9.3 The Spinning Top and Gyroscope 287 Substituting with dφ from this relation into the angular velocity of precession = dφ/dt and using τext = dL/dt, we get: = dL τext dφ = = dt L sin θ dt L sin θ (9.17) But |τ→ext | = |→ r × M→ g | = rMg sin(180◦ − θ ) = rMg sin θ, then = becomes: Mgr L (9.18) Using Eq 9.12, we can write L = Iω, where I and ω are the moment of inertia and angular speed of the spinning top about its axis of symmetry Then the top’s precessional angular speed becomes: = Mgr Iω (9.19) This relation is valid only when ω, and this condition is satisfied if ω is large If this condition is not fulfilled, the motion of the top becomes much more complicated Using = 2π/Tp and ω = 2π/Ts , where Tp is the precession period and Ts is the spinning period, we find that the period of precession Tp is given by: Tp = 4π I MgrTs (9.20) In fact, Eqs 9.19 and 9.20 also apply to gyroscopes A gyroscope is a device for measuring or maintaining orientation, based on the concept of conservation of angular momentum The toy gyroscope shown in Fig 9.14 has one end of its axle resting on a support (assumed to be a frictionless pivot), while the other end is free and precessing horizontally with angular speed A symmetric wheel attached to this axle spins rapidly about its axis with a large angular speed ω (like the top of Fig 9.13) Based on our findings for the spinning top, let us analyze the behavior of the toy gyroscope of Fig 9.14 The wheel is rotating about its axis of symmetry with an → angular speed ω and has an initial angular momentum L along the x-axis Since → → τ→ext and dL are along the y-axis and perpendicular to L , this causes the direction → → of L to change, but not its magnitude Therefore, the changes dL are always in → the horizontal xy-plane Consequently, the angular momentum L and the wheel axis 288 Angular Momentum with which the wheel moves are always horizontal This means that the axis of the wheel does not fall, but will precess with the angular speed Fig 9.14 A toy gyroscope is given by Eq 9.19 z a wheel rotating with an angular speed ω about an axis ext y N dL supported at one end while the other is free During time dt, Pivot the torque → τ ext and the change O x r L → in angular momentum d L are → Wheel perpendicular to L , which Mg rotates in the xy-plane with a precessional angular speed Example 9.11 Assume that the cylindrical wheel of the gyroscope of Fig 9.14 has a radius R = cm and a center of mass located cm from the pivot O If the gyroscope takes s for completing one revolution of precession, what is the spinning angular speed of the wheel and its period? Solution: The precessional angular speed about the z-axis is: = 2π rad rev = = 0.4 π rad/s = 1.257 rad/s 5s 5s The moment of inertia of a cylindrical wheel about its axis of symmetry is I = 21 MR2 and its weight is Mg From Eq 9.19: ω= = Mgr Mgr = I ( MR2 ) = 2gr R2 2(9.8 m/s2 )(0.03 m) = 292.4 rad/s = 46.5 rev/s (0.04 m)2 (1.257 rad/s) Thus, the spinning period is: Ts = 2π rad 2π = = 2.15 × 10−2 s ω 292.4 rad/s 9.4 Exercises 9.4 289 Exercises Section 9.1 Angular Momentum of Rotating Systems (1) Calculate the angular momentum of a particle of mass m = kg that has a → → → → velocity v→ = (2 i + j ) m/s when its position vector is → r = (3 i − j ) m (2) Two cars, each having a mass m = 1,500 kg, are moving in a horizontal circle of radius r = 10 m with the same speed v = 10 m/s The circle is centered at the origin O in the xy-plane, and the positive z-axis is directed upwards If one of them is moving clockwise and the other counterclockwise, see Fig 9.15, find the angular momentum of each car about O Fig 9.15 See Exercise (2) y m r x O m (3) A particle of mass m = kg has a position vector that depends on time t and → → is given by → r = (3t i − 4t j ) m Find the angular momentum of the particle as a function of time (4) A particle of mass m is moving horizontally with constant velocity v→ as shown in Fig 9.16 Find the magnitude and direction of the angular momentum of → the particle, Li , (i = 1, 2, , 8), respectively about the eight points Oi , (i = 1, 2, , 8) Fig 9.16 See Exercise (4) O6 O7 d O5 m O8 O4 d O1 d O2 d O3 290 Angular Momentum (5) A ball of mass m = 0.5 kg is moving horizontally with a speed v = 10 m/s at the instant when its position is identified in Fig 9.17 (a) What is the angular momentum of the ball about O at this instant? (b) Neglecting air resistance, find the rate of change of its angular momentum about O at this instant Fig 9.17 See Exercise (5) m r mg 6m 30 O (6) By definition, kinetic energy K = 21 mv , where m and v are the mass and speed of a particle, respectively Show that the kinetic energy of a particle moving in a circular path is K = L /2I, where L and I are, respectively, the angular momentum and moment of inertia of the particle about the center of the circle (7) A canonical pendulum consists of a bob of mass m attached to the end of a cord of length The bob whirls around in a horizontal circle of radius r at a constant speed v while the cord always makes an angle θ with the vertical, see Fig 9.18 Show that the magnitude of the angular momentum of the bob about its point of support O is given by: L= m2 g sin θ tan θ Fig 9.18 See Exercise (7) O m 9.4 Exercises 291 (8) Two identical particles and have respective position vectors → r and → r with respect to an arbitrary origin O The two particles have equal and opposite linear momenta → p and −→ p as shown in Fig 9.19 Show that the total angular momentum of this system is independent of the choice of the origin and independent of where the traveling particles are located Fig 9.19 See Exercise (8) p p r2 r1 O (9) Two particles of masses m1 = kg and m2 = kg are joined by a rod of mass M = 0.5 kg and length d = 0.75 m The assembly rotates freely in the xy-plane about a pivot through the center of the rod, as shown in Fig 9.20 Find the angular momentum of the system when the speed of each particle is v = m/s Fig 9.20 See Exercise (9) y m1 M x O d m2 (10) Consider the seconds hand of a particular analog clock to be a thin rod of length = 14 cm and mass m = g, see Fig 9.21 (a) If the seconds hand rotates constantly, what is its angular speed? (b) Find the magnitude of the angular momentum of the seconds hand about an axis perpendicular to the center of the clock’s face 292 Angular Momentum Ho urs Fig 9.21 See Exercise (10) tes s nd co Se Minu m (11) Three identical particles, each of mass m = 0.5 kg, are attached at equal distances from one end of a rod of length = m and mass M = kg, see Fig 9.22 The system is rotating with angular speed ω = rad/s about an axis perpendicular to the rod through the free end at O (a) What is the moment of inertia of the system about O? (b) What is the angular momentum of the system about O? Fig 9.22 See Exercise (11) m m M m O a a a Pivot (12) Each of two identical particles of mass m = 0.5 kg attached at one end of two identical rods each of length a = 0.2 m and mass M = 0.3 kg The other ends of the two rods are mounted perpendicular to a lightweight axle such that the distance between the rods is d = 0.6 m, see Fig 9.23 The axle rotates at ω = rad/s (a) What is the total angular momentum of the two particles about the CM of the system? (b) What is the total angular momentum of the two rods about the axle? (c) What angle does the total angular momentum of the whole system make with the axle? (13) Three identical thin rods, each of mass m and length R, are fastened together to form the letter H A circular hoop, of mass m and radius R, is fastened to the rods to form the rigid structure shown in Fig 9.24 The rigid structure rotates 9.4 Exercises 293 with a constant angular speed about a vertical axis with a period of rotation T (a) Find an expression for the structure’s moment of inertia and angular momentum about the axis of rotation (b) Evaluate the two expressions of part (a) when m = 0.5 kg, R = 0.1 m, and T = s Fig 9.23 See Exercise (12) Axle m M a y CM d x M m a Fig 9.24 See Exercise (13) Axle z R R R m m m m (14) A block of mass m1 located on a smooth horizontal surface is connected by a light non-stretchable cord that passes over a pulley to a second block of mass m2 , which hangs vertically, see Fig 9.25 The pulley is a uniform cylinder of mass M and radius R, and it rotates freely about its axle (a) Find an expression for the net external torque about the pulley’s axle (b) Find an expression for the net angular momentum about the pulley’s axle (c) Find an expression for the magnitude of the acceleration of the two blocks and its value if m1 = kg, m2 = kg, M = kg, R = 0.1 m, and g = 10 m/s2 (15) A disk has a moment of inertia I = kg.m2 about its axis of symmetry The angular speed of the disk depends on the time t by ω = (12 rad/s3 ) t (a) Find 294 Angular Momentum the angular acceleration α and the angular momentum L of the disk as a function of time, and find their values at t = s (b) Show that using the expressions for α and L leads to the same expression for the net torque on the disk as a function of time, and find its value at t = s Fig 9.25 See Exercise (14) M m1 R m2 (16) A uniform solid sphere of mass M = 10 kg and radius R = 10 cm turns counterclockwise with an angular speed ω = rad/s about a vertical axis that touches its surface, see Fig 9.26 What is the magnitude and direction of its angular momentum about this axis? Fig 9.26 See Exercise (16) z M R R (17) A boy of mass m = 40 kg is standing on the rim of a merry-go-round that is rotating with angular speed ω = 0.5 rev/s about an axis through its center The merry-go-round is a uniform disk of mass M = 120 kg and radius R = 3.5 m Find the total angular momentum of the boy-disk system by treating the boy as a point (18) Two wheels of radii Ra and Rb are connected by a non-stretchable belt that does not slip on their circumferences, see Fig 9.27 The radius Ra is four times 9.4 Exercises 295 the radius Rb Find the ratio of the moment of inertia Ia /Ib and mass Ma /Mb if both wheels have: (a) the same angular momentum about their central axis, and (b) the same rotational kinetic energy Fig 9.27 See Exercise (18) a Ra b Rb (19) If an impulsive force F(t) with moment arm R acts on a rigid body of moment of inertia I for a short time t, then show that the angular speed of the body will change from an initial value ωi to a final value ωf according to the angular impulse formula: JR = τ dt = FR t = I(ωf − ωi ) where F is the average value of the force during the time it acts on the body [Hint: It is the rotational analogy of Eq 7.9] (20) A wheel of radius Ra and moment of inertia Ia is rotating about its central axle with angular speed ωa Another small wheel is stationary and has a radius Rb and moment of inertia Ib about its central axle The smaller wheel is moved until it touches the larger wheel and rotates due to the friction between them, as in the upper part of Fig 9.28 After the initial slipping period is over, the two wheels rotate at constant angular speeds ωa and ωb , see the lower part of Fig 9.28 By applying the angular impulse relationship of Exercise 19, find the final angular speed ωb of the small wheel (21) A block of mass m1 located on a rough horizontal surface is connected by a light non-stretchable cord that passes over a pulley to a second block of mass m2 , which is allowed to move on a rough inclined plane of angle θ, as shown in Fig 9.29 The pulley is a uniform cylinder of mass M and radius R, and rotates freely about its axle The coefficients of kinetic friction for the two blocks on the horizontal and inclined planes are μk1 = 0.35 and μk2 = 0.5, respectively (a) Draw free-body diagrams of the two blocks and the pulley (b) Find the acceleration of the two blocks and the tensions in the two sections 296 Angular Momentum of the cord when m1 = kg, m2 = kg, M = 10 kg, R = 0.1 m, sin θ = 4/5, cos θ = 3/5, and g = 10 m/s2 (c) If the system starts from rest, find the angular momentum of the pulley about its axis as a function of time Fig 9.28 See Exercise (20) Time t F a b Ra Rb Ib Ia F Time t t F a b Ra Ib Ia Fig 9.29 See Exercise (21) F a, m1 Rb M,R a, m2 (22) Determine the angular momentum of the Earth: (a) about its rotational axis (assume that Earth is a uniform sphere of mass M = 6.0 × 1024 kg and radius R = 6.4 × 106 m), and (b) about the Sun (assume Earth to be a particle at 1.5 × 1011 m from the Sun) (23) Two blocks having masses m1 and m2 (m2 > m1 ) are connected to each other by a light non-stretchable cord that passes over two identical pulleys; each pulley is a uniform cylinder with a mass M and radius R, which rotates freely about its axle, as shown in Fig 9.30 Assume no slipping happens between the cord and 302 Angular Momentum his stomach at 0.2 m from the axis of the turntable The moment of inertia of the student with his arms outstretched is kg.m2 , but it is 3.2 kg.m2 with his hands at his stomach (b) Find the initial and final kinetic energy of the system Explain the meaning if they are different Fig 9.38 See Exercise (40) m d/4 CM M d Section 9.3 The Spinning Top and Gyroscope (42) To form a top, a uniform disk of mass M = 50 g and radius R = cm is rigidly attached to an axial rod of negligible mass The top spins on a frictionless surface about its axis of symmetry with angular speed ω = 6,000 rev/min How much work was done to get the top to spin at that rate? (43) The center of the disk in exercise 42 is at r = cm from the tip of the top at the surface of contact to the disk What is the angular speed of precession of the top about the vertical axis? (44) To form a toy gyroscope, a disk of mass M = 150 g and radius R = cm is mounted at the center of a thin axle of 20 cm length The disk spins at ω = 50 rev/s when one end of the axle rests on a stand and the other end precesses horizontally What is the angular speed of precession of the top about the vertical axis? (45) A top of mass M = 200 g spins about its axis of symmetry with angular speed ω = 18 rev/s and makes an angle θ = 25◦ with the vertical It experiences precession at a rate of rev every s The center of mass of the top is cm from its tip (a) What is the moment of inertia of the top? (b) Find the torque on the top 10 Mechanical Properties of Matter The physical states of matter can generally be divided into three broad classes: solids, liquids, and gases, see Fig 10.1 A solid maintains its shape: it resists the action of external forces that tend to change its shape or volume Liquids and gases are fluids A fluid can easily change shape, and flows when subjected to a force The three states of matter are distinguishable at the microscopic level as follows: Solid Maintans its Shape Fixed Volume Liquid Shape of the Container Fixed Volume Gas Shape of the Container Volume of the Container Fig 10.1 The three states of matter: solid, liquid, and gas A solid is a highly ordered array of atoms or molecules that are bound securely by mutual electrical forces A liquid is a crowded assembly of mobile atoms or molecules Each atom or molecule is in contact with several neighbors, but is not bound securely to any of them As an atom or molecule moves about in a liquid, it collides frequently with its neighbors H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_10, © Springer-Verlag Berlin Heidelberg 2013 303 304 10 Mechanical Properties of Matter A gas consists of atoms or molecules that are far apart and consequently move independently, with no forces keeping them together or pushing them apart Collisions of atoms or molecules in gases are infrequent in comparison to those in liquids 10.1 Density and Relative Density The density (or mass density) of a material is defined as the mass per unit volume If a mass m is distributed uniformly over a volume V, the density will be given by the following equation: ρ= m V (10.1) The SI unit of density is kg/m3 If the mass is not uniformly distributed, then Eq 10.1 defines the average density The densities of several materials are listed in Table 10.1 Table 10.1 Density and relative density comparison (approximates) Material type Material name Density (kg/m3 ) Relative density Gas Helium 0.179 1.79 × 10−4 Air 1.29 1.29 × 10−3 Carbon dioxide 1.98 1.98 × 10−3 Alcohol 7.9 × 102 0.79 Gasoline 8.6 × 102 0.86 Water × 103 Mercury 13.6 × 103 13.6 Glass (common) 2.4 − 2.8 × 103 2.5 Aluminum 2.7 × 103 2.7 Iron 7.86 × 103 7.86 Copper 8.92 × 103 8.92 Silver 10.5 × 103 10.5 Lead 11.36 × 103 11.36 Uranium 19.07 × 103 19.07 Gold 19.3 × 103 19.3 Liquid Solid 10.1 Density and Relative Density 305 The relative density of a substance tells us how many times more dense the substance is than pure water, see Table 10.1 Sometimes we refer to it as the specific gravity (SG) Thus: SG = ρ ρwater (10.2) Example 10.1 Calculate the average density of both the Earth and the Sun, given that the mass and radius of the Earth are mE = 5.98 × 1024 kg and RE = 6.37 × 106 m, respectively, and the mass and radius of the Sun are mS = 1.99 × 1030 kg and RS = 6.95 × 108 m, respectively Compare between the resulting average densities Solution: Matter in the Earth and the Sun is not uniform In spite of this fact, we can use Eq 10.1 to calculate their average density Using the given radius, we calculate the volume of the Earth to be: VE = 43 π RE3 = 1.08 × 1021 m3 Thus, the average density of the Earth is: ρE = mE 5.98 × 1024 kg = = 5.54 × 103 kg/m3 VE 1.08 × 1021 m3 In comparison to water, the average density of the Earth is 5.54 times more dense than water Similarly, we use the given radius of the Sun to calculate its volume: VS = 43 π RS3 = 1.41 × 1027 m3 Thus, the average density of the Sun is: ρS = mS 1.99 × 1030 kg = = 1.41 × 103 kg/m3 VS 1.41 × 1027 m3 In comparison to water, the average density of the Sun is 1.41 times more dense than water Although the mass and volume of the Earth are much smaller than the mass and volume of the Sun, the average density of the Earth is nearly four times the average density of the Sun That is: ρE 4ρS 306 10.2 10 Mechanical Properties of Matter Elastic Properties of Solids All solids are to some extent elastic This means that we can change their dimensions slightly by pulling, pushing, twisting, and/or compressing them We shall discuss the elastic properties of solids by introducing the concepts of stress and strain Stress: Stress is the magnitude of the applied external force that acts perpendicularly on a unit area of the object Strain: Strain is a measure of the degree of deformation of the object It is found that for small stresses, stress is proportional to strain The proportionality constant is called the elastic modulus and it depends on the material being deformed, as well as on the nature of the deformation Therefore: Elastic modulus = Stress Strain (10.3) This relation is equivalent to Hooke’s law that states: Stress ∝ Strain In this chapter, we introduce the three most famous types of deformations and their elastic moduli: Young’s Modulus: Measures the resistance of a solid to a change in its length Shear Modulus: Measures the resistance to motion of the planes of solids when sliding over each other Bulk Modulus: Measures the resistance of a solid (or a liquid) to a change in its volume 10.2 Elastic Properties of Solids 307 10.2.1 Young’s Modulus: Elasticity in Length Young’s Modulus measures the resistance of a solid to a change in its length, which indicates its stiffness Consider a metallic long rod of original length L and crosssectional area A When an external force F⊥ is applied perpendicularly to the crosssectional area A of a rod, its internal forces resist its distortion As a final result, the rod attains equilibrium when its length increases to a new length L + L and the magnitude of the perpendicular external force F⊥ exactly balances the internal forces, see Fig 10.2a In light of this, we define the tensile stress and the tensile strain as follows: Tensile stress = Tensile strain = F⊥ A (N/m2 ) (10.4) L L (10.5) The relation between the tensile stress and the tensile strain is linear when the rod is in its elastic range When the stress exceeds what is called the elastic limit, the rod is permanently distorted and will not return to its original shape after the stress is removed As the stress is increased even further, the rod will ultimately break, see Fig 10.2b F Stress Breaking point Elasti climit A L L L Range of permanent deformation Elastic behaviour Strain F (a) (b) Fig 10.2 (a) A rod of height L and cross-sectional area A The rod stretches by an amount application of a tensile stress (b) The stress versus strain curve for an elastic solid L after 308 10 Mechanical Properties of Matter We use Eqs 10.4 and 10.5 to define Young’s modulus, Y, as: Y= Tensile stress F⊥ /A = Tensile strain L/L (N/m2 ) (10.6) This quantity is used to characterize a solid that is stressed under either tension or compression Table 10.2 depicts some values for Y Table 10.2 Young’s modulus for different materials (approximates) Young’s modulus Y × (109 N/m2 ) Material name Rubber 0.004 Lead 16 Glass 65 – 78 Aluminum 70 Brass 91 Copper 110 Steel 200 Tungsten 350 Example 10.2 A pendulum consists of a big sphere of mass m = 30 kg from the end of a steel wire that has a length L = 15 m, a cross-sectional area A = × 10−6 m2 , and Young’s modulus Y = 200 × 109 N/m2 Find the tensile stress on the wire and the increase in its length Solution: The applied force on the wire must equal to the weight of the sphere, i.e F⊥ = m g Thus, the tensile stress will be: Tensile stress = mg (30 kg) × (9.8 m/s2 ) F⊥ = = = 3.27 × 107 N/m2 A A × 10−6 m2 Using the value of the Young’s modulus Y = 20 × 1010 N/m2 and the length of the steel wire before stretching L = 15 m, we get: Y= F⊥ /A L/L ⇒ L= F⊥ /A 3.27 × 107 N/m2 L= × 15 m = 2.45 × 10−3 m Y 200 × 109 N/m2 Note that this large stress produces a relatively small change in L 10.2 Elastic Properties of Solids 309 When we carefully study the deformation of the rod, we find that the rod’s length L increases by L in the direction of the force while it radius r decreases by | r|, where r is negative, in a direction perpendicular to the force, see Fig 10.3 The tensile strain L/L of the rod is called the linear strain The strain − r/r is called the lateral strain, and Poisson’s ratio μ is defined as: μ= r/r L Lateral strain =− =− Linear strain L/L r r L ⇒ μ=− L dr r dL (10.7) The minus sign is inserted in this definition to make μ positive A r r - |Δr| F⊥ F⊥ L 39.58 After Before Fig 10.3 The length of the rod will increase by L + ΔL L and its radius will decrease by r (exaggerated scale) after applying a tensile stress F⊥ /A Example 10.3 A cylindrical steel rod has a length of m and a radius of 0.5 cm A force of magnitude × 104 N is acting normally on each of its ends Find the change in its length and radius, if the Young’s modulus Y is 200 × 109 N/m2 and the Poisson’s ratio μ is 0.25 Solution: Using F⊥ = × 104 N, A = π r = π × (0.5 × 10−2 m)2 = 7.9 × 10−5 m2 , L = m, and Y = 200 × 109 N/m2 in Eq 10.6, we have: L= F⊥ L (2 × 104 N)(2 m) = = 2.5 × 10−3 m = 0.25 cm YA (200 × 109 N/m2 )(7.9 × 10−5 ) From the definition of the Poisson’s ratio μ, we have: r=− Note that μr L 0.25 × (0.5 × 10−2 m)(2.5 × 10−3 m) =− = −1.56 × 10−6 m L (2 m) L ≈ 1,600 | r|, i.e | r| is extremely small compared to L 310 10 Mechanical Properties of Matter 10.2.2 Shear Modulus: Elasticity of Shape Another type of deformation occurs when a solid is subject to a force applied parallel to one of its surfaces while the opposite surface is kept fixed Figure 10.4 shows a cylindrical rod subjected to a linear or torsional shear stress deforming it by an amount x due to a force F parallel to the surface area A As a final result, the shape of the rod will attain equilibrium when the effect of the shear force F balances exactly the internal shear forces For linear shearing, we define the shearing stress and the shearing strain as follows: Shearing stress = F Tangential acting force = Area of surface being sheared A Shearing strain = x Distance sheared = = tan θ Distance between surfaces h A x F (N/m2 ) (10.8) θ (10.9) Top view F F h x F Fixed end Linear shearing Torsional shearing Fig 10.4 The left part shows a cylindrical rod of height h The middle part shows a linear shear where the rod is subject to a shearing force F parallel to each of its surface areas The rod is deformed through an angle θ which is defined as the shearing strain The right part shows a torsional shear when one end of the rod is kept fixed The approximation tan θ θ is valid for small strains We use Eqs 10.8 and 10.9 to define the shear modulus, S, as follows: S= F /A Shearing stress = Shearing strain x/h (N/m2 ) (10.10) S is also called the modulus of rigidity or the torsion modulus and is significant only for solids Table 10.3 depicts some values for S 10.2 Elastic Properties of Solids 311 Table 10.3 Shear modulus for different materials (approximates) Material name Shear modulus S × (109 N/m2 ) Rubber 0.001 Lead Glass 23 Aluminum 23 Brass 36 Copper 42 Steel 80 Tungsten 120 Example 10.4 Assume that the rod in Fig 10.4 has a cross-sectional area A = × 10−3 m2 , length h = m, and is made of brass with a shear modulus S = 36 × 109 N/m2 How large should the shear force F exerted on each edge of the rod be if the displacement x is 0.02 cm? Solution: The shearing stress on each edge is: Shearing stress = F F = = 500 F m−2 A × 10−3 m2 The shearing strain is: Shearing strain = × 10−4 m x = = × 10−4 h 1m From the definition of the shearing modulus Eq 10.10 and the last two results, we have: S= 500F m−2 Shearing stress = Shearing strain × 10−4 Using the given shear modulus value for brass, we get: 36 × 109 N/m2 = 500F m−2 × 10−4 Thus: F = (36 × 109 N/m2 )(2 × 10−4 )/(500 m−2 ) = 14,400 N = 1.44 × 104 N 312 10 Mechanical Properties of Matter 10.2.3 Bulk Modulus: Volume Elasticity Another type of deformation occurs when an object is subject to an equal increase in normal forces acting on all its faces For such a study, it is appropriate to define the pressure P as the force acting perpendicularly on a unit area of the object That is: P= F⊥ A (N/m2 ) (10.11) Hence, we can study the deformation of an object subject to an equal increase in pressure on all its faces Figure 10.5 shows a cube of original volume V and face area A under uniform pressure P When the force F⊥ on each face increases to F⊥ + F⊥ , the pressure will increase to P + P and consequently the volume V will decrease to V = V − | V |, where V is negative, see Fig 10.5 V V- P V P+ P P P P P P+ P P+ P P+ P P+ P P P+ P Fig 10.5 When the uniform pressure P on each face of a cube of volume V increases to P + P, its volume will decrease to V − | V | In light of this, we define the volume stress and the volume strain as: Volume stress = P= Volume strain = − F⊥ A (N/m2 ) (10.12) V V (10.13) We use Eqs 10.12 and 10.13 to define the bulk modulus, B, as: B= Volume stress F⊥ /A P =− =− Volume strain V /V V /V ⇒ B = −V dP dV (10.14) 10.2 Elastic Properties of Solids 313 The minus sign is inserted in Eq 10.13 to make B a positive number, because an increase/decrease of pressure always causes a decrease/increase in volume Example 10.5 A sphere of lead has a volume V = 0.5 m3 when placed in atmospheric pressure (Pa 105 N/m2 ) The sphere is lowered to a particular depth in the ocean where the water pressure is P = 108 N/m2 = 1,000 Pa The bulk modulus B of lead is × 109 N/m2 (a) What is the change in volume of the sphere? (b) What is the relative density change in lead? Solution: (a) From the definition of the bulk modulus, we have: B=− P V /V The change in pressure is: P = P − Pa = 108 N/m2 − 105 N/m2 = 9.99 × 107 N/m2 Using V = 0.5 m3 , we can find the change in volume as follows: V =− V P (0.5 m3 )(9.99 × 107 N/m2 ) = −6.2 × 10−3 m3 =− B × 109 N/m2 The negative sign indicates a decrease in volume (b) We can find the new volume V of lead as follows: V = V −| V | = 0.5 m3 −|−6.2×10−3 m3 | = 0.4938 m3 ⇒ V = 0.9876 V If the original density of lead is denoted by ρ = m/V, then the new density ρ will be: ρ = m m m = 1.0126 = 1.0126 ρ = V 0.9876 V V Thus, a thousand times increase in pressure on the surfaces of a sphere of lead causes a decrease in its volume by about 1.3% and consequently an increase in density by the same percentage 314 10 Mechanical Properties of Matter Table 10.4 depicts some values for the bulk modulus B Table 10.4 Bulk modulus for different materials (approximates) Material name Bulk modulus B × (109 N/m2 ) Rubber Lead Glass 37 Aluminum 70 Brass 61 Copper 140 Steel 160 Tungsten 200 10.3 Fluids What is a Fluid? Liquids and gases are fluids The liquid state of any substance always exists at a higher temperature than its solid state The reason for lumping liquids and gases together and calling them fluids is because neither liquids nor gases (such as liquid water and steam, for example) have a fairly rigid three-dimensional array of atoms/molecules as compared to solids (such as ice, for example) In contrast to solids, fluids can flow and conform to the boundaries of any container in which they are placed This is because a fluid cannot sustain a force that is tangent to its surface In the language of the previous section, a fluid flows because it cannot withstand a shearing stress On the other hand, a fluid can exert a force in a direction perpendicular to its surface Pressure in Fluids Figure 10.6 shows a pressure device inside a fluid-filled vessel The device consists of a light piston of area A fitting in a vacuumed cylinder and resting on a light spring As we insert the device into the fluid, the fluid will compress the piston due to the effect of a normal force of magnitude F⊥ Using Eq 10.11, after replacing F⊥ by F, we define the average pressure exerted by the fluid on the piston by the following relation: 10.3 Fluids 315 P= F A (N/m2 ) (10.15) Fig 10.6 A pressure device inside a fluid-filled vessel The pressure is measured by the relative position of a movable F A Pressure Device piston in the device Vacuum The pressure at any point in the fluid is the limit of the above ratio as piston, centered on that point, approaches zero That is: P = lim A→0 F dF = A dA (N/m2 ) A of the (10.16) It was found by experiment that at a given point in a static fluid, the pressure P given by Eq 10.16 has the same value no matter how the pressure device is oriented Moreover, all points at the same depth from a liquid surface have the same value of pressure The SI unit of pressure is N/m2 , which is given the special name pascal (Pa) That is: Pa = N/m2 (10.17) Atmospheric pressure at sea level Pa (abbreviated by atm) is: atm = 1.01 × 105 Pa 105 Pa (10.18) The pascal is also related to the torr, the bar, and the pound per square inch, which are other common (non-SI) pressure units The torr unit (named after Evangelista Torricelli who invented the mercury barometer) was formerly called millimeter of mercury (mm Hg) The bar unit is usually used in meteorological sciences Finally, the pound per square inch (lb/in2 ) is often abbreviated by psi Note that: ⎧ ⎪ ⎨ 760 torr atm = or = 14.7 psi ⎪ ⎩ 760 mm Hg and bar = 105 Pa (Exactly) (10.19) 316 10 Mechanical Properties of Matter Table 10.5 depicts some approximate pressure values Table 10.5 Some approximate pressures Locations Pressure (Pa) × 1016 Pressure (atm) × 1011 Center of the Sun Center of the Earth × 1011 × 106 Highest laboratory pressure × 1010 × 105 Deepest Ocean 1.1 × 108 1.1 × 103 Automobile tire (excess of atm) × 105 × 105 Atmosphere at sea level 1.0 Normal blood pressure (excess of atm)a 0.16 × 105 0.16 Best laboratory vacuum 10−12 10−17 a The systolic pressure that corresponds to 120 mm Hg on the physician’s pressure gauge To study the mechanics of fluids, we need to deal with: Fluids at rest, or fluid statics (hydrostatics) Fluids in motion, or fluid dynamics (hydrodynamics) 10.4 Fluid Statics Variation of Pressure with Depth As indicated in the previous section, all points at the same depth from a liquid surface have the same value of pressure The variation of pressure P with depth h in a liquid of density ρ open to the atmosphere can be found by considering a small horizontal area dA at that depth, as shown in Fig 10.7 The force dF that acts downwards on dA must be equal to the weight of the liquid column of height h plus the weight of the atmospheric air column Accordingly, we have: Volume of the liquid column = h dA Mass of the liquid column = h dAρ Weight of the liquid column = h dAρg Weight of the atmospheric air column = Pa dA Total force dF on the horizontal area d A = Pa dA + h dAρg Thus, from Eq 10.16, the pressure P = d F/d A at depth h gives: P = Pa + ρgh ⇒ dP = P − Pa = ρgh (10.20)