1 2 1.1 Linear Momentum and its Conservation Momentum Momentum can be defined as "mass in motion." The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. The momentum of an object is equal to the mass of the object times the velocity of the object. Momentum = mass • velocity p=m•v where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. Momentum is a vector quantity. 3 1.1 Linear Momentum and its Conservation In 3 dimensions Px = m vx Py = m vy Pz = m vz Units of momentum (kg.m/s) From Newton’s second law F = ma F = m(dv/dt) F = (mdv/dt) F = (dp/dt) 4 1.1 Linear Momentum and its Conservation If two particles interact while isolated from their surroundings, the particles exert forces on each other where Newton’s third law can be applied. F21 = dp1/dt F12 = dp2/dt F21 + F12 =0 dp1/dt + dp2/dt= 0 p1i + p2i = p1f + p2f Momentum of an isolated system is conserved 5 1.2 Impulse and Momentum From Newton’s second law F = (dp/dt) dp = F dt ∆p = ∫ F dt This is called the impulse which is the change in momentum. I = ∆p = ∫ F dt I = F ∆t 6 1.3 Collisions The total momentum of a system before a collision is equal to the total momentum of a system after a collision. 7 1.4 Elastics and Inelastic Collisions in one Dimension A collision between two objects is considered elastic if the total kinetic energy as well as the momentum before and after the collision is conserved. Where it is considered inelastic if the total kinetic energy is not conserved even if the momentum is conserved before and after the collision. 8 1.4 Elastics and Inelastic Collisions in one Dimension Elastic collisions Momentum is conserved m1v1i + m2v2i = m1v1f + m2v2f Kinetic energy is conserved (1/2)m1v1i2 + (1/2)m2v2i2 = (1/2)m1v1f2 + (1/2)m2v2f2 We can find that v1f = ((m1-m2)/(m1+m2)) v1i v2f = ((2m1)/(m1+m2)) v1i 9 1.5 Two Dimensional collisions We use the same conservation laws of kinetic energy and momentum but we analyze the equations in all dimensions. You should remember how to find vector components using the angles !! In order to solve a problem with collisions: Choose the system. If it is complex, subsystems may be chosen where one or more conservation laws apply. Draw diagrams of the initial and final situations, with momentum vectors labeled. Define a coordinate system Apply momentum conservation; there will be one equation for each dimension. If the collision is elastic, apply conservation of kinetic energy as well. Solve. 10 1.6 The Center of Mass The Center of Mass is the average position of the mass of an object. It is a uniquely interesting point. We may consider an object as a hollow, mass less shell with all its mass located at this Center of Mass. The general motion of an object can be considered as the sum of the translational motion of the CM, plus rotational, vibrational, or other forms of motion about the CM. In the first picture, the diver’s motion is pure translation; where in the second picture it is translation plus rotation. There is one point that moves in the same path a particle would take if subjected to the same force as the diver. This point is called the center of mass 11 1.6 The Center of Mass If the particles in the system have masses m1, m2, ...mN, with total mass ∑iN mi = m1+m2+···+mN ≡ M and respective positions r1, r2, ...,rN, then the center of mass position rCM is: rCM = (1/M) ∑iN mi ri which means that the x, y and z coordinates of the center of mass are xCM = (1/M) ∑iN mi xi yCM = (1/M) ∑iN mi yi zCM = (1/M) ∑iN mi zi For an extended object rCM is given by an integral over the mass elements of the object: rCM = (1/M) ∫ r dm which means that the x, y and z coordinates of the center of mass are xCM = (1/M) ∫ x dm yCM = (1/M) ∫ y dm zCM = (1/M) ∫ z dm 12 1.7 Motion of System of Particles The velocity of the center of mass of a system vCM = drCM/dt = (1/M) ∑i mi (dri/dt) = (1/M) ∑i mi vi Which gives the momentum of the system MvCM = ∑i mi vi = ∑i pi = ptot The acceleration of the center of mass aCM = dvCM/dt = (1/M) ∑i mi (dvi/dt) = (1/M) ∑i mi ai Which gives the force MaCM = ∑i mi ai = ∑i Fi which leads to ∑ Fext = MaCM = (d ptot / dt) P is constant when the external force is zero 13 1.8 Rocket Propulsion The operation of a rocket depends on the law of conservation of linear momentum as applied to a system of particles where the system is the rocket plus its ejected fuel. Applying the law of conservation of momentum (M+∆m)v = M(v+ ∆v) + ∆m(v-ve) Where the velocity of the rocket changes after its ejects the fuel. The fuel is ejected with speed ve relative to the rocket. We get M ∆v = ve∆m Mdv = vedm = - vedM Since dm = - dM ∫ dv = - ve ∫(dM/M) vf – vi = ve ln (Mi/Mf) We call the force exerted on the rocket by the ejected fuel as the thrust Thrust = M (dv/dt) 14 Examples A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.2 m/s in 0.832 s. What linear impulse and average force does a 70 kg passenger in the car experience? Assume the car accelerates in the x-direction. The final momentum of the passenger is pxf = (70 kg)(5.2 m/s) = 364 kgm/s. The initial momentum is zero. ∆px = pxf - pxi = 364 kgm/s = Favg∆t. Favg = (364 kgm/s)/(0.832 s) = 437.5 N in the xdirection. A 3 kg steel ball strikes a wall with a speed of 10.0 m/s at an angle of 60o with the surface. It bounces off with the same speed and angle. If the ball is in contact with the wall for 0.2 s, what is the average force exerted on the ball by the wall? The balls initial momentum is pi = pxii + pyij = (3 kg 10 m/s)sin60oi+(3 kg 10 m/s)cos60oj. Its final momentum is pf = pxfi + pyfj = -(3 kg 10 m/s)sin60oi+(3 kg 10 m/s)cos60oj. ∆p = pf - pi = -2(30 kgm/s)sin60oi = -(51.96 kgm/s)i. ∆p = Favg ∆t. Favg = -(51.96 kgm/s)i/(0.2 s) = -259.8 Ni. 15 Examples A 0.1 kg ball is thrown straight up into the air with an initial speed of 15 m/s. Find the momentum of the ball (a) at its maximum height and (b) At half rise time. (a) At the ball's maximum height its velocity is zero, and therefore its momentum is zero. (b) First we find maximum rise time. Vf = vi-gt which gives that t=1.5s which means that half rise time is 0.75s. Then Vf = 7.5 m/s and momentum is 1.125 kgm/s. A 10 g bullet is stopped in a block of wood (m = 5 kg). The speed of the bulletwood combination immediately after the collision is 0.6 m/s. What was the original speed of the bullet? m1v1 + m2v2 = (m1 + m2)vf. m1 is the mass of the bullet, m2 is the mass of the block. Initially the block is at rest, v2 = 0. Therefore (0.01 kg)v1 = (5.01 kg)(0.6 m/s). v1 = 300.6 m/s. 16 Examples A car of mass 1000 kg travelling west at 20 m s−1 crashes into the rear of a stationary bus of mass 5000 kg. The vehicles lock together on impact. Assume that road friction is negligible. a What is their joint velocity immediately after the collision? b What is the total kinetic energy of the system before the collision? c What is the total kinetic energy of the system after the collision? d Is this an elastic or inelastic collision? Explain. a For this problem, west will be taken as the positive direction. Conservation of momentum must be used to determine the final common velocity v' of the vehicles. Σpi=Σpf pi(car) +pi(bus) =pf(bus and car) (1000 × 20) + 0 = (1000 + 5000) v' v' = 20 000 ÷ 6000 = 3.3 m/s towards the west b Since the bus is stationary before the collision, the initial kinetic energy of the system is the initial kinetic energy of the car. It has a mass of 1000 kg and was moving at 20 m/s before the collision. KE= 12mv 2= 0.5 × 1000 × 202 = 2.0 × 105 J c After the collision, the car and bus are locked together and moving at 3.3 m/s. Their total kinetic energy is: KE= 12mv 2= 0.5 × (1000 + 5000) × 3.32= 3.3 × 104 J d This is an inelastic collision. 17 A large amount of kinetic energy has been lost, so the collision is inelastic. Questions An 8 N force acts on a 5 kg object for 3 sec. What impulse is given the object? What change in momentum does this impulse cause? If the object’s initial velocity was 25 m/s. what is its final velocity? I = F ∆t = 8 (3) = 24 N.s I = ∆p = 24 kg.m/s ∆p = mvf - mvi = 5 (vf – 25) = 24 vf = 29.8 m/s A 6 N force acts on a 3 kg object for 10 sec. What will be the final velocity of the object if its initial velocity was 10 m/s? ∆p = F ∆t = 6 (10) = 60 kg.m/s 60 = 3 (vf - 10) vf = 30 m/s A 20,000 N truck is acted upon by a force that decreases its speed from 10 m/s to 5 m/s in 5 sec. What is the magnitude of the force? m = 20000/9.8 = 2040.8 kg F = ∆p/ ∆t = 2040.8(5-10)/5 = - 2040.8 N 18 Questions A 2000 N force acts on a rocket of mass 1000 kg, increasing its speed from rest to 200 m/s. How long did the force act? F = ∆p/ ∆t 2000 = 1000(200-0)/t t = 100 s A 700 kg car moving at 20 m/s collides with a stationary truck with mass 1400 kg. The two vehicles interlock as a result of the collision. What is the velocity of the car/truck? m1v1i + m2v2i = m1v1f + m2v2f 700(20) + 0 = (700+1400) vf Vf = 6.67 m/s A rocket sits on a launch pad loaded with fuel. The igniter causes the rocket to rise off the pad, oxidizing 100 g of fuel and ejecting it out the back of the rocket at -650 m/s. After launch, the rocket has a mass of 3 kg. What is its speed after launch? M∆v = m ve 3 ∆v = 0.1 (650) ∆v = 21.67 m/s 19 Questions A 0.4 kg ball is thrown straight up into the air with an initial speed of 15 m/s. Find the momentum of the ball (a) at its maximum height and (b) At half rise time. (c) At half maximum height. (a) At max height P=0 (b) First we find rise time vf = vi – gt 0 = 15 – 9.8t t = 1.53 s t ½ = 0.77 s Velocity at half rise time vf = vi – gt = 15 – 9.8 (0.77) = 7.45 m/s P = 2.98 kg.m/s (c) The maximum height ∆y = vi t+ (1/2)gt2 ∆y = 15 (1.53) + (1/2)(-9.8)(1.53)2 = 22.95 - 11.47 = 11.48 m The half height ∆y1/2 = 5.74 m vf2 = vi2 + 2g ∆y = (15)2 - 2 (9.8) (5.74) = 225 - 112.504 = 112.496 vf = 10.6 m/s P = 4.24 kg.m/s 20 Questions A 0.2 kg ball is thrown straight up into the air with an initial speed of 30 m/s. Find the momentum of the ball (a) at its maximum height and (b) At half rise time. (c) At half maximum height. (a) At max height P=0 (b) First we find rise time vf = vi – gt 0 = 30 – 9.8t t = 3.06 s t ½ = 1.53 s Velocity at half rise time vf = vi – gt = 30 – 9.8 (1.53) = 15 m/s P = 3 kg.m/s (c) The maximum height ∆y = vi t+ (1/2)gt2 ∆y = 30(3.06) + (1/2)(-9.8)(3.06)2 = 91.8 - 45.88 = 45.92 m The half height ∆y1/2 = 22.96 m vf2 = vi2 + 2g ∆y = (30)2 - 2 (9.8) (22.96) = 449.984 vf = 21.21 m/s P = 4.24 kg.m/s 21 Questions A 0.5 kg ball is thrown straight up into the air with an initial speed of 20 m/s. Find the momentum of the ball (a) at its maximum height and (b) At half rise time. (c) At half maximum height. (a) At max height P=0 (b) First we find rise time vf = vi – gt 0 = 20 – 9.8t t = 2.04 s t ½ = 1.02 s Velocity at half rise time vf = vi – gt = 20 – 9.8 (1.02) = 10 m/s P = 5 kg.m/s (c) The maximum height ∆y = vi t+ (1/2)gt2 ∆y = 20(2.04) + (1/2)(-9.8)(2.04)2 = 40.8 - 20.4= 20.4 m The half height ∆y1/2 = 10.2 m vf2 = vi2 + 2g ∆y = (20)2 - 2 (9.8) (10.2) = 200.08 vf = 14.14 m/s P = 7.07 kg.m/s 22 Questions A 15 g pellet moving at 600 m/s hits a stationary 100 g block and becomes embedded in the block. With what speed do they move off together? m1v1i + m2v2i = m1v1f + m2v2f 0.015(600) + 0.1 (0) = (0.015 + 0.1) vf vf = 78.26 m/s A 0.5 kg ball traveling at 6 m/s collides head-on with a 1 kg ball moving in the opposite direction at 12 m/s. The 0.5 kg ball moves away at 14 m/s in the opposite direction after the collision. Find the velocity of the 1 kg ball after the collision. m1v1i + m2v2i = m1v1f + m2v2f 0.5 (6) + 1 (-12) = 0.5 (-14) + 1 v2f v2f = 1.29 m/s A plastic ball of mass 0.2 kg moves at 0.6 m/s, east. It collides with a 0.1 kg ball moving at 0.8 m/s, west. After the collision, the velocity of the 0.1 kg ball is 0.6 m/s, east. What is the velocity of the 0.2 kg ball after the collision? 0.2 (0.6) + 0.1 (-0.8) = 0.2 v1f + 0.1 (0.6) v1f = -0.1 m/s 23 Questions A machine gun fires 35.0g bullets at a speed of 750.0m/s. If the gun can fire 200bullets/min, what is the average force the shooter must exert to keep the gun from moving? P = mv = 0.035 (750) = 26.25 kg.m/s Time for one bullet t = 60/200 = 0.3 s F = p/t = 26.25/0.3 = 86.5 N A 3.00kg particle is located on the x axis at x =−5.00m and a 4.00kg particle is on the x axis at x =3.00m. Find the center of mass of this two–particle system. The center of mss is at Xcm = (-15 + 12)/(3+4) = -0.43 m A 2.0kg particle has a velocity of v1 = (2.0i−3.0j) m/s, and a 3.0kg particle has a velocity (1.0i+6.0j)m/s. Find (a) the velocity of the center of mass and (b) the total momentum of the system. Vcm = (2i-6j +3i+18j)/(5) = (5i +12j)/5 = (1i + 2.4j) m/s 24 25 26 2.1 Angular Position, velocity and acceleration Rigid Object A rigid object is one that is “non deformable” that is, the relative posed remains constant. Rotation When a disc rotates about a fixed axis with an angle Ɵ, it moves along an arc s where s = r Ɵ and Ɵ is measured by radians. We can convert between degrees and radians using: Ɵ (rad) = (π/180) Ɵ (degree) 27 2.1 Angular Position, velocity and acceleration Angular Displacement As a result of rotation the point at position p1 moves to position p2. The angular displacement is expressed by ∆Ɵ = Ɵ2 – Ɵ1 Angular Speed The angular speed w is the ratio of the angular displacement to the time interval and has the units of (s-1). w = ∆Ɵ/ ∆t w = dƟ/dt 28 2.1 Angular Position, velocity and acceleration Angular Acceleration The angular acceleration α is the change of angular velocity in time. α = ∆w/ ∆t α = dw/ dt 29 2.2 Rotational Kinematics The equations of rotational motion are 30 2.3 The Relation Between Linear and Rotational There is a relation between linear and rotational quantities. Angular velocity and Tangential velocity Every point on the rotating object moves in the same direction and travels the same distance s which means its velocity is v = ds/dt But s = r Ɵ Which means that v = d(rƟ)/dt = r dƟ/dt v = rw Which means that all points of the object have the same angular speed but different tangential speed because they have different distances from the center of the object. 31 2.3 The Relation Between Linear and Rotational Angular acceleration and Linear Acceleration The tangential acceleration of any point is resulted from change in tangential velocity. at = dv/dt = d(rw)/dt = r dw/dt at = rα Where the rotational acceleration is the result from changing the direction of the velocity with time. ar = v2/r = r2w2/r ar = rw2 The total acceleration a = (at2 + ar2)1/2 = (r 2 α2 + r 2 w4 )1/2 = r(α2 + w4 )1/2 32 2.4 Rotational Kinetic Energy The kinetic energy of the ith particle of the rigid object KEi = (1/2)mivi2 The total kinetic energy of the object KE = ∑iKEi The moment of inertia I is I = ∑imiri2 Which means KE = ∑iKEi = ∑i (1/2)mivi2 = ∑i (1/2) miri2wi2 = ∑i (1/2)Iwi2 33 2.5 Moment of Inertia The moment of inertia of an object can be found by dividing the object into small elements each of mass ∆m then taking the limit where ∆m approaches zero. I = lim∆m0∑imiri2 = ∫r2dm Uniform Thin Ring The ring is thin we could neglect the thickness which means that all elements have the same distance from the axis. I = ∫r2dm = R2 ∫dm = MR2 34 2.5 Moment of Inertia Long Thin Rod Rotating about an axis passing through the center The mass dm is related to the mass per unit length λ dm = λ dx = (M/L)dx I = ∫r2dm = (1/12)ML2 Long Thin Rod Rotating about an axis passing through the end I = ∫r2dm = (1/3)ML2 35 2.5 Moment of Inertia 36 2.6 Torque Torque, also called moment or moment of force is the tendency of a force to rotate an object about an axis. Just as a force is a push or a pull, a torque can be thought of as a twist. A torque is a force exerted at a distance from the axis of rotation; the easiest way to think of torque is to consider a door. When you open a door, where do you push? If you exert a force at the hinge, the door will not move; the easiest way to open a door is to exert a force on the side of the door opposite the hinge, and to push or pull with a force perpendicular to the door. This maximizes the torque you exert. 37 2.6 Torque Torque is the product of the distance from the point of rotation to where the force is applied multiplied by the force multiplied by the sine of the angle between the line you measure distance along and the line of the force. τ = r F sinƟ The unit of torque is N.m. If an object is under multiple torques then the net torque is the summation of all torques. τ = ∑iτ i 38 2.6 Rigid Object under Torque If a particle of mass m rotating in a circle of radius r under the influence of a tangential force Ft and a radial force Fr. The tangential force provides a tangential acceleration a Ft = mat The torque about the center of the circle τ = rFt = rmat = rmrα = mr2α = I α 39 2.7 Work and Energy in Rotational Motion The work done on an object that is under a translational and rotational motion is given by the change in both translation kinetic energy and the rotational kinetic energy. W = ∆KEt + ∆KEr = [(1/2)mvf2 – (1/2)mvi2] + [(1/2)Iwf2 – (1/2)Iwi2] 40 Examples -A disc rotates at 1200 rev/min about its central axis. What is the linear speed of a point 3.0 cm from its center? ∆θ = 1200 x 2π = 7536 rad ω = 7536 / 60 = 125.6 rad/s vt = 3.768 m/s -During a certain period of time, the angular position of a swinging door is described by θ = 5.00 + 10.0 t + 2.00 t², where θ is in radians and t is in seconds. Determine the angular position, angular speed, and angular acceleration of the door (a) at t = 0 and (b) t = 3.00 s. ω = 10 + 4 t α=4 At t = 0 then θ = 5 , ω = 10 , α = 4 At t = 3 then θ = 53 , ω = 22 , α = 4 -A racing car travels on a circular track of radius 250 m. If the car moves with a constant linear speed of 45.0 m/s, find (a) its angular speed and (b) the magnitude of its acceleration. ω = 0.18 rad/s , at = 0 , ar = 8.1 Total acceleration a = 8.1 41 Examples -A bicycle travels 141 m along a circular track of radius 15 m. What is the angular displacement in radians of the bicycle from its starting position? ∆θ = s / r = 9.4 rad - A grindstone, originally rotating at 126 rad/s undergoes a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. What is its angular acceleration? ∆θ = 125.6 rad , ∆θ = ωit + (1/2) αt2 α = -27.6 rad/ s2 - A ball attached to a string starts at rest and undergoes a constant angular acceleration as it travels in a horizontal circle of radius 0.30 m. After 0.65 sec, the angular speed of the ball is 9.7 rad/s, what is the tangential acceleration of the ball? ωi = 0 , ωf = 9.7 rad/s , t = 0.65 s , ωf = ωi + αt , α = 14.9 rad/ s2 a = 4.97 m/s2 - Two motorcycles are riding around a circular track at the same angular velocity. One motorcycle is at a radius of 15 m; and the second is at a radius of 18 m. What is the ratio of their linear speeds, v2/v1? ω1 = ω2 , v1 r1 = v2 r2 42 v1 15 = v2 18 , v2 /v1 = 1.2 Examples - A 100 cm meter rule is pivoted at its middle point (that is, at the 50 cm point). If a weight of 10 N is hanged from the 30 cm mark and a weight of 20 N is hanged from its 60 cm mark, find out whether the meter rule will remain balanced about its pivot or not? τ1 = 10 (0.2) = 2 N.m τ2 = 20(0.1) (-) = - 2 N.m The meter rule will balance -The net torque on the pulley about the axle is the torque due to the 30 N force plus the torque due to the 20 N force τ1 = 20(r)sin(90) = 20r N.m τ2 = 30(r)sin(-90) =-30r N.m τ = -10r 43 Examples -Find the total torque at point A then B Around A: τ = 0 + 80(0.5 L)sin(37) + 70(L)sin(120) + 0 + 60(0.5L) sin(-90) τ = 0.3L + 60.6L – 30L = 60.9 L mN Around B: τ = 90Lsin(-140) + 80(0.5 L)sin(-143) + 0 + 0 + 60(0.5L) sin(90) τ = -57.9L - 24.1L + 30L = -52L mN 44 Questions A propeller (arm length 1.2 m) starts from rest and begins to rotate counterclockwise with a constant angular acceleration of size 2.7 rad/s2. a. How long does it take for the propeller's angular speed to reach 5.7 rad/s? b. How many revolutions does it take for the propeller's angular speed to reach 5.7 rad/s? c. What is the linear speed of the tip of the propeller at 5.7 rad/s? d. What is the linear, rotational and total acceleration of the tip of the propeller at this point? wi = 0 α = 2.7 rad/s2 wf = 5.7 rad/s r = 1.2 m wf = wi + αt 5.7 = 2.7 t t = 2.11 s ∆θ = wi t + (1/2) αt2 ∆θ = 6.01 rad Number of revolutions = 0.96 v = rw v = 6.84 m/s a = rα a = 3.24 The total acceleration a = 39.12 m/s2 ar = rw2 ar = 38.99 45 Questions An airplane propeller is rotating at 1900 rev/min. a. Compute the propeller's angular velocity in rad/s. b. How long in seconds does it take for the propeller to turn through 30 degrees? w = [1900 (2)(3.14)]/60 = 198.87 rad/s ∆θ = 30(2)(3.14)/180 = 1.05 rad t = 0.0053 s A 500 N rock rests on an 8.0-m-long board that weighs 100 N. The board is supported at each end. The support force at the right end is 3 times the support force at the left end. How far from the right end is the rock standing? F1 + F2 – 100 – 500 = 0 F1 + F2 = 600 4 F2 = 600 F2 = 150 N The pivot point at the center ∑τ = 0 4F2sin(-90) + (4-x)(500)sin(-90) + F1(4)sin(90) = 0 -4F2 - 2000 + 500x + 12 F2 = 0 X = 1.6 m 46 Questions A uniform horizontal beam 6.0 m long and weighing 4 kg is supported at points p and q each 1.0 m from opposite ends of the beam. Masses of 10 kg and 8 kg are placed near p and q respectively, one on each end of the beam. Calculate the reaction forces at p and q. Fp + Fq – (8)(9.8) – (10)(9.8) – (4)(9.8) = 0 Fp + Fq = 215.6 The pivot point is q ∑τ = 0 (1)(8)(9.8) – 2(4)(9.8) + 2Fp – 3(10)(9.8) = 0 78.4 – 78.4 + 2Fp – 294 = 0 Fp = 147 N Fq = 68.6 N 47 Questions A spot of paint on a tire moves in a circular path of radius 0.42 m. Through what angle must the tire rotate for the spot to travel 1.48 m? s = rθ θ = 3.523 rad A disk with a 1.0-m radius reaches a maximum angular speed of 18 rad/s before it stops 35 revolutions (220 rad) after attaining the maximum speed. how long did it take the disk to stop? t = 24.44 s How many revolutions does a disc make in in 5.44 seconds accelerating from 0 to 7.598 thousand rpm? Number of revolutions = 688.9 Two 0.51 kg cartons are at one end of a bin that is 0.69 m long. Where should a 1.8 kg carton be placed so that the center of mass of the three cartons is at the center of the bin? 48 Should be at 0.098 m Questions An Oxygen molecule rotating in the xy plane about an axis passing through the center of the molecule perpendicular to its length. The mass of each atom is 2.66x10-26 kg and at room temperature the average seperation between the two atoms is 1.21x10-10 m. Calculate the moment of inertia of the molecule about the z axis. If the angular speed of the molecule about the z axis is 4.6x1012 rad/s. what is its kinetic energy? I = ∑mr2 = m(d/2)2 + m(d/2)2 = 1.95x10-46 kgm2 KE =(1/2)Iw2 KE =(1/2)(1.95x10-46)(4.6x1012) = 2.06x10-21 J 49 Questions If A = 3N, determine the weight of B,C,D,E, and F 50 51 2 3.1 Vector Product vector product is the product of two vectors in three-dimensional space. It results in a vector which is perpendicular to both of the vectors being multiplied and normal to the plane containing them. Consider two vectors A and B in the standard unit-vector notation The cross product A x B is a vector perpendicular on both vectors A and B where its magnitude could be calculated by A x B = |A | | B | sinθ Where θ is the angle between the two vectors. The vector product could be found by finding the determinant 53 3.1 Vector Product The direction of the resulted vector could be found using the right hand rule. The unit vectors i, j, and k from the given orthogonal coordinate system satisfy the following equalities: i×i=j×j=k×k=0 i×j=k j×k=i j × i = −k k × j = −i k×i=j i × k = −j 54 3.2 Angular Momentum Angular momentum (L) is a measure of an object's tendency to keep rotating and to maintain its orientation. Mathematically it depends on the object's mass, m, radius, r, and rotational velocity, v, and is proportional to mvr. L=rxp L = r m v sinθ Where θ is the angel between r and p The total torque of a particle is ∑τ = r x ∑F = r x (dp/dt) = d(rxp)/dt = dL/dt 55 3.3 Angular Momentum of a Rotating Rigid Object The angular momentum of a rotating object is L = m r2 w = I w Where I is the moment of inertia of the object The total external torque is the change of angular momentum in time is ∑τext = dL/dt = d(Iw)/dt = Iα 56 3.4 Conservation of Angular Momentum If the total external torque acting on a system equals zero then the angular momentum (magnitude and direction) is conserved. ∑τext = 0 Li = Lf 57 3.5 The Motion of Gyroscopes and Tops A gyroscope is a fascinating mechanical device with a lot of technological applications. You can easily make a gyroscope from a wheel with a handle put through the middle. If you dangle the end of the handle from a string, the wheel doesn't flop over but stays up straight! This might seem a bit odd because the weight is pointing straight down. Instead it slowly moves around in a circle, that is called precess. The angular momentum of the wheel points along its axis of rotation. 58 3.5 The Motion of Gyroscopes and Tops We know that Then τ = dL/dt dL = L(dθ/dt) τ = L(dθ/dt) = Iw(dθ/dt) The torque due to gravity is mgh mgh = Iw(dθ/dt) (dθ/dt) = mgh/Iw wp = mgh/Iw Where h is the distance between the pivot point and the center of mass and (dθ/dt) is called the precessional frequency wp This is valid when w<<wp 59 3.7 Angular Momentum as a Fundamental Quantity The angular momentum is an intrinsic property of atoms, molecules, and their constituents, a property that is a part of their very nature. The angular momentum of these systems have discrete values which are multiples of the fundamental unit of angular momentum ћ = h/2π, where h is called Planck’s constant. Fundamental unit of angular momentum = ћ=1.054 × 10-34 kg.m2/s. 60 Examples A light rigid rod 1.00 m in length rotates in the xy plane about a pivot through the rod's center. Two particles of masses 4.00 kg and 3.00 kg are connected to its ends. Determine the angular momentum of the system about the origin at the time the speed of each particle is 5.00 m/s. LTot = Lrod + L1 + L2 We are told the rod is a "lightweight" rod, which is another way of saying its mass and its moment of inertia are small enough they may be ignored. LTot = L1 + L2 The angular momentum for a "point particle" is L = m r vt L1 = (3.0 kg) (0.5 m) (5.0 m/s) = 7.5 kg m2/s L2 = (4.0 kg) (0.5 m) (5.0 m/s) = 10.0 kg m2/s 61 LTot = L1 + L2 = 17.5 kg m2/s Examples A playground merry-go-round of radius R = 2.0 m has a moment of inertia I = 250 kg m2 and is rotating at 10 rev/min. A 25-kg child jumps onto the edge of the merry-go-round. What is the new angular speed of the merry-go-round? Lf = Li Li = Iiwi Ii = 250 kg m2 wi = 10 rev/min Li = (250 kg m2) ( 10 rev/min) = 2 500 (kg m2 rev / min) Lf = If wf If = IMgR + Ichild Ichild = m r2 = (25 kg) (2.0 m)2 = 100 kg m2 If = (250 + 100) kg m2 = 350 kg m2 62 wf = [ 2 500 kg m2 rev / min ] / [ 350 kg m2 ] = 7.14 rev / min Examples A uniform solid disk of mass 3.00 kg and radius 0.200 m rotates about a fixed axis perpendicular to its face. If the angular frequency of rotation is 6.00 rad/s, calculate the angular momentum of the disk when the axis of rotation passes through its center of mass ICM = (1/2) M R2 ICM = (1/2) (3.00 kg) (0.200 m)2 ICM = 0.06 kg m2 L = I = ( 0.06 kg m2) (6.0 1/s) L = 0.54 kg m2 / s 63 Examples A 1000 kg car has four 10 kg wheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the wheels about their axles? Assume that the wheels have the same rotational inertial as uniform disks of the same mass and size. Why do you not need to know the radius of the wheels? 64 Examples A solid cylinder of radius 10 cm and mass 12 kg starts from rest and rolls without slipping a distance L=6 m down a roof that is inclined at angle θ=30°. What is the angular speed of the cylinder about its center as it leaves the roof? 65 Examples Two disks of moments of inertia I1 and I2 and having angular velocities ω 1 and ω 2 respectively are brought in contact with each other face to face such that their axis of rotation coincides. The situation is as shown in the figure just before the contact. What is the angular velocity of the combined system of two rotating disks, if they acquire a common angular velocity? 66 Questions 67 Questions Li = Lf I1w1i + I2w2i = I1w1f + I2w2f I1w1i + 0 = (I1+ I2)wf wf = I1w1i/(I1+ I2) KEi= (1/2) I1w1i2 KEf= (1/2) (I1+ I2) wf2 = (1/2) (I1+ I2) [I1w1i/(I1+ I2)]2 KEf= (1/2) (I1w1i) 2 /(I1+ I2)= KEi (I1 /(I1+ I2)) Which means tha the kinetic energy decreased 68 Questions 69 Questions Li = Lf Li = mvl Lf = (m+M)vfl mvl = (m+M)vfl vf=mv/(m+M) KEi = (1/2)mv2 KEf = (1/2)(m+M)vf2 Fraction lost of kinetic energy =[(1/2)mv2- (1/2)(m+M)vf2]/ (1/2)mv2 =[mv2- (m+M)vf2]/ mv2 =[mv2- (m+M) m2v2/(m+M)2]/ mv2 =M/(m+M) 70 Questions v=dr/dt v=5j P=mv P=10j L=rxP L=(6i+5tj)x(10j)=60k 71 Questions Lgyro=Lspace Igyrowgyro=Ispacewspace 20(100)=5x105(dθ/dt) 20(100)=5x105(30/dt) dt=131 s 72 73 74 4.1 Equilibrium Conditions Equilibrium means that the object is at rest or that the center of mass is moving with constant velocity relative to the observer. We will deal here with objects in Static Equilibrium where the object is at rest. Static Equilibrium Conditions An object is in static equilibrium (it is not moving) IF: 1) it is not translating (not moving up, down, left, or right) AND 2) it is not rotating (not spinning clockwise or counter clockwise) (We are talking about motion in a 2D plane (x,y) here.) 75 4.1 Equilibrium Conditions If a stationary mass is acted on by several forces, then in order to NOT translate, the net force must be zero. Fnet = 0 ∑Fx = 0 and ∑Fy = 0 This is not enough!! Look at the figure below, the two forces are equal and the net force is zero but the object will rotate. Which means we need to add another condition for static equilibrium. The net torque must be zero as well. ∑τ = 0 76 4.2 Center of Gravity The center of gravity is located at the center of mass of an object as long as the gravitational acceleration is uniform along the entire object. 77 4.3 Elastic Properties of solids Deformation of objects happens when acted on by external forces. The deformation could be change in shape, size or both. Stress is the net force acting on an object per unit area. Strain is the measure of degree of deformation. Modulus describes the behavior of an object under stress or how easily does it deforms 78 4.3 Elastic Properties of solids Young Modulus is the length deformation modulus. The stress F/A The strain ∆L/L0 Y = stress / strain 79 4.3 Elastic Properties of solids Shear Modulus is the shape deformation modulus. It measures the resistance to motion in parallel planes within a solid When an object is subjected to a force parallel to one of its faces while the other face is held fixed by another force, the object is called deformed in shape. The stress F/A The strain ∆x/h S = stress / strain 80 4.3 Elastic Properties of solids Bulk Modulus is the volume deformation modulus. It measures the resistance of solids and liquids to change their volumes. The object undergoes a change in volume when it is subjected to equal forces applied perpendicular over its entire surface. The stress F/A The strain ∆V/Vo B = stress / strain 81 Examples The system is at static equilibrium, find T1, T2, T3 and θ The total force = zero T1cos35 - 40=0 T1 = 40/cos35 = 40/0.82 = 48.78 N T2 - T1sin35=0 T2 = 28 N T3cosθ - 50=0 T3sinθ - T2 =0 T3sinθ - 28 =0 tan θ = 28/50 θ = 29.3 T3 = 57.3 N 82 Examples At what minimum angle can the ladder lean without slipping? The wall is frictionless and there is friction between the floor and the ladder. Forces: 1. Normal force at bottom of ladder 2. Friction force at bottom of ladder 3. Ladder’s weight 4. Normal force at top of ladder Pivot: Choose bottom of ladder, Why? Torques: 1. Due to ladder’s weight 2. Due to normal force at top of ladder m n1 – n2 = 0 Force, y: n1 – mg = 0 Force, x: Torque: Ln2sinf – (L/2) mg cosf = 0 From the force equations we get n2= m mg. Therefore, msinf – (1/2)cosf = 0 and so, tanf = 1/(2m) 83 Examples What is the force on the elbow? m = 6 kg Assume biceps force acts 3.4 cm from pivot point O. The force we know least about is the force on the elbow. So, let’s take the elbow (O) as the pivot. Net torque: 1 L Fm mh m g 2 d Fua , x 0 0 0 0 L mh Fmd mgL 0 2 Fua , y Fm mh g mg 0 Fua , y (m mh ) g Fm 84 Examples A steel wire of 2 mm in a diameter is stretched by applying a force of 72N. Find the stress in the wire. Diameter = 2 mm; radius r = diameter / 2 = 2/2 = 1 mm = 1 x 10 -3 m ; F = 72 N The Stress = F/A = F/πr2 = 2.292 x 107 N m-2 A brass wire of length 5m and cross section 1mm2 is hung from a rigid support, with a brass weight of volume 1000cm3 hanging from the other end. Find the decrease in the length of the wire, when the brass weight is completely immersed in water. (Ybrass = 1011 NM-2 ; G = 9.8 ms2 ; water = 1 g cm-3 . ) When the weight is hung in water, it loses weight which is equal to the weight of the liquid displaced, according to Archimedes, principle . Weight of the body hung in a liquid or water = weight of the body in air weight of the liquid displaced . The decrease in length = 0.49 mm 85 Examples A rectangular plastic block of length 600 mm, width 20 mm and height 400 mm has its lower face fixed to a bench and a shearing force of 200 N is applied to the upper face. If the upper face is displaced by 20 mm, find the shear stress, the shear strain and the shear modulus. ∆x=20 mm = 0.02 m h=400mm = 0.4 m A= 20x600 = 12000 mm2 = 0.012 m2 Stress = F/A = 200/0.012 = 16666.7 N/ m2 Strain = ∆x/h = 0.02/0.4 = 0.05 S = stress/strain = 16666.7/0.05 = 333333.3 N/ m2 86 Questions Copper has a tensile strength of about 3.0 × 108 N/m2. (a) What is the maximum load that can be hung from a copper wire of diameter 0.42 mm? (b) If half this maximum load is hung from the copper wire, by what percentage of its length will it stretch? F/A = 3x108 N/m2 r = 0.21 x 10-3 m The maximum load Fmax = 41.5422 N F = 20.77 N Y=(F/A)/(∆L/L ) Y=12x1010 ∆L/L = 0.00125 = 0.125% 87 Questions An elevator cable is to be made of a new type of composite developed by Acme Laboratories. In the lab, a sample of the cable that is 2.00 m long and has a cross-sectional area of 0.200 mm2 fails under a load of 1000 N. The actual cable used to support the elevator will be 20.0 m long and have a cross-sectional area of 1.20 mm2. It will need to support a load of 20,000 N safely. Will it? Stress = F/A = 1000/(0.2) = 500 N/mm2 Stress = F/A = 20000/(1.2) = 16666.7 N/mm2 It will fail to support the load, the stress is larger 88 Questions A ladder with length L and mass m rests against a wall. Its upper end is a distance h above the ground. The center of gravity of the ladder is one-third of the way up the ladder. A firefighter with mass M climbs halfway up the ladder. Assume that the wall, but not the ground, is frictionless. What is the force exerted on the ladder by the wall and by the ground ? 89 Questions ∑F=0 Fw – Fgx = 0 Fw = Fgx Fgy – mg – Mg = 0 Fgy = (m+M)g ∑ τ=0 mg (a/3) + Mg (a/2) – Fwh = 0 Fwh = ag ((m/3)+(M/2)) Fw = (g/h) ((m/3)+(M/2)) 90 91 92 5.1 Newton’s Law of Universal Gravitation Before Newton, scientists studied the motion of planets but did not have any idea about the forces between them. Then Newton discovered gravity when an apple fell on his head and he suggested that there are forces between planets that keep them together and moving in the same path. Newton’s Law of Universal Gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. where 93 5.2 Free Fall Acceleration In the same way, Earth attracts objects with a gravitational force. When the object is on the surface of the Earth Fg mg g GMe m re2 GMe m re2 GMe re2 If the object is not on the surface. g GMe re h 2 94 5.3 Kepler’s Laws and Motion of Planets Kepler's first law The path of each planet around the sun is an ellipse with the sun at one focus. If F1 and F2 are the two foci, P is the planet and S is the sun: F1P + F2P = constant. Kepler's second law A line joining the sun to the planet sweeps out equal areas in equal intervals of time. If the time taken by a planet to travel from P1 to Q1 is equal to the time taken to travel from P2 to Q2, the areas covered are equal (shaded region). Kepler's third law A planet moves around the sun in such a way that the square of its time period is proportional to the cube of the semi-major axis of its elliptical orbit. If T is the time period of revolution and 'a' the semi-major axis then, for circular orbits 95 2 3 a = r (radius) , T =Kr 5.3 Kepler’s Laws and Motion of Planets Deriving Kepler's third law from Newton’s law Assuming the path of the planet around the sun is circular (in real life they are almost circular except for Mercury and Pluto) We know that ac=v2/r And Newton’s law is F= ma Where Ms is the mass of the sun and Mp is the mass of the planet. The orbital speed is 2лr/T Where T is the period or the time to complete a whole revolution F GM s M p M pac r2 GM s M p v2 GM s 2 r r r2 GM s 2r r T 4 2 3 2 T r GM s 2 T Kr 2 3 96 5.4 Gravitational Field The field lines are directed radically inwards, because at any point in the Earth’s field, a body will feel a force directed toward the centre of the Earth. The field lines become more spread out as the distance from the Earth increases. GMe g 2 9.8 N / kg re 97 5.5 Gravitational Potential Energy The general expression for gravitational potential energy arises from the law of gravity and is equal to the work done against gravity to bring a mass to a given point in space. happens when acted on by external forces. U U f U i F (r )dr rf ri U f Ui GMe m r r 2 dr i rf U f U i GMe m rf ri When ri =infinity then Ui=0 Uf dr r2 GMe m r In the case of multiple objects U is the work needed to separate those particles, for example for 3 particles: Utot=U12+U13+U23 98 5.6 Energy Considerations in Planetary and Satellite Motion E K U The total energy of a planet orbiting is Where GMm mv2 Fg 2 ma r r Which leads to E Then 1 2 GMm mv 2 r K GMm 2r E K U E GMm GMm r 2r GMm E 2r In case of elliptical orbits where a is the semi major axis E GMm 2a 99 Examples The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. If the radius of the Moon is about 0.250 RE, find the ratio of their average densities. gmoon/gearth = 1/6 6GMmoon/rmoon2= GMearth/rearth2 M=Vρ ρmoon/ ρearth = 2/3 100 Examples Plaskett’s binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This means that the masses of the two stars are equal . Assume the orbital speed of each star is 220 km/s and the orbital period of each is 14.4 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.) V=s/T=2πr/T r=4.359x1010 m V2/r=GM/(2r)2 M=1.265x1032 kg 101 Examples Three uniform spheres of mass 2.00 kg, 4.00 kg, and 6.00 kg are placed at the corners of a right triangle. Calculate the resultant gravitational force on the 4.00-kg object, assuming the spheres are isolated from the rest of the Universe. F64=Gm6m4/r264 F64=G(6)(4)/16 F64=10.005x10-11N F24=Gm2m4/r224 F24=G(2)(4)/9 F24=5.93x10-11N Ftot=-10.005x10-11i+5.93x10-11j 102 Examples Three objects of equal mass are located at three corners of a square of edge length. Find the gravitational field at the fourth corner due to these objects. gx=(Gm/l2)+(Gmcos45/(2l2)) gx=(Gm/l2)+ (0.707Gm/2l2) gx=(Gm/l2)+0.354 (Gm/l2)=9.03x10-11(m/l2) gy=(Gm/l2)+(Gmcos45/(2l2)) gy=(Gm/l2)+(0.707Gm/2l2) gy =(Gm/l2)+0.354 (Gm/l2)=9.03x10-11(m/l2) 103 Questions If the mass of Mars is 0.107 ME and its radius is 0.53 RE, estimate the gravitational field g at the surface of Mars. gmars = GMmars/rmars2 gmars = G0.107Mearth/(0.53rearth)2 gmars = G0.107Mearth/(0.53rearth)2 gmars = 0.381GMearth/rearth2 gmars = 0.381gearth gmars = 0.381(9.8)=3.733 N/kg 104 Questions During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between Earth and the Sun. (a) What force is exerted on the Moon by the Sun? (b) What force is exerted on the Moon by Earth? (c) What force is exerted on Earth by the Sun? Where mass of sun =1.991x1030 kg, mass of moon =7.36x1022 kg, mass of earth =5.98x1024 kg, Distance between moon and sun = 1.496x1011 m Distance between moon and earth = 3.81x1010 m Fmoon-sun=Gmmms/rms2 Fmoon-sun=G(7.36x1022)(1.991x1030)/(1.496x1011)2 Fmoon-sun=43.67x1019N Fmoon-earth=20.22x1015N Fearth-sun=22.54x1023N 105 Questions A satellite of Earth has a mass of 100 kg and is at an altitude of 2.0 x 106 m. (a) What is the potential energy of the satellite-Earth system? (b) What is the magnitude of the gravitational force exerted by Earth on the satellite? U=GMm/r U=GME (100)/(RE+(2.0 x 106 )) U=G(5.98x1024 (100)/(6371000+(2.0 x 106 )) U=4.76x109 J F=GMm/r2 F=568.6 N 106 107 108 6.1 Periodic and Simple Harmonic Motion Periodic motion is the motion which is repeated in equal intervals of time. Simple harmonic motion is a special type of periodic motion where a force acting on the object always acting towards the position of equilibrium. 109 6.2 Motion of an Object Attached to a Spring When a pulling force is applied to a spring it stretches and undergoes a displacement of x from its original, or “unstrained”. Also when a pushing force is applied to the spring, and it again undergoes a displacement from its unstrained length. Experiment reveals that for relatively small displacements, the applied force required to stretch or compress a spring is directly proportional to the displacement x. F=-kx Where the constant k is called the spring constant. 110 6.2 Motion of an Object Attached to a Spring When the restoring force has the mathematical form given by F=–kx, the type of friction-free motion illustrated in the Figure below and is called as “simple harmonic motion.” By attaching a pen to the object and moving a strip of paper past it at a steady rate, we can record the position of the vibrating object as time passes. F = ma F = - kx ma = - kx ax = - (k/m) x 111 6.3 The Particle in Simple Harmonic Motion Simple harmonic motion, like any motion, can be described in terms of displacement, velocity, and acceleration. Using a reference circle (radius=A) that indicates how to determine the displacement. The ball starts on the x axis at x=+A and moves through the angle θ in a time t. Since the circular motion is uniform, the ball moves with a constant angular speed w (in rad/s). Therefore, the angle has a value (in rad) of θ =wt. The displacement x can be written by: x = A cos θt = A cos wt If the object doesn’t start at zero angle we add the phase φ x = A cos (wt+ φ) As time passes, the shadow of the ball oscillates between the values of x=+A and x=–A, corresponding to the limiting values of +1 and –1 for the cosine of an angle. The radius A of the reference circle, then, is the amplitude of the simple harmonic motion. 112 6.3 The Particle in Simple Harmonic Motion As the ball moves one revolution or cycle around the reference circle, its shadow executes one cycle of back-and-forth motion. For any object in simple harmonic motion, the time required to complete one cycle is the period T. w = 2л/T Often, instead of the period, it is more convenient to speak of the frequency f of the motion, the frequency being just the number of cycles of the motion per second. f = 1/T The velocity can be found from the displacement v = dx/dt v = -Aw sin (wt+ φ) vmax = Aw The acceleration can be found from the velocity a= a = dv/dt cos(wt+ φ)= w2 x amax = Aw2 Aw2 113 6.4 Energy in Simple Harmonic Motion The object's kinetic energy is K = (1/2)mv2 = (1/2)mω2A2sin2(ωt + φ) Its potential energy is elastic potential energy. The elastic potential energy stored in a spring displaced a distance x from its equilibrium position is U=(1/2)kx2. The object's potential energy therefore is U = (1/2)kx2 = (1/2)mω2x2 = (1/2)mω2A2cos2(ωt + φ) The total mechanical energy of the object is E = K+U = (1/2)mω2A2(sin2(ωt + φ)+cos2(ωt + φ)) = (1/2)mω2A2 The energy E in the system is proportional to the square of the amplitude E = (1/2)kA2 It is a continuously changing mixture of kinetic energy and potential energy. 114 6.5 The Pendulum An ideal simple pendulum consists of a point mass m suspended from a support by a massless string of length L. If the mass is displaced from its equilibrium position while keeping the string taut, it exhibits periodic motion, moving in a vertical plane along a circular arc. F = -mg (sinθ) = ma = md2s/dt2 And for small angles sin θ = θ , s=Lθ Then d2 θ /dt2 = -g θ / L Which is equivalent to the simple harmonic motion equation θ(t) = θmaxcos(ωt + φ) Where w2=g/L 115 6.5 The Pendulum A physical pendulum is an object suspended in a uniform gravitational field from a point other than its center of mass. The object can rotate about an axis through the suspension point. When the CM is displaced from its stable equilibrium point under the support, the gravitational force exerts a torque about the support, resulting in angular acceleration. The CM accelerates towards its equilibrium position. The object exhibits periodic motion. For small displacements, when sinθ ~ θ, the motion is simple harmonic, θ(t) = θmaxcos(ωt + φ), with ω2 = (mgd)/I. Here I is the moment of inertia of the object about the axis of rotation through the support and d is the perpendicular distance of the CM from the axis of rotation through the support. 116 Examples A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find (a) the displacement of the particle,(b) its velocity, and(c) its acceleration. (d) Find the period and amplitude of the motion. (a) The displacement as a function of time is x(t) = Acos(ωt + φ). Here ω = 2/s, φ = π/6, and A = 5 cm. The displacement at t = 0 is x(0) = (5 cm)cos(π/6) = 4.33 cm. (b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s. (c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2. (d) The period of the motion is T = πs, and the amplitude is 5 cm. 117 Examples A 20 g particle moves in simple harmonic motion with a frequency of 3 oscillations per second and an amplitude of 5cm. (a) Through what total distance does the particle move during one cycle of its motion? (b) What is its maximum speed? Where does that occur? (c) Find the maximum acceleration of the particle. Where in the motion does the maximum acceleration occur? (a) The total distance d the particle moves during one cycle is from x = -A to x = +A and back to x = -A, so d = 4A = 20 cm. (b) The maximum speed of the particle is vmax = ωA = 2πfA = 2π 15 cm/s = 0.94 m/s. The particle has maximum speed when it passes through the equilibrium position. (c) The maximum acceleration of the particle is amax = ω2A = (2πf)2A = 17.8 m/s2. The particle has maximum acceleration at the turning points, where it has maximum displacement. 118 Examples A 1kg mass attached to a spring of force constant 25N/m oscillates on a horizontal frictionless track. At t = 0 the mass is released from rest at x = 3cm, that is the spring is compressed by 3cm. Neglect the mass of the spring. Find (a) The period of its motion, (b) the maximum value of its speed and acceleration, and (c) the displacement, velocity and acceleration as a function of time. (a) The period is T = 2πSQRT(m/k) = 2πSQRT(1 s2/25) = 1.26 s. (b) The angular acceleration is ω = SQRT(k/m) = 5/s. The maximum speed is vmax = ωA = 15 cm/s. The maximum acceleration of the particle is amax = ω2A = 0.75 m/s2. (c) x(t) = Acos(ωt + φ) = (3 cm)cos((5/s)t + π) = -(3 cm)cos((5/s)t) v(t) = -ωAsin(ωt + φ) = (15 cm/s)sin((5/s)t) a(t) = -ω2Acos(ωt + φ) = (0.75 m/s2)cos((5/s)t) 119 Examples A child swings on a playground swing with a 2.5 m long chain. a) What is the period of the child’s motion? b) What is the frequency of the vibration? a) for a pendulum T = 2*π*sqrt (l/g) = 2*3.14*sqrt(2.5/9.8) T=3.1735 sec/swing b) f = 1/T = 1/3.17=.315 swings/sec=.315 Hz By what factor should the length of a simple pendulum be changed if the period of vibration were to be tripled? T2=3T1 2π (l2/g)(1/2)=(3) 2π (l1/g)(1/2) (l2)(1/2)=(3) (l1)(1/2) (l2)=(9) (l1) 120 Questions The position of a particle is given by x = 2.5 cos πt, where x is in meters and t is in seconds. (a) Find the maximum speed and maximum acceleration of the particle. (b) Find the speed and acceleration of the particle when x = 1.5 m. 121 Questions A particle moves in a circle of radius 15 cm, making 1 revolution every 3 s. (a) What is the speed of the particle? (b) What is its angular velocity ω? (c) Write an equation for the x component of the position of the particle as a function of time t, assuming that the particle is on the positive x axis at time t = 0. A 2.4-kg object is attached to a horizontal spring of force constant k = 4.5 kN/m. The spring is stretched 10 cm from equilibrium and released. Find its total energy. 122 Questions Find the total energy of a 3-kg object oscillating on a horizontal spring with an amplitude of 10 cm and a frequency of 2.4 Hz. A 3-kg object oscillating on a spring of force constant 2 kN/m has a total energy of 0.9 J. (a) What is the amplitude of the motion? (b) What is the maximum speed? A 2.4-kg object is attached to a horizontal spring of force constant k = 4.5 kN/m. The spring is stretched 10 cm from equilibrium and released. Find (a) the frequency of the motion, (b) the period, (c) the amplitude, (d) the maximum speed, and (e) the maximum acceleration. (f) When does the object first reach its equilibrium position? What is its acceleration at this time? 123 Questions A 3-kg object attached to a horizontal spring oscillates with an amplitude A = 10 cm and a frequency f = 2.4 Hz. (a) What is the force constant of the spring? (b) What is the period of the motion? (c) What is the maximum speed of the object? (d) What is the maximum acceleration of the object? A simple pendulum of length L is released from rest from an angle φ0. (a) Assuming that the pendulum undergoes simple harmonic motion, find its speed as it passes through φ = 0. (b) Using the conservation of energy, find this speed exactly. (c) Show that your results for (a) and (b) are the same when φ0 is small. (d) Find the difference in your results for φ0 = 0.20 rad and L = 1 m. 124 Questions A 3-kg plane figure is suspended at a point 10 cm from its center of mass. When it is oscillating with small amplitude, the period of oscillation is 2.6 s. Find the moment of inertia I about an axis perpendicular to the plane of the figure through the pivot point. The displacement of a particle at t = 0.25 s is given by the expression x = (4.0 m) cos (3.0 p t + p), where x is in meters and t is in seconds. Determine (a) the frequency and period of the motion, (b) the amplitude of the motion, (c) the phase constant, and (d) the displacement of the particle at t = 0.25 s x=Acos(wt+ φ) w=3π T=0.66 s f=1.5 Hz A=4 m φ= π x at t=0.25 x=2.83 m 125 Questions A block of unknown mass is attached to a spring of spring constant 6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be + 30.0 cm/s. Calculate (a) the mass of the block, (b) the period of the motion, and (c) the maximum acceleration of the block. When x = A/2 = 5cm = 0.05 m then v=0.3 m/s x = Acos(wt) then cos(wt) = 0.5 w=(k/m)(1/2) m=0.5 kg T=1.8 s amax=Aw2 126 127