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notes_8 - Center of Mass So far we have considered all...

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Unformatted text preview: Center of Mass So far we have considered all objects as “point” masses; CHAPTER 8 what happens to the equations of motion, etc., if we have SYSTEMS OF PARTICLES & CONSERVATION OF MOMENTUM a large or “extended” system (like a collection of masses • Center of Mass o finding the center of mass or an object with a complicated shape)? Examples : (1) 2-d molecule ( H2O): • Motion of the center of mass (2) dumbell: • Conservation of momentum • Collisions o impulse o inelastic collisions o elastic collisions o coefficient of restitution (3) hammer: (4) pool balls: Then the concept of a center of mass becomes important because we can relate the motion of the object ( or objects) to the motion of the center of mass. If we have a collection of masses, we choose some origin, then the center of mass (cm) relative to that origin is at: r rcm = ( x cm , y cm , zcm ), y y r rcm cm mi (xi , yi , zi ) r rcm cm r ri mi (xi , yi , zi ) x O x O where: x cm = m1x1 + m2 x 2 + ... ∑i mi x i ∑i mi x i = = , m1 + m2 + ... M ∑i mi y cm = m1y1 + m2 y 2 + ... ∑i mi y i ∑i mi y i = = , m1 + m2 + ... M ∑i mi zcm = m1z1 + m2 z2 + ... ∑i mi zi ∑i mi zi = = . m1 + m2 + ... M ∑i mi and r r r ∑i mi ri ∑i mi ri i.e., rcm = = M ∑i mi r r or M rcm = ∑i mi ri where M ( = ∑i mi ) is the total mass of the ensemble. y 1m So, how can I find the center of mass? 1m 2m 3m 1 1. Calculation (possibly using symmetry). 2. Experimentally. origin (0,0) 3 1m 3m 2 1m 1m x 1m Consider the sheet as 3 separate object (masses), then mass 1 + mass 2 + mass 3 = 35kg. Question 1: Find the center of mass of the sheet of metal of uniform thickness, shown below, relative to the lower left hand corner. The mass of the sheet is 35kg. 1m 1m Area 1 + Area 2 + Area 3 = (3 + 1 + 3)m2 = 7 m2 . Since the sheet has constant thickness, mass ∝ area . ∴ mass 2 = 35 7 = 5.00 kg 2 . m m mass 1 = mass 3 = 3 × 5.00 = 15.0 kg, 2m 3m 1m 3m and mass 2 = 1 × 5.00 = 5.00 kg Choose axes and an origin (0,0). By symmetry, the positions of the individual cm’s are: 3m 1 (0.5,1.5): 2 (1.5,0.5): 3 (2.5,1.5) The coordinates of the cm are given by: ∑ mx ∑ my x cm = i i i and y cm = i i i M M (15.0 × 0.5) + (5.00 × 1.5) + (15.0 × 2.5) ∴ x cm = , 35.0 52.5 = 1.50 m . i.e., x cm = 35.0 ~ Surprise, surprise ?? The gravitational potential energy of an ensemble of particles is easily found: m3 m1 y3 y1 y2 m2 also y cm = (15.0 × 1.5) + (5.00 × 0.5) + (15.0 × 1.5) 35.0 47.5 = 1.36 m . i.e., y cm = 35.0 r ∴ rcm = (1.50 m,1.36 m) , from the origin chosen. DISCUSSION PROBLEM [8.1]: If the y-direction is vertical then: UG = ∑i Ui = ∑i mi gy i = g∑i mi y i . ∑ my But y cm = i i i . M ∴ UG = Mgy cm, i.e., UG depends on the total mass and the vertical coordinate of the cm only. Can you think of another way with less calculations? (HINT: not by choosing another origin but what about NOTE: We have assumed that the value of g is the same negative mass?) for all particles, i.e., g is constant. The center of mass of an object can also be located by integration. The x- component for the object shown is 1 x cm = ∑i x i ( ∆mi ), M y yi r ri where ∆mi is an ∆mi element located at x O xi ( x i , y i ) and M is the mass of the object. If the number of elements → ∞, then ∆mi → 0 , and 1 1 x cm = Limit ∑i x i ( ∆mi ) = ∫ x.dm. M ∆mi →0M Similarly, for the y- and z-components: 1 1 y cm = ∫ y.dm and zcm = ∫ z.dm. M M Also, generalizing an earlier expression, we can write: 1r r rcm = ∫ r .dm. M Here are two examples of using integration to locate the center of mass of an object of uniform thickness. Question 2: Show that the center of mass of a uniform rod is half-way along its length. Choose the x-axis along the length of the rod and let its cross-sectional area be A. y dm x dx x x=l Then, from before, we have x cm = triangular sign shown below, relative to the left hand 1 ∫ x.dm. M If the density of the rod is ρ, then dm = ρAdx and M = ρAl , so x cm = 1l 1l xρAdx = ∫ x.dx ∫ ρAl 0 l0 l 1 x2 1 = = l, 2 l 2 0 i.e., half-way along the rod. Notice that providing the rod is uniform, the position of the cm is independent of the cross-sectional area. Question 3: Find the position of the center of mass of a corner. Dr. Robin’s English tea shop a Assume the sign is of uniform thickness. b We take the left hand corner as the origin. Assume the thickness of the sign is t and its density is ρ. y ˆ j a dm r r y ˆ i b a O ( )() x i j = 2a 3 ˆ + b 3 ˆ . dx x 1r r We start with the expression rcm = ∫ r .dm, M r ˆ j where r = xi + y 2 ˆ , dm = ρty.dx and M = ρt ab 2 . () () 2 2 ˆ ˆ1 ˆ j = ∫ ( xi + ( y 2 )ˆ ).ydx = ∫ xydxi + ∫ y dxj . ab ab 2 a b 0 r ∴ rcm a 2 x3 ˆ b x3 ˆ r i.e., rcm = 2 i + 3 j a 3 0 a 3 0 r rcm × xcm = 2a 3 0 2 Wow! Is there a simpler method?? () But y = b a x , a 2 2 a r ˆ1 ˆ ∴ rcm = ∫ b a x 2dxi + ∫ b a x 2dxj ab 0 20 () () ycm = b 3 The center of mass of an object can be found also by experiment ... by suspending the object from two different points ... ... the center of mass lies on the vertical line through each suspension point. Therefore, the cm is where two such lines intersect. Center of Mass × 1-dimensional motion. 2-dimensional motion. Despite the complicated overall motion of the wrenches, the motion of the cm is simple if the gravitational force is the only external force, e.g., no friction, etc. Then Hey ... why the interest in the center of mass? the translational motion of the cm is described by the normal equations of motion, e.g., r r Fext = Ma cm. × × × cm cm If a rocket explodes in flight, providing that the gravitational force is the only external force, i.e., neglecting air resistance, the cm will continue on a See the article I wrote ... “Center of Mass, binary stars and new planets” parabolic trajectory, as will the individual fragments. on the web-site A similar effect is seen with exploding fireworks: Newton’s Laws for the center of mass: DISCUSSION PROBLEM [8.2]: First Law: If the center of mass of a system is at rest (or is moving with constant velocity) it will remain at rest (or moving with constant velocity) unless a net external force acts on the system. Second Law: A canoeist drops a paddle in the lake. Assuming both If a net external force acts on a system of particles, the the canoe and paddle are stationary, can he retrieve the center of mass is given an acceleration that is paddle by moving towards the bow? ... Why ... or why proportional to, and in the same direction as, the force, r r i.e, Fnet = Ma cm. not? 20m/s 16m/s 1500kg 3000kg East West What happens to the cm if the various parts of a system What’s happening to the cm? are moving relative to each other ... ? By definition, the x position of the cm is ∑ mx x cm = i i i . ∑i mi But v cm = Question 4: A 1500kg car is moving westward with a dx cm = dt speed of 20m/s, and a 3000kg truck is traveling east with a speed of 16m/s. What is the velocity of the center of mass? i.e., v cm = dx i dt = ∑i mi v i = ∑i mi v i , M ∑i mi ∑i mi ∑i mi 3000 × 16 + 1500 × ( −20) = +4 m/s 3000 + 1500 East Even though the car is traveling faster to the West than the truck, the greater product of mass and velocity of the truck means the motion of the cm is to the East. For a collection of objects, the momentum of the In chapter 4, we stated Newton’s 2nd Law in the form: r r r dp d ( mv ) = Fnet = dt dt r r where p = mv is called the momentum of the particle of r mass m and velocity v . (This is a more general expression for the 2nd Law and is more like the way Newton stated it, i.e., ensemble is the vector sum of the individual momenta, i.e., r r r P = ∑i pi = ∑i mi v i We take the time derivative r r rr dpi dP d r = ∑i ( mi v i ) = ∑i = ∑i Fi = Fnet , dt dt dt Newton’s 2nd law for a collection of objects The net applied force equals the rate of change of i.e., a net force acting on a collection of objects produces momentum. r r As we saw earlier, F = ma is applicable only when m is a change in the total momentum. But, conversely, if r r there is no net external force, i.e., Fnet = 0, then P is constant.) constant in time. Dimension: p ⇒ [M] Units: kg.m/s [L ] [ T] (vector) ~ Conservation of momentum ~ If there is zero net external force on a system, the total momentum of the system remains constant. 55kg +2.5m/s v 100kg x Take the woman and the boat as the system ... since there are no external forces, e.g., wind or a push, the total momentum is conserved. Before she jumped, initial momentum Pi = 0 , since the woman and the boat are stationary. Question 5: A woman, of mass 55kg, jumps from the After she jumped, final momentum Pf = 0. bow of a boat, of mass 100kg, onto a dock. If the boat is Taking the + x -direction as the + ve direction and the initially at rest and her velocity is 2.5m/s to the right, velocity of the boat = v , we have: what is the velocity of the boat after she jumps? Pi + Pf = 0 i.e., (55 × 2.5) + (100 × v ) = 0 ∴ v = −1.38 m/s. Since v < 0 it is directed to the left. Let’s look at some “collisions” ... i.e., when a force is applied to change the motion. Examples: DISCUSSION PROBLEM [8.3]: Force ( ×105N) 3 2 (a) (b) 1 A cheetah crouches, motionless, behind a bush (a). It Time (ms) 0 50 100 “Crushing a Beetle” spots an antelope and accelerates after its prey (b). Clearly, at (a) the cheetah’s momentum is zero whereas at (b) it has finite momentum. I thought momentum was supposed to be conserved ... does this scenario violate the law of conservation of momentum? Explain. “Hard” collision F “Soft” collision t r r r dp r From earlier we have: F = , i.e., Fdt = dp. dt If the force is applied for a time ∆t ( = t 2 − t1 ), we define the impulse of the force as: 2r r r t2 r r r I = ∫ Fdt = ∫ dp = p2 − p1 = ∆p , t1 1 i.e., the impulse is the change in momentum. The average force for the time period ∆t is: r r r r 1 t2 r I ∆p ∆( mv ) = = Fav = Fdt = . ∫ ∆t t ∆ t ∆t ∆t 1 Equal areas Fav ∆t Whoops ... a mistake ? ~ Can you spot it? ~ r rr The impulse I = ∆p = Fav ∆t (vector) [L ] Dimension: same as momentum ⇒ [M] . [ T] Units: N ⋅ s (alternatively kg ⋅ m/s). Fav Fav ′ t1 ∆t t2 Note: for a given change in ∆p: 1 Fav ∝ , i.e., Fav ∆t = constant. ∆t (Check web-site notes for “averages”.) The impulse relation: r r F.∆t = ∆p is rather like the work-kinetic energy theorem: r r F • ∆ s = ∆K . Let’s take a closer look at ... Fav .∆t = ∆p . If an object (truck, ball, you, etc.) is brought to a stop, then the longer the time ( ∆t ) it takes, the smaller the required force ( Fav ) .... many examples in everyday life. vo Same ∆p v=0 Large ∆t : small Fav (little damage) Question 6: A 60g handball moving with a speed of 5.0m/s strikes a wall at an angle of 40o and bounces off with the same speed at the same angle. If it is in contact vo v=0 with the wall for 2.0ms, (a) what is the average force exerted by the ball on the wall? If you catch the ball and Small ∆t : large Fav (lots of damage!) your hand moves 15cm, (b) what is the impulse of the ball on your hand, and (c) what is the average force exerted by the ball on your hand? (a) (b) Catching a hard ball (same ∆p ): (a) Small ∆t : large Fav (ouch!) (b) Large ∆t : small Fav (no problem!) r IWB r p1 5m/s 40o 40o r r r r ∆p ( = p 2 − p1 = IWB ) r p2 r − p1 5m/s ˆ i r rr r (a) Change in momentum is ∆p = p2 − p1 = IWB: r ∆p = (0.06 × 5 × cos 40)ˆ − (0.06 × −5 × cos 40)ˆ i i = 0.460 kg.m/s in the ˆ direction. i rrr But ∆p = I = Fav ∆t , r 0.460 ∆p ∴ Fav = = = 230 N . ∆t 2 × 10 −3 r r (b) Note: FBH = −FHB r r ∴ IBH = − IHB, r r i.e., IBH = IHB . r r Since IBH = ∆p r IBH = 0 − (0.06 × 5) = 0.30 N ⋅ s. r IBH 0.15m v o = 5m/s v=0 (c) If it takes ∆τ seconds to bring the ball to a stop, then I BH = FBH ∆τ = ∆p = 0.30 kg ⋅ m/s. But, what is ∆τ ? • v 2 = v o2 + 2a∆x . v2 52 ∴a = − o = = −83.3m/s2. 2 ∆x 2 × 0.15 r IBH 0.15m v o = 5m/s v=0 r IHB r IHB • v = v o + a∆τ . v ∴ ∆τ = − o = 0.060s. a ∆p 0.30 ∴ FBH = = = 5.0 N. ∆τ 0.060 Ballistic pendulum: Closer look at collisions: ~ Momentum is always conserved ~ But there are two types of collisions ... p1 = m1v 1 • non-elastic (inelastic) collisions m2 (Kinetic energy not conserved). • elastic collisions If the bullet stays in the target, conservation of (Kinetic energy conserved). momentum tells us initial momentum = final momentum 1. Non-elastic (inelastic) collisions. (e.g., when objects “stick” together) m1 m2 r p1 r p2 r pf ( m1 + m2 ) r r r p1 + p 2 = pf r r ∴ v f ( = v cm ) = h p2 = 0 r pf . m1 + m2 Examples: • bullet lodges in a target • book falling on a table ~ What happens to the “lost” KE ? ~ i.e., m1v1 = ( m1 + m2 ) v f m + m2 ∴ v1 = 1 vf . m1 With conservation of mechanical energy, ∆K = − ∆U, 1 ( m1 + m2 ) v f 2 = ( m1 + m2 ) gh 2 ∴ v f = 2 gh . ( m + m2 ) 2 gh . i.e., v1 = 1 m1 Can be used to determine the speed of a bullet. What about the kinetic energy? 2. Elastic collisions: momentum and kinetic energy are both conserved. r p1i p1 = m1v 1 m2 r p 2i h p2 = 0 Kinetic energy immediately after collision: 1 p 22 p12 K f = ( m1 + m2 ) v f 2 = = . 2 2( m1 + m2 ) 2( m1 + m2 ) r v 2i r v 1f r v 2f r p 2f r r r r ∴ p1i + p2i = p1f + p2 f Kinetic energy before collision: 1 p2 Ki = m1v12 = 1 . 2 2 m1 r v 1i r p1f and K1i + K2i = K1f + K2 f The analysis is a little more difficult because one needs to know three of the velocities involved in order to determine the fourth. The change in kinetic energy: p12 p2 ∆K = K f − K i = − 1 < 0. 2( m1 + m2 ) 2 m1 Since ( m1 + m2 ) > m1, ∆K is always negative, i.e., kinetic energy is never conserved. What happens to the “lost” energy? Elastic collisions in 1-dimension m1 v 1i m2 v 2i v 2i < v 1i Before collision m1 v 1f m2 v 2f > v 1f After collision v 2f Using conservation of momentum and kinetic energy: m1v1i + m2 v 2i = m1v1f + m2 v 2 f Note that this expression −( v 2i − v1i ) = v 2 f − v1f is true only for a 1-dimensional perfectly elastic and 1 1 1 1 m1v1i2 + m2 v 2i2 = m1v1f 2 + m2 v 2 f 2. 2 2 2 2 Re-arranging the two equations gives, collision, i.e., one in which kinetic energy is conserved. In the macroscopic world, collisions are somewhere m1 ( v1i − v1f ) = m2 ( v 2 f − v 2i ) between perfectly inelastic and perfectly elastic; at the and microscopic level elastic collisions are common, e.g., m1 ( v1i2 − v1f 2 ) = m2 ( v 2 f 2 − v 2i2 ) . between atoms. Dividing the second equation by the first, we get ( v1i2 − v1f 2 ) ( v 2 f 2 − v 2i2 ) = , ( v1i − v1f ) ( v 2 f − v 2i ) i.e., We define a coefficient of restitution (e) to tell what type of a collision we have ... ( v1i + v1f )( v1i − v1f ) ( v 2 f + v 2i )( v 2 f − v 2i ) = . ( v1i − v1f ) ( v 2 f − v 2i ) ∴ v1i + v1f = v 2 f + v 2i , i.e., −( v 2i − v1i ) = v 2 f − v1f speed of approach m1 v 1i m2 v 2i v 2i < v 1i Note that ... • if e = 1 the collision is perfectly elastic. speed of recession m1 v 1f v − v1f speed of recession = e = 2f . speed of approach v1i − v 2i m2 v 2f > v 1f • if e = 0 the collision is perfectly inelastic*. v 2f * In a perfectly inelastic collision the two colliding objects stick together. For a bouncing ball, if the speed at the instant before the collision is v1i and the speed immediately after the collision is v 1i v1f , then v 1f v e = 1f , v1i since the speed of the second object (the ground) is Question 7: A 2kg object moving at 6m/s collides with unchanged. Clearly, if the ball does not reach the 4kg object that is originally at rest. After the collision, original height, v1f < v1i , so e < 1. the 2kg object moves backward at 1m/s. Find (a) the velocity of the 4kg object after the collision, (b) the Typical values (on concrete): ebaseball = 0.57 . ebasketball = 0.57. eracquetball = 0.86 . egolfball = 0.89. Show for yourselves: if a ball is dropped from a height h onto a solid floor, the height of the first bounce is h1 = e2 h . coefficient of restitution, and (c) the energy “lost” in the collision. We don’t know whether it is elastic or not! Before collision ... m2 = 4kg m1 = 2kg v 2i = 0 v 1i = +6m/s After collision ... v 1f = −1m/s v 2f = ? Momentum before : p1i + p 2i = 12kg ⋅ m/s m r v 1i r v 1f θ m φ r v 2f Momentum after: p1f + p 2f = ( −2 + m2 v 2f ) Consider the special case of an off-center elastic (a) Since: p1f + p2 f = p1i + p2i , ( −2 + m2 v 2 f ) = 12 . Elastic collisions in 2-dimensions 14 ∴ v 2 f = = +3.5m/s . 4 v − v1f 3.5 − ( −1) (b) e = 2 f = = 0.75. Therefore, the 6−0 v1i − v 2i collison is non-elastic; kinetic energy is not conserved. (c) Change in kinetic energy ( ∆K = K f − Ki ): 1 1 1 1 ∆K = m1v1f 2 + m2 v 2 f 2 − m1v1i2 + m2 v 2i2 2 2 2 2 = (1 + 24.5) − (36) = −10.5J . “lost” collision between two equal masses, with one initially at rest. Then we have: r r r r r r mv1i = mv1f + mv 2 f , i.e., v1i = v1f + v 2 f , and 1 1 1 mv1i2 = mv1f 2 + mv 2 f 2, i.e., v1i2 = v1f 2 + v 2 f 2 . 2 2 2 r r r r r r But v1i2 = v1i • v1i = ( v1f + v 2 f ) • ( v1f + v 2 f ) r r = v1f 2 + v 2 f 2 + 2 v1f • v 2 f . r r ∴ 2 v1f • v 2 f = 0, r r i.e., v1f and v 2 f are perpendicular, so θ + φ = 90o . r v 1f r v 2i r v 1i 30o 60o ˆ j ˆ i r v 2f Question 8: A 170g hockey puck moving at 10m/s (a) Since momentum is conserved, we have: mv ˆ = mv cos 30ˆ + mv cos 60ˆ , i i i 1i 1f 2f mass that is initially at rest. The incoming puck is i.e., 10 = deflected at an angle of 30o from its original direction of 3 1 v1f + v 2 f . 2 2 ∴ 3v1f + v 2 f = 5 makes an off-center collision with another puck of equal motion while the second puck is deflected at an angle of 60o. (a) What is the velocity of each puck after the collision? (b) Show that the collision is elastic. ... ... ... (1) and 0 = mv1f sin 30ˆ − mv 2 f sin 60ˆ , j j 1 3 v. i.e., 0 = v1f − 2 2 2f ∴ v1f − 3v 2 f = 0 ... ... ... (2). Multiply (2) by 3 , then 3v1f − 3v 2 f = 0 ... ... ... (3). Subtract (3) from (1) and we get 4 v 2 f = 10 , i.e., v 2 f = 2.50 m/s. Substituting in (2) we find v1f = 3 × v 2 f = 4.33m/s. (b) To show the collision is elastic we must show that kinetic energy is conserved. The initial kinetic energy is 1 1 K1i = mv1i2 = × 0.170 × (5.0)2 = 2.125J . 2 2 The final kinetic energy is 1 1 K1f + K2 f = mv1f 2 + mv 2 f 2 2 2 1 1 = × 0.170 × (4.33)2 + × 0.170 × (2.50)2 2 2 = 1.594 + 0.531 = 2.125J . Thus, kinetic energy is conserved. ...
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