on the flywheel rim moves over the time interval from t 1 = 2.0 s to t 2 = 5.0 s. c) Find the average angular velocity, in rad/s and in rev/min over that interval. d) Find the instantaneous angular velocity at t 1 = 2.0 s and t 2 = 5.0 s. e) Find the average angular acceleration between t 1 = 2.0 s and t 2 = 5.0 s. f) Find the instantaneous angular acceleration at t 1 = 2.0 s and t 2 = 5.0 s.

7 SECTION 9.2: CONSTANT ANGULAR ACCELERATION Angular velocity and acceleration are vectors pointing along the axis of rotation and direction determined by right-hand rule. There is an equivalence of linear and angular quantities x θ v x ω z a x α z. and so for motion with constant acceleration we have a(t) = a α(t) = α z v(t) = v 0 + a x t ω(t) = ω 0z + α z t x = x 0 + v 0x t a xt 2 θ(t) = θ 0 + ω 0z t α zt 2 as well useful relations v x (t) 2 = v 2 0x + 2a x (x x 0 ) ω z (t) 2 = ω 2 0z + 2α z (θ(t) θ 0 ) x(t) x 0 = 1 2 (v 0x + v x (t)) t θ(t) θ 0 = 1 2 (ω 0z + ω z (t)) t.

8 SECTION 9.2: CONSTANT ANGULAR ACCELERATION Example 9.3. You have finished watching a movie Blu-ray and the disk is slowing to a stop. The disc s angular velocity at t = 0 is 27.5 rad/s, and its angular acceleration is a constant 10.0 rad/s 2. A line PQ on the disc s surface lies along the +x-axis at t = 0 s. a) What is the disc s angular velocity at t = s? b) What angle does the line PQ make with the +x-axis at this time?

9 SECTION 9.2: CONSTANT ANGULAR ACCELERATION Example. Supposed the disc in Example 9.3 was initially spinning at twice the rate (i.e rad/s) and slowed down at twice the rate (i.e rad/s 2 ). a) Compared to the situation in Example 9.3 how long would it take the disc to come to a stop? (i) the same amount of time; (ii) twice as much time; (iii) four times as much time; (iv) one half as much time; (v) one fourth as much time b) Compared to the situation in Example 9.3 through how many revolutions would the disc rotate before coming to a stop? (i) the same number of revolutions; (ii) twice as many revolutions; (iii) four times as many revolutions; (iv) one half as many revolutions; (v) one fourth as many revolutions.

10 SECTION 9.3: RELATING LINEAR AND ANGULAR KINEMATICS The linear displacement of a given point P on rotating body is s = rθ, (9) where θ must be measured in radians.

11 SECTION 9.3: RELATING LINEAR AND ANGULAR KINEMATICS This suggests that the (magnitude of) linear velocity of the point P is v = ds dt = d(rθ) dt = r dθ dt (10) or v = rω. (11) and the tangential (or parallel) component of linear acceleration a = dv dt = rdω = rα. (12) dt and the centripetal (or perpendicular) component a = v2 r = ω2 r. (13)

12 SECTION 9.3: RELATING LINEAR AND ANGULAR KINEMATICS Example 9.4. An athlete whirls a discus in a circle of radius 80.0 cm. At a certain instant, the athlete is rotating at 10.0 rad/s and the angular speed is increasing at 50.0 rad/s 2. At this instant, find the tangential and centripetal components of the acceleration of the discus and the magnitude of the acceleration.

13 SECTION 9.3: RELATING LINEAR AND ANGULAR KINEMATICS Example 9.5. You are designing an airplane propeller that is to turn at 2400 rpm (or revolutions per minute). The forward airspeed of the plane is to be 75.0 m/s, and the speed of the tips of the propeller blades through the air must not exceed 270 m/s. (This is about 80% of the speed of the sound in air. If the speed of the propeller tips were greater than this, they would produce a lot of noise.) (a) What is the maximum possible propeller radius? (b) With this radius, what is the acceleration of the propeller tip?

14 SECTION 9.3: RELATING LINEAR AND ANGULAR KINEMATICS Example. Information is stored on a disc in a coded pattern of tiny pits. The pits are arranged in a track that spirals outwards towards the rim of the disc. As the disc spins inside a player, the track is scanned at a constant linear speed. How must the rotation speed of the disc change as the player s scanning head moves over the track? (i) The rotation speed must increase. (ii) The rotation speed must decrease. (iii) The rotation speed must stay the same.

15 SECTION 9.4: ENERGY IN ROTATIONAL MOTION When a rigid body rotates around a fixed axis different parts of it move with different linear velocities and so have different kinetic energies: K i = 1 2 m iv 2 i = 1 2 m ir 2 i ω 2. (14) The total (rotational) kinetic energy is then where K = i K i = 1 2 m ir 2 i ω 2 = 1 2 is known as the moment of inertia. ( mi r 2 i ) ω 2 = 1 2 Iω2 (15) I m i r 2 i (16) Note that the rotational kinetic energy looks very similar to the translational kinetic energy 1 2 Iω2 1 2 mv2. (17)

16 SECTION 9.4: ENERGY IN ROTATIONAL MOTION Example 9.6. A machine part consists of three discs linked by light-weight struts. (a) What is this body s moment of inertia about an axis through the center of the disc A, perpendicular to the plane of the diagram? (b) What is its moment of inertia about an axis through the centers of disks B and C? (c) What is the body s kinetic energy if it rotates about the axis through A with angular speed ω = 4.0 rad/s?

17 SECTION 9.4: ENERGY IN ROTATIONAL MOTION For more complicated objects the moment of inertia must be obtained (as we shall see) by volume integration. Here we summarize some useful results:

18 SECTION 9.4: ENERGY IN ROTATIONAL MOTION Example 9.7. We wrap a light, nonstretching cable around a solid cylinder of mass 50 kg and diameter m, which rotates in frictionless bearings about a stationary horizontal axis. We pull the free end of the cable with a constant 9.0 N force for a distance of 2.0 m; it turns the cylinder as it unwinds without slipping. The cylinder is initially at rest. Find its final angular speed and the final speed of the cable.

19 SECTION 9.4: ENERGY IN ROTATIONAL MOTION Example 9.8. We wrap a light, non-stretching cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free and of the cable to a block of mass m and release the block from rest at distance h above the floor. As the block falls, the cable unwinds without stretching or slipping. Find expression for the speed of the cylinder as the block strikes the floor.

20 SECTION 9.5: PARALLEL-AXIS THEOREM There are (infinitely) many axis passing through a given object, and thus there are infinitely many moments of inertia. However, some of them are related to each other. Consider two axis of rotation: axis O: passing through the center of mass and axis D: parallel to the first one, but displaced by distance d. Let s choose z-axis to point along the direction of the first axis and with origin at the center of mass. Then the O axis of rotation has coordinates and the D axis of rotation has coordinates (x, y, z) = (0, 0, z) (18) (x, y, z) = (a, b, z) (19) where d = a 2 + b 2. (20)

21 SECTION 9.5: PARALLEL-AXIS THEOREM There are (infinitely) many axis passing through a given object, and thus there are infinitely many moments of inertia. However, some of them are related to each other. Consider two axis of rotation: axis O: passing through the center of mass and axis D: parallel to the first one, but displaced by distance d. Let s choose z-axis to point along the direction of the first axis and with origin at the center of mass. Then the O axis of rotation has coordinates and the D axis of rotation has coordinates (x, y, z) = (0, 0, z) (21) (x, y, z) = (a, b, z) (22) where d = a 2 + b 2. (23)

22 SECTION 9.5: PARALLEL-AXIS THEOREM Then the two moments of inertia are given by I O = ( m i x 2 i + y 2 ) i i (24) and I D = i m i ((x i a) 2 + (y i b) 2). (25) The second equation can be expanded as I D = ( m i x 2 i + y 2 ) i 2a m i x i 2a m i x i + ( a 2 + b 2) i i i i where there third and forth terms vanish. If we denote by total mass M = i m i then the final expression is I D = I O + Md 2. (26) also known as the parallel-axis theorem. m i

23 SECTION 9.5: PARALLEL-AXIS THEOREM Example 9.9. A part of mechanical linkage has a mass of m = 3.6 kg. Its moment of inertia I P about an axis l = 0.15 m from its center of mass is I P = kg m 2. What is the moment of inertia I cm about a parallel axis through the center of mass?

24 SECTION 9.6: MOMENT-OF-INERTIA CALCULATIONS In general moment-of-inertia is defined by replacing summation with integration, i.e. I = r 2 m i I = r 2 dm. (27) i In the case when the density (or mass per unit volume) ρ = dm dv (28) is independent of position the moment of inertia I = r 2 ρdv (29) or I = r 2 ρ dx dy dz (30) where the limits of integration are set by the geometry of the body.

25 SECTION 9.6: MOMENT-OF-INERTIA CALCULATIONS Example Consider a hollow cylinder of uniform mass density ρ with length L, inner radius R 1, and outer radius R 2. Find its moment of inertia about its axis of symmetry.

26 SECTION 9.6: MOMENT-OF-INERTIA CALCULATIONS Example Find the moment of inertia of a solid sphere of uniform mass density ρ (like a billiard ball) about an axis through its center.

27 CHAPTER 10: DYNAMICS OF ROTATIONAL MOTION Section 10.1: Torque Section 10.2: Torque and Angular Acceleration for Rigid Body Section 10.3: Rigid-body Rotation About a Moving Axis Section 10.4: Work and Power in Rotational Motion Section 10.5: Angular Momentum Section 10.6: Conservation of Angular Momentum

28 SECTION 10.1: TORQUE When force acts on an object it can change its translational as well as rotational motion. The effect on the rotational motion depends not only on the magnitude of the applied force, but also to which point the force is applied. For example:

29 SECTION 10.1: TORQUE The relevant physical quantity which measures the twisting effort of some force F applied to point A with respect to some other point B is called torque. If we place origin of coordinates to point B then the torque can be defined as τ r F. (31) where F force applied to object at point B r position of B measured with origin at A. It is assume that both vectors F and r lie in the plane orthogonal to the axis of rotation, then torque would point along the axis of rotation determined by right-hand rule. It is a convention to use right hand coordinate system and so the torque is positive if the force causes counterclockwise rotation and negative if the force cause clockwise rotation.

30 SECTION 10.1: TORQUE Moreover the magnitude of torque is τ = rf sin θ. (32) First Interpretation: If we define lever-arm as distance between line of action of force, i.e. l r sin θ (33) where θ is the angle between vectors F and r then the torque is product of force and lever-arm τ = Fl. (34) Secondly Interpretation: If we define tangential component of the force as F F sin θ (35) then torque is product of tangential force to radius of rotation τ = F r. (36)

31 SECTION 10.1: TORQUE Example To loosen a pipe fitting, a weekend plumber slips a piece of scrap pipe (a cheater ) over his wrench handle. He stands on the end of the cheater, applying his full 900 N weight at a point 0.80 m from the center of the fitting. the wrench and cheater make an angle of 19 with horizontal. Find the magnitude and direction of the torque he applies about the center of the fitting.

32 SECTION 10.2: TORQUE FOR RIGID BODY Consider a single particle in a rigid body whose displacement from axis of rotation is r 1, mass is m 1 and force acting on it F 1. Since the only component of the force which contributes to rotation is parallel, the second law for this force gives But since the parallel acceleration is and torque is we get F 1, = m 1 a 1,. (37) a 1, = r 1 α z (38) τ 1,z = F 1, r 1 (39) τ 1,z = m 1 r 2 1α z. (40) The same applies to every point-like particle in a rigid body, i.e. τ i,z = m i r 2 i α z = Iα z. (41) i i

33 SECTION 10.2: TORQUE FOR RIGID BODY Example (setup as in Example 9.7) We wrap a light, nonstretching cable around a solid cylinder of mass 50 kg and diameter m, which rotates in frictionless bearings about a stationary horizontal axis. We pull the free end of the cable with a constant 9.0 N force for a distance of 2.0 m; it turns the cylinder as it unwinds without slipping. The cylinder is initially at rest. What is the cable s acceleration?

34 SECTION 10.2: TORQUE FOR RIGID BODY Example (setup as in Example 9.8) We wrap a light, non-stretching cable around a solid cylinder with mass M and radius R. The cylinder rotates with negligible friction about a stationary horizontal axis. We tie the free and of the cable to a block of mass m and release the block from rest at distance h above the floor. As the block falls, the cable unwinds without stretching or slipping. What are the acceleration of falling block and the tension in the cable?

35 SECTION 10.3: ROTATION ABOUT A MOVING AXIS More generally a given rigid body can have both rotational motion (about some axis passing through center of mass) and translation motion (of the center of mass). In this case the total kinetic energy is a sum of rotational and translational kinetic energies, i.e. K = 1 2 Mv2 cm I cmω 2. (42) This can be shown by imagining the rigid body made up of many particles with masses m i moving with velocities v i so that the total kinetic energy is K = 1 m i ( v i v i ). (43) 2 i If we now define a relative velocity of each particle with respect to center of mass as v i = v i v cm (44) then (42) follows.

36 SECTION 10.3: ROTATION ABOUT A MOVING AXIS An important example is rolling without slipping. Then the center of mass velocity is related to angular velocity v cm = Rω (45) Then the motion is purely rotational about point 1, or a rotational around point 0 and translational motion of 0.

37 SECTION 10.3: ROTATION ABOUT A MOVING AXIS To see that two views give the same kinetic energy we can use the parallel-axis theorem, i.e. K = 1 2 I 1ω 2 = 1 2 ( Icm + MR 2) ω 2 = 1 2 I cmω MR2 ω 2 = 1 2 I cmω Mv2 cm. (46) If in addition the potential energy changes (as in the case of gravitational potential energy), then the total mechanical energy K = 1 2 Mv2 cm I cmω 2 + Mgy cm (47) and thus one might think of potential energy coming from a point particle with mass M located at the body s center of mass.

38 SECTION 10.3: TORQUE FOR RIGID BODY Example You make a primitive yo-yo by wrapping a massless string around a solid cylinder with mass M and radius R. You hold the free end of the string stationary and release the cylinder from rest. The string unwinds but does not slip or stretches the cylinder descends and rotates. Using energy considerations, find the speed of the center of mass of the cylinder after it has descended a distance h.

39 SECTION 10.3: ROTATION ABOUT A MOVING AXIS In a special case when the axis of rotation is the axis of symmetry and does not change direction, the combined rotational and translational dynamics can be described in context of second law, i.e. FI = M a cm I τ zi = I cm α z (48) I where the sum in both equations goes only external quantities (i.e. forces and torques).

40 SECTION 10.3: TORQUE FOR RIGID BODY Example You make a primitive yo-yo by wrapping a massless string around a solid cylinder with mass M and radius R. You hold the free end of the string stationary and release the cylinder from rest. The string unwinds but does not slip or stretches the cylinder descends and rotates. Find downward acceleration of the cylinder and the tension in the string.

41 SECTION 10.4: WORK AND POWER IN ROTATIONAL MOTION Similarly to how translational work is given by an integral W = xf the rotational work is given by W = F ds = F Rdθ = x i F x (x)dx (49) θf θ i τ z (θ)dθ (50) In the case of force independent of angle we have W = τ z (θ f θ i ) = τ z θ. (51)

42 SECTION 10.4: WORK AND POWER IN ROTATIONAL MOTION Moreover, the power associated with rotational work is P = dw dt = τ z dθ dt = τ zω z. (52) And the integral of (50) can also be rewritten by applying the (rotational) second law, W = 1 2 Iω2 f 1 2 Iω2 i (53) i.e expressing work as a change of rotational kinetic energy.

43 SECTION 10.4: WORK AND POWER IN ROTATIONAL MOTION Example An electric motor exerts a constant τ = 10 N m torque on grindstone which has moment of inertia of I = 2.0 kg m 2 about its shaft. The system starts from rest. Find the work W done by the motor in t = 8.0 s and the grindstone kinetic energy K at this time. What average power P av is delivered by the motor?

44 SECTION 10.5: ANGULAR MOMENTUM A rotation analog of linear momentum is angular momentum L r p = r (m v). (54) Then the rate of change change of angular momentum d L = d r ( (m v) + r m d v ) dt dt dt = r (m a) = r F = τ (55) is nothing but torque of the net force acting on a particle.

45 SECTION 10.5: ANGULAR MOMENTUM For a rigid body the same is true for every particle and if we choose the axis of rotation to point along z axis, then L z = i L zi = i r i (m i v i ) = i r i (m i r i ω z ) = ω z. (56) If the axis of rotation is also a symmetry axis, then L = I ω. (57) If we apply (55) to every particle in a rigid body, then I τ I = d L dt where the sum on the left-hand-side is over all external torques. Moreover if the rigid body rotates around a symmetry axis then τz = Iα z.

46 SECTION 10.5: ANGULAR MOMENTUM Example A turbine fan in a jet engine has a moment of inertia of I = 2.5 kg m 2 about its axis of rotation. As the turbine starts up, its angular velocity is given by ( ω z = 40 rad/s 3) t 2. (a) Find the fan s angular momentum as a function of time and find its value at t = 3.0s. (b) Find the net torque on the fan as a function of time, and find its value at t = 3.0 s.

47 SECTION 10.6: CONSERVATION OF ANGULAR MOMENTUM When the net external torque acting on a system is zero, the total angular momentum of the system is constant. L f = L i This does not mean that the change of angular momentum of part of a given body must not changes.

48 SECTION 10.5: ANGULAR MOMENTUM Example A physics professor stands at the center of a frictionless turntable with arms outstretches and a m = 5.0 kg dumbbell in each hand. He is set rotating about the vertical axis, making one revolution in T = 2.0 s. Find his final angular velocity is he pulls the dumbbells in to his stomach. His moment of inertia (without the dumbbells) is I i = 3.0 kg m 2 with arms outstretched and I f = 2.2 kg m 2 with his hands at his stomach. The dumbbells are r i = 1.0 m from the axis initially and r f = 0.20 m at the end.

49 CHAPTER 11: EQUILIBRIUM AND ELASTICITY Section 11.1: Conditions for Equilibrium Section 11.2: Center of Gravity Section 11.3: Solving Rigid-Body Equilibrium Problems Section 11.4: Stress, Strain and Elastic Moduli Section 11.5: Elasticity and Plasticity

50 SECTION 11.1: CONDITIONS FOR EQUILIBRIUM An object is said to be in equilibrium if two conditions are satisfied: First condition for equilibrium. The sum of all external forces acting on the body is zero: F = 0. (58) Second condition for equilibrium. The sum of all torques (due to external forces about any point) on the body is zero: τ = 0. (59) Although we shall only apply these equilibrium conditions to bodies at rest, the same conditions applies to bodies in uniform translational motion (without rotation).

51 SECTION 11.1: CONDITIONS FOR EQUILIBRIUM

52 SECTION 11.1: CONDITIONS FOR EQUILIBRIUM Example. Which situation satisfies both the first and second conditions for equilibrium? (i) a seagull gliding at a constant angle below the horizontal and at constant speed; (ii) an automobile crankshaft turning at an increasing angular speed in the engine of a parked car; (iii) a thrown baseball that does not rotate as it sails through the air.

53 SECTION 11.2: CENTER OF GRAVITY Consider a collection of particles which make up a given object, then their center of mass is then given by r cm = m 1 r 1 + m 2 r 2 + m 3 r i = m i r i m 1 + m 2 + m M, (60) where M = i m i. (61) Now let us consider gravitational torque caused by gravitational forces with axis of rotation in the origin τ = i = i = i τ i r i (m i g i ) ( ) ri m i (M g i ) (62) M

54 SECTION 11.2: CENTER OF GRAVITY In general the gravitational acceleration changes with attitude. For example, the center of gravity of 452-m-tall Petronas Towers in Malaysia is 2 cm below the center of mass. Most of the time the gravitational acceleration can be assumed to be constant and then the torque is given by ( τ = i m i r i M ) (M g) = r cm (M g). (63) However. it turns out that even in a more general case when the gravitational acceleration decreases with altitude where r cg is the center of gravity. τ = r cg (M g). (64)

55 SECTION 11.1: CENTER OF GRAVITY Finding the center of gravity (theoretically) is easy when the object is symmetric, but what if the object has irregular shape? To find the center of gravity experimentally one can suspend an object from a given point.. By repeating the same experiment for two distinct points a pair of such lines can be constructed and then the point where the lines intersect should be the center of gravity.

56 SECTION 11.2: CENTER OF GRAVITY Example A uniform plank of length L = 6.0 m and mass M = 90 kg rests on sawhorses separated by D = 1.5 m and equidistant from the center of the plank. Cousin Throckmorton wants to stand on the right-hand end of the plank. If the plank is to remain at rest, how massive can Throckmorton be?

57 SECTION 11.2: CENTER OF GRAVITY Example. A rock is attached to the left end of a uniform meter stick that has the same mass as the rock. In order for the combination of rock in meter stick to balance atop the triangular object (see figure below), how far from the left end of the stick should the triangular object be placed?

58 SECTION 11.3: RIGID-BODY EQUILIBRIUM PROBLEMS For problem in two dimensions we can choose a coordinate system so that the vectorial equilibrium conditions (58) and (59) are replaced by three scalar conditions F i,x = 0 (65) i F i,y = 0 (66) i τ i,z = 0. (67) i

59 SECTION 11.3: RIGID-BODY EQUILIBRIUM PROBLEMS Example An auto magazine reports that a certain sports car has 53% of its weight on the front wheel. The distance between the axles is D = 2.46 m. How far in front of the rear axle is the car s center of gravity? (Let s denote is by L cg )

60 SECTION 11.3: RIGID-BODY EQUILIBRIUM PROBLEMS Example Sir Lancelot, who weights 800 N, is assaulting a castle by climbing a uniform ladder that is 5.0 m long and weights 180 N. The bottom of the ladder rests on a ledge and leans across the moat in equilibrium against a frictionless, vertical castle wall. The ladder makes an angle of 53.1 with the horizontal. Lancelot pauses one-third of the way up the ladder. (a) Find normal and friction forces on the base of the ladder. (b) Find the minimum coefficient of static friction needed to prevent slipping at the base. (c) Find the magnitude and direction of the contact force on the base of the ladder.

61 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI So far we considered (mostly) rigid bodies. In reality all objects can deform when external forces are applied to them, e.g. stretching, squeezing, twisting For each type of deformation there are two relevant quantities: stress which characterizes the cause of the deformation strain which characterizes the effect of the deformation Often the two quantities are directly proportional to each other. Coefficient of proportionality is known as elastic modulus, Stress = Elastic modulus (68) Strain We have already seen an example of this (or Hooke s) law F = k. (69) x

62 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Tensile/compressive stress. For stretching/squeezing deformations the tensile/compressive stresses are defined as σ = F A (70) and the tensile/compressive strains are defined as ε = l l 0 l = l l. (71) The units of the stress is [Stress] = [Force] [Distance] 2 and in SI system a unit of stress is measures in Pascals, but the strain is dimensionless. 1 Pa 1 N/m 2.

63 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Young s modulus. In the case of tensile/compressive stress the elastic modulus is known as Young s modulus Y = σ ε = F /A l/l 0 = F l 0 A l. (72) This definition insures that Young s modulus is positive if the perpendicular component of force F is chosen to point outwards: Tensile: F > 0 and l > 0 Y > 0 Compressive: F < 0 and l < 0 Y > 0 (73) (Note that since the strain was dimensionless, the Young s modulus has the same units as stress.) For many materials the tensile and compressive Young s modulus are the same and the Hooke s law works very well.

64 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Here is a table of the modulus for some of the materials However for composite materials (such as concrete or stone) the compressive stress can be quite large, while the tensile stress easily breaks the object and thus the Hooke s law does not hold.

65 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Example A steel 2.0 m long has a cross-sectional area of 0.30 cm 2. It is hung by one end from a support, and a 550 kg milling machine is hung from its other end. Determine the stress on the rod and the resulting strain and elongation.

66 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Bulk stress. There situations when the same tensile/compressive stress is applied to all spatial directions simultaneously. Such an isotropic (i.e. the same in all directions) stress is called (sometimes bulk stress) or pressure p = F A and an isotropic strain is called bulk or volume strain (74) ε bulk = V V 0. (75) where V = V f V i. Units of pressure 1 Pa 1 N/m 2 1 atm 10 5 Pa 1 psi 6900 Pa. (76)

67 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Confusing Point #1: The pressure is usually positive which is in conflict with our choice for perpendicular direction to point outwards of an object. This can be fixed by either stating that p = F A. (77) Confusing Point #2: The bulk stress is defined not as pressure but as a negative change in pressure as an object is moved from one substance (e.g. air) to another (e.g. water), i.e. where p = p f p i σ bulk = p. (78) Confusing Point #3: The corresponding elastic modulus (known as bulk modulus) is defined with a minus sign B = σ bulk ε bulk = p V 0 V. (79) Two out of three confusion (i.e. #1 and #3) could have been avoided if the pressure was defined with a minus sign as in (77).

68 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Example A hydraulic press contains V 0 = 0.25 m 3 (250 L) of oil. Find the decrease in the volume V of the oil when it is subjected to a pressure increase p = Pa (about 160 atm). The bulk modulus of the oil is B = Pa (about atm) and its compressibility is k = 1/B = atm 1.

69 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Shear stress. In addition to tensile/compressive deformations due to perpendicular component of the applied force, a parallel component of the applied force can cause a deformation. For such deformations we defined shear stress as shear strain as and shear modulus as σ shear = F A ε shear = x h (80) (81) S = σ shear = F /A ε shear x/h = F h Ax. (82) If we imagine a shear deformation of rectangular parallelepiped of dimensions x, y, z then S = F xy xz x. (83)

70 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Example Suppose a brass base plate of an outdoor sculpture experiences shear force in an earthquake. The vertically oriented square plate is x = 0.5 cm thick. Whats is the force exerted on each of its edges is the resulting displacement is x = 0.16 mm?

71 SECTION 11.4: STRESS, STAIN, AND ELASTIC MODULI Example A copper rod of cross-sectional area A = cm 2 and length 1.00 m is elongated by mm, and a steel rod of the same cross-sectional area but m in length is elongated by mm. a) Which rod has greater tensile strain? the copper rod the steel rod the stain is the same b) Which rod is under greater tensile stress? the copper rod the steel rod the stain is the same

72 SECTION 11.5: ELASTICITY AND PLASTICITY For larger deformations the linear dependence of stress as functions of stain (68) does not hold. A typical dependence for metals under tension is illustrated in and for rubber under tension is illustrated by

73 CHAPTER 9: ROTATION OF RIGID BODIES Angular quantities Angular position in terms of revs, degrees and radians Average angular velocity and acceleration Instantaneous angular velocity and acceleration Constant angular acceleration Angular velocity and position as functions of time Angular velocity as function of angular position Angular position as function of angular velocity Relating angular and linear quantities Relating linear velocity from angular velocity Relating tangential acceleration and angular acceleration Relating centripetal acceleration and angular velocity Energy in rotational motion Definition of rotational kinetic energy Applications of parallel axis theorem Calculation of moment of inertia

74 CHAPTER 10: DYNAMICS OF ROTATIONAL MOTION Torque Scalar and vectorial definitions of torque Lever-arm interpretation of torque Tangential interpretation of torque Torque for rigid body Rotational Newton s second law Rotational and translational kinetic energy Relation of velocities for motion without slipping Application of second law for rotations about a moving axis Definitions and calculations of rotational work and power Angular momentum Definition of angular momentum Conservation of angular momentum of closed systems Change of angular momentum due to external torques

75 CHAPTER 11: EQUILIBRIUM AND ELASTICITY Equilibrium conditions Translational equilibrium conditions Rotational equilibrium conditions Identification of of equilibrium conditions Center of gravity Definition of center of mass Calculation of torque due to gravitational force Finding center of gravity (theoretical/experimental) 2D equilibrium problems Drawing free-body diagram for an equilibrium object Applying two translational equilibrium conditions Applying a rotational equilibrium condition Stress/Stain/Modulus Tensile,compressive, bulk and shear stresses Tensile,compressive, bulk and shear strains Tensile,compressive, bulk and shear moduli