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11-Apr-16 Example: A 4.00 kg particle moves along the x-axis. Its position varies with time according to x(t) = t + 2t3, where x is measured in meters and t is in seconds. Example continued How much additional work must be done to stretch the spring to 0.80 m? (a) What is the particle’s kinetic energy? 2 W F dr 0.80 m Fdx 0.40 m 0.80 m K 1 2 kx 2 0.40 m 2 1 2 1 dx mv m 2 1 6t 2 J 2 2 dt (b) What is the particle’s acceleration? What is the net force acting on the particle? 1 20.0 N/m0.80 m 2 0.40 m 2 4.8 J 2 a Note that the result is not 1.6 J. d 2x 12t m/s2 dt 2 Fnet ma 48t N 57 58 Example: A box of mass m is towed up a frictionless incline at constant speed. The applied force F is parallel to the incline. What is the net work done on the box? Example continued (c) The power delivered to the particle at time t? y F P Fv 48t 1 6t 2 48t 288t 3 W Nrb Fpb x (d) The work done on the particle from t = 0 to t = 2 sec? xf tf 2 t f dx W F dr Fdx F dt Fv dt dt xi ti ti web Apply Newton’s 2nd law: 48t 288t3 dt 1248 J 0 59 F F x Fpb web sin 0 y N rb web cos 0 60 15 11-Apr-16 Example continued Example continued Fpb F mg sin The magnitude of F is: The work by the normal force is WN N rb x cos 90 0 If the box travels along the ramp a distance of x then the work by the force F is WF Fx cos0 mgx sin The net work done on the box is Wnet WF Wg WN The work by gravity is mgx sin mgx sin 0 0 Wg web x cos 90 mgx sin 61 Example: A raindrop of mass 3.3510-5 kg falls vertically at constant speed under the influence of the forces of gravity and drag. In falling through 100 m… Example: What is the net work done on the box in the previous example if the box is not pulled at constant speed? F F w sin ma F ma w sin x 62 eb (a) What is the work done by gravity? eb y Proceeding as before: W F dr Far Wnet WF Wg WN x ma mg sin x mgx sin 0 ma x Fnet x yf mg yˆ dy yˆ yi wer 63 mgy 0.033 J 64 16 11-Apr-16 Example: A 2.00 kg mass hangs suspended from a string. The ceiling is 3.00 m above the floor and the length of the string is 1.00 m. Example continued (b) What is the work done by drag? ceiling (a) What is the gravitational potential energy of the mass if the ceiling is the reference point? Wnet Wdrag Wg 0 Wdrag Wg mgy 0.033 Joules U g mgy 2.00 kg 9.8 m/s2 1.00 m 19.6 J floor 65 66 Example: A cart starts from position 4 with v = 15.0 m/s to the left. Find the speed of the cart at positions 1, 2, and 3. Ignore friction. Example continued (b) What is the gravitational potential energy of the mass if the floor is the reference point? U g mgy 2.00 kg 9.8 m/s2 2.00 m 39.2 J E4 E3 (c) What is the gravitational potential energy of the mass if the ball’s location is the reference point? U g mgy 1 1 mgy4 mv42 mgy3 mv32 2 2 2.00 kg 9.8 m/s 0 m 0J 2 U 4 K 4 U 3 K3 v3 v42 2 g y4 y3 20.5 m/s 67 68 17 11-Apr-16 Example continued E4 E2 U 4 K4 U 2 K2 The work-energy theorem can be used to solve problems instead of using Newton’s laws. Or use E3=E2 Determine the speed of a mass m that starts from rest and slides along a frictionless ramp for some distance d using (a) Newton’s laws and (b) the work-energy theorem. 1 1 mgy4 mv42 mgy2 mv22 2 2 v2 v 2 g y4 y2 18.0 m/s 2 4 E4 E1 U 4 K 4 U 1 K1 1 1 mgy4 mv42 mgy1 mv12 2 2 Answer: v 2gd sin Or use E3=E1 Repeat the calculation but take the coefficient of kinetic friction to be k. E2=E1 Answer: v1 v42 2 g y4 y1 24.8 m/s v 2 gd sin k cos 69 70 Example: Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of 10.0 grams. Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.00 cm, 5.00 cm)? Momentum and Collisions y Use conservation principles to calculate momentum and energy transfer between two or more interacting bodies. x A 71 x xcm ma xa mb xb mb xb ma mb ma mb ycm ma ya mb yb mb yb ma mb ma mb 72 18 11-Apr-16 Example: The positions of three particles are (4.00 m, 0.00 m); (2.00 m, 4.00 m); and (-1.00 m, -2.00 m). The masses are 4.00 kg, 6.00 kg, and 3.00 kg respectively. What is the location of the center of mass? Example continued 10.0 g xb 2.00 cm 10.0 g 30.0 g xb 8.00 cm xcm y 2 10.0 g yb 5.00 cm 30.0 g 10.0 g yb 20.0 cm ycm 1 x 3 rb 8.00 xˆ 20.0 yˆ cm 73 Example: In a railroad freight yard an empty freight car of mass m rolls along a straight level track at 1.00 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.00m. The two cars couple together upon collision. Example continued xcm m1 x1 m2 x2 m3 x3 m1 m2 m3 4.00 kg 4.00 m 6.00 kg 2.00 m 3.00 kg 1.00 m 4.00 6.00 3.00 kg (a) What is the speed of the two cars after the collision? 1.92 m ycm 74 pi p f p1i p2i p1 f p2 f m y m2 y2 m3 y3 1 1 m1 m2 m3 m1v1 0 m1v m2 v m1 m2 v 4.00 kg 0.00 m 6.00 kg 4.00 m 3.00 kg 2.00 m 4.00 6.00 3.00 kg m1 v v1 0.20 m/s m1 m2 1.38 m 75 76 19 11-Apr-16 Example: Body 1 of mass M has an original velocity of 6.00 m/s in the +xdirection toward a stationary body 2 of the same mass. After the collision, body 1 has vx=+1.00 m/s and vy=+2.00 m/s. What is the magnitude of body 2’s velocity after the collision? Example continued (b) Suppose instead that both cars are at rest after the collision with what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.00 m/s. pi p f Final Initial 1 p1i p2i p1 f p2 f m1v1i m2 v2i 0 0 1 v1i m v2i 1 v1i 0.25 m/s m2 2 2 77 Example: Head-on 1D elastic collision. The values of v1i and v2i are known. Determine the velocity of the pucks after the collision. Example continued y momentum: x momentum: pix p fx piy p fy p1ix p2ix p1 fx p2 fx p1iy p2iy p1 fy p2 fy m1v1ix 0 m1v1 fx m2 v2 fx Initial 1 0 0 m1v1 fy m2 v2 fy v1i Final v2i v2f 1 v1f 2 2 Solve for v2fy: Solve for v2fx: v2 fx 78 Take Fext = 0 so that pi = pf. m1v1ix m1v1 fx v2 fy m2 v1ix v1 fx m1v1 fy m2 pi p f v1 fy p1i p2i p1 f p2 f 2.00 m/s m1v1i m2 v2i m1v1 f m2 v2 f 5.00 m/s The mag. of v2 is v2 f v 2 2 fy v 2 2 fx 5.40 m/s m1 v1i v1 f m2 v2 f v2i 79 (1) 80 20 11-Apr-16 Example continued Example continued The collision is elastic so Ki = Kf.. v1i v1 f v2i v2 f Divide (3) by (1) to get Ki K f K1i K 2i K1 f K 2 f Take mv m v mv m v 2 1 1i 2 2 2i 2 1 1f 2 2 2f v2 f v1i v1 f v2i substitute into (1) and solve for v1f m m2 2m2 v1i v2i v1 f 1 m1 m2 m1 m2 (2) Rewrite (2): m1v12i m1v12f m2v22 f m2v22i v m v m1 v12i v12f m2 v22 f v22i m1 v1i v1 f (3) 1i v1 f 2 2i v2 f v Then solve for v2f: 2i 2m1 m2 m1 v2 f v1i m m v2i 2 m1 m2 1 v2 f 81 Example: An elastic collision in 2D. Take particles A and B to be protons and vAi = 3.50105 m/s and vBi = 0. After the collision, particle A travels at an angle = 37 with the x-axis and particle B travels at an unknown angle . Determine the velocities of the protons after the collision and also the value of . 82 Take Fext = 0 so that pi = pf. pix p fx x-component: p Ai , x pBi , x p Af , x p Bf , x m Av Ai 0 m Av Af cos mB vBf cos piy p fy Final Initial A A y-component: vAi p Ai , y p Bi , y p Af , y p Bf , y 0 0 m Av Af sin mB vBf sin B Ki K f B Kinetic energy: 83 K Ai K Bi K Af K Bf 2 2 mAv Ai mAv Af mB vBf2 84 21 11-Apr-16 Example continued Example continued vAi vAf cos vBf cos (1) 0 vAf sin vBf sin (2) Summary, so far: 2 2 v Ai v Af vBf2 (3) Now use (3) to determine vBf: 2 2 vBf v Ai v Af vAi vAf cos vBf cos Rewrite (1) and (2): 2 v Ai 1 cos2 v Ai sin 2.11105 m/s vAf sin vBf sin Now use (2) to solve for : Square each and add together (and use a trig identity): v v 2v Ai v Af cos v 2 Ai 2 Af 2 Bf v Af v Ai cos 2.80 105 m/s Use (3) to simplify to: sin v Af vBf sin 0.8 53 85 86 Example continued Example: The Ballistic Pendulum. A bullet of mass mb is fired at a block of wood of mass mp. The bullet embeds in the wood and the block rises a height h before coming to rest. What was the speed of the bullet as it entered the block? What is v? Even though energy is not conserved during the collision, it will be conserved during the upward swing. This is in inelastic collision. h vb mb mp pi p f Ei E f pbi p pi pbf p pf Ki U i K f U f mb vb 0 mb m p v 1 mb m p v 2 0 mb m p gh 2 v 2 gh mb m p v vb mb 87 88 22 11-Apr-16 Example continued Example: A 3.00 gram particle is moving toward a 7.00 gram particle with a speed of 3.00 m/s. How much kinetic energy is lost in the collision? (a) With what speed does each particle approach the center of mass? K K f K i 1 mb m p v 2 1 mbvb2 2 2 vcm 2 1 1 mb m p 2 v mb m p v 2 mb 2 2 mb v1m1 v2 m2 0.900 m/s m1 m2 v1 v1 vcm 1 mb m p m p v 2 2 mb 3.00 m/s 0.900 m/s 2.10 m/s v2 v2 vcm 0.00 m/s 0.900 m/s 0.900 m/s 89 Example continued 90 Momentum in the center of mass frame (b) What is the momentum of each particle relative to the center of mass? ptotal p1 p2 m1v1 m2v2 m1 v1 vcm m2 v2 vcm p1 m1v1 0.0063kg m/s p2 m2v2 0.0063kg m/s Note: the total momentum relative to the center of mass is zero. m1m2 v1 v2 m1m2 v2 v1 0 m1 m2 m1 m2 The momentum of a system in the center of mass frame is zero. 91 92 23 11-Apr-16 Example: A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find the constant angular acceleration of the centrifuge. Rotational Momentum and Inertia 3600 Systems of particles that involve rotation require separate equations to account for total system energy and momentum. rev 2 rad 1 min 120 rad/sec min 1 rev 60 sec 2 rad 50 rev 100 rad 1 rev 2 02 2 0 93 Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the rate of 1.2610-5 sec/year. 72 rad/sec2 94 Example continued (b) If the pulsar’s angular acceleration is constant, how many years from now will the pulsar stop rotating? 0 t 0 2 T0 t 0 9 1010 sec 2800 years (a) What is the pulsar’s angular acceleration? 02 2 (c) The pulsar originated in a supernova explosion in the year 1054 A.D. What was the initial period for the pulsar? d d 2 2 dT 2 2 109 rad/sec2 dt dt T T dt dT sec 1 year -13 sec 1.26 105 4 10 dt year 3.15107 sec sec 95 t 953 years 3.0 1010 sec 2 0 t t 250 rad/sec T 2 T 0.025 sec 25 msec 0 96 24 11-Apr-16 Example continued The KE of the pulsar decreases over time. The rate of energy loss is (d) What is the rotational kinetic energy of the pulsar? K rot 1 2 I 2 Assume the pulsar to be a solid sphere of uniform density. M = 1.4Msun and R = 10 km. P 2 MR 2 5 1 1038 kg m 2 For comparison, the luminosity of the sun is 3.861026 W. I The moment of inertia for a sphere is dE d 1 2 d I I 4 1031 W dt dt 2 dt 190 rad/sec K rot 1 2 I 2 1042 J 2 97 98 Example: (a) Find the moment of inertia of the system below. The masses are m 1 and m2 and they are separated by a distance r. Assume the rod connecting the masses is massless. Example: What is the moment of inertia of a thin disk when you add a small mass near the rim? The disk has a rotation axis though its center and points out of the page. R r1 and r2 are the distances between mass 1 and the rotation axis and mass 2 and the rotation axis (the dashed, vertical line) respectively. I new I disk mr 2 1 M disk R 2 mR 2 2 m1 99 r1 r2 m2 100 25 11-Apr-16 Parallel Axis theorem Example continued Take m1 = 2.00 kg, m2 = 1.00 kg, r1= 0.33 m , and r2 = 0.67 m. I new I cm md 2 2 I mi ri 2 m1r12 m2 r22 0.67 kg m 2 Where m is the mass of the object and d is the distance the axis is moved. Note: the new axis must be parallel to the original axis! i 1 (b) What is the moment of inertia if the axis is moved so that is passes through m1? 2 I mi ri 2 m2 r22 m2 r 2 1.00 kg m 2 i 1 101 Example: A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min. When switched off, it rotates 50 times before coming to rest. Find the constant angular acceleration of the centrifuge. Example: Rank the I1, I2, I3, and I4 from greatest to smallest. a 2 1 102 I new I cm md 2 3 4 I 3 I cm b a I 4 I cm m 2 2 1 m a2 b2 12 a b I 2 I cm m 2 2 2 2 3600 rev 2 rad 1 min 120 rad/sec min 1 rev 60 sec 2 rad 50 rev 100 rad 1 rev 2 02 2 0 I1 I cm mL2 Ranking: I1 > I2 > I4 > I3 103 02 2 72 rad/sec2 104 26 11-Apr-16 Example: A pulsar is a rapidly spinning neutron star that emits a radio beam like a lighthouse emits a light beam. We receive a radio pulse for each rotation of the star. The period T of rotation is found by measuring the time between pulses. The pulsar in the Crab Nebula has a period of T = 0.033 sec that is increasing at the rate of 1.2610-5 sec/year. Example continued (b) If the pulsar’s angular acceleration is constant, how many years from now will the pulsar stop rotating? 0 t 0 2 T0 t 0 9 1010 sec 2800 years (a) What is the pulsar’s angular acceleration? (c) The pulsar originated in a supernova explosion in the year 1054 A.D. What was the initial period for the pulsar? d d 2 2 dT 2 2 109 rad/sec2 dt dt T T dt dT sec 1 year -13 sec 1.26 105 4 10 dt year 3.15107 sec sec 105 t 953 years 3.0 1010 sec 2 0 t t 250 rad/sec T 2 T 0.025 sec 25 msec 0 106 Example continued The KE of the pulsar decreases over time. The rate of energy loss is (d) What is the rotational kinetic energy of the pulsar? K rot 1 2 I 2 Assume the pulsar to be a solid sphere of uniform density. M = 1.4Msun and R = 10 km. The moment of inertia for a sphere is P 2 MR 2 5 1 1038 kg m 2 dE d 1 2 d I I 4 1031 W dt dt 2 dt For comparison, the luminosity of the sun is 3.861026 W. I 190 rad/sec K rot 1 2 I 2 1042 J 2 107 108 27 11-Apr-16 Example: Calculate the torque due to the three forces shown about the left end of the bar (the red X). The length of the bar is 4m and F2 acts in the middle of the bar. Top view of door F r Hinge end F2=30 N 30 F3=20 N Line of action of the force X Lever arm r r r r sin sin The torque is: F1=25 N r F rF sin 10 45 Same as before 109 Example continued 110 Example continued Lever arm for F2 F2=30 N 30 F3=20 N The torques are: 10 X 45 1 0 2 1.73 m 30 N 51.9 Nm 3 0.695 m 20 N 13.9 Nm F1=25 N Lever arm for F3 The lever arms are: The net torque is + 65.8 Nm and is the sum of the above results. r1 0 r2 2m sin 60 1.73 m r3 4m sin10 0.695 m 111 112 28