MOMENTUM! Momentum Impulse Conservation of Momentum in 1 Dimension Conservation of Momentum in 2 Dimensions Angular Momentum Torque Moment of Inertia

Momentum Defined p = m v p = momentum m = mass v = velocity

Momentum Facts p = m v SI unit for momentum: kg · m /s (no special name). Momentum is a conserved quantity (this will be proven later). A net force is required to change a body’s momentum. Momentum is directly proportional to both mass and speed. Something big and slow could have the same momentum as something small and fast.

Momentum Example 3 m /s 10 kg Calculate the momentum of the block. What is known? m = 10 kg v = 3 m/s What is missing? momentum  p = ? Equation p = m·v Solve p = (10 kg)(3 m/s) = 30 kg·m/s

Calculating Momentum Car: m = 1800 kg; v = 60 m /s Bus: m = 9000 kg; v = 100 m /s Train: m = 3.6 ·104 kg; v = 150 m /s

Momentum of the Train Bus: m = 9000 kg; v = 100 m /s Calculate the momentum of the bus. What is known? m = 9000 kg v = 100 m/s What is missing? momentum  p = ? Equation p = m·v Solve p = (9000 kg)(100 m/s) = 900,000 kg·m/s

Momentum of the Car Car: m = 1800 kg; v = 60 m /s Calculate the momentum of the car. What is known? m = 1800 kg v = 60 m/s What is missing? momentum  p = ? Equation p = m·v Solve p = (1800 kg)(60 m/s) = 108,000 kg·m/s

Momentum of the Train Calculate the momentum of the train. What is known? m = 3.6 ·104 kg v = 150 m/s What is missing? momentum  p = ? Equation p = m·v Solve p = (3.6 x 104 kg)(150 m/s) = 5,400,000 kg·m/s Train: m = 3.6 ·104 kg; v = 150 m /s

Comparing Momentum p = 108,000 kg·m/s p = 900,000 kg·m/s Car: m = 1800 kg; v = 60 m /s p = 108,000 kg·m/s Bus: m = 9000 kg; v = 100 m /s p = 900,000 kg·m/s Train: m = 3.6 ·104 kg; v = 150 m /s p = 5,400,000 kg·m/s

Calculating Velocity from Momentum Bus: m = 9000 kg; Car: p = 108,000 kg·m/s What velocity does the bus need to travel at in order to have the same momentum as the car? Known: m = 9000 kg p = 108,000 kg·m/s Missing: velocity = v = ? Equation: v = p/m = Solve: v = (108,000 kg·m/s) / (9000 kg) = 12 m/s

Calculating Velocity from Momentum Car: m = 1800 kg; Train: p = 5,400,000 kg·m/s What velocity does the car need to travel at in order to have the same momentum as the train? Known: m = 1800 kg p = 5,400,000 kg·m/s Missing: velocity = v = ? Equation: v = p/m = Solve: v = (5,400,000 kg·m/s) / (1800 kg) = 3000 m/s

Calculating Mass Car #1: p = 108,000 kg·m/s; Car #2: v = 50 m/s Another car has the same momentum as the first car, but a larger mass. If it is traveling at 50 m/s, what is its mass? Known: v = 50 m/s p = 108,000 kg·m/s Missing: mass = m = ? Equation: m = p/v = Solve: m = (108,000 kg·m/s) / (50 m/s) = 2160 kg

Conservation of Momentum Momentum in a system is ALWAYS conserved. The total momentum two or more objects have prior to a collision is equal to the total momentum after the collision.

Conservation of Momentum in 1-D The total momentum of the objects is the same before and after the collision. Positive is defined as the right direction. before: pT = m1 v1 - m2 v2 v1 v2 m1 m2 m1 v1 - m2 v2 = - m1 va + m2 vb after: pT = - m1 va + m2 vb va vb m1 m2

Directions after a collision Some collisions cause the objects to go in opposite directions. Other collisions cause the objects to go in similar directions. m1 v1 v2 m2 m1 va vb m2

Elastic vs. Inelastic Elastic collisions occur when the object “bounce” off each other and no energy is lost to changes in shape. Inelastic collision occur when the objects get “stuck” to each other, or energy is lost to changes in shape. m1 v1 v2 m2 m1 vb m2

Sample Problem 1 35 g 7 kg 700 m/s v = 0 A rifle fires a bullet into a giant slab of butter on a frictionless surface. The bullet penetrates the butter, but while passing through it, the bullet pushes the butter to the left, and the butter pushes the bullet just as hard to the right, slowing the bullet down. If the butter skids off at 4 cm/s after the bullet passes through it, what is the final speed of the bullet? (The mass of the rifle matters not.) 35 g 7 kg v = ? 4 cm/s continued on next slide

Sample Problem 1 (cont.) Let’s choose left to be the + direction & use conservation of momentum, converting all units to meters and kilograms. 35 g 7 kg p before = 7 (0) + (0.035) (700) = 24.5 kg · m /s 700 m/s v = 0 35 g 7 kg p after = 7 (0.04) + 0.035 v = 0.28 + 0.035 v v = ? 4 cm/s p before = p after 24.5 = 0.28 + 0.035 v v = 692 m/s v came out positive. This means we chose the correct direction of the bullet in the “after” picture.

Sample Problem 2 (0.035) (700) = 7.035 v v = 3.48 m/s 35 g 7 kg Same as the last problem except this time it’s a block of wood rather than butter, and the bullet does not pass all the way through it. How fast do they move together after impact? v 7. 035 kg (0.035) (700) = 7.035 v v = 3.48 m/s Note: Once again we’re assuming a frictionless surface, otherwise there would be a frictional force on the wood in addition to that of the bullet, and the “system” would have to include the table as well.

Proof of Conservation of Momentum The proof is based on Newton’s 3rd Law. Whenever two objects collide (or exert forces on each other from a distance), the forces involved are an action-reaction pair, equal in strength, opposite in direction. This means the net force on the system (the two objects together) is zero, since these forces cancel out. F M F m force on M due to m force on m due to M For each object, F = (mass) (a) = (mass) (v / t ) = (mass v) / t = p / t. Since the force applied and the contact time is the same for each mass, they each undergo the same change in momentum, but in opposite directions. The result is that even though the momenta of the individual objects changes, p for the system is zero. The momentum that one mass gains, the other loses. Hence, the momentum of the system before equals the momentum of the system after.

Conservation of Momentum applies only in the absence of external forces! In the first two sample problems, we dealt with a frictionless surface. We couldn’t simply conserve momentum if friction had been present because, as the proof on the last slide shows, there would be another force (friction) in addition to the contact forces. Friction wouldn’t cancel out, and it would be a net force on the system. The only way to conserve momentum with an external force like friction is to make it internal by including the tabletop, floor, or the entire Earth as part of the system. For example, if a rubber ball hits a brick wall, p for the ball is not conserved, neither is p for the ball-wall system, since the wall is connected to the ground and subject to force by it. However, p for the ball-Earth system is conserved!

Sample Problem 3 An apple is originally at rest and then dropped. After falling a short time, it’s moving pretty fast, say at a speed V. Obviously, momentum is not conserved for the apple, since it didn’t have any at first. How can this be? answer: Gravity is an external force on the apple, so momentum for it alone is not conserved. To make gravity “internal,” we must define a system that includes the other object responsible for the gravitational force--Earth. The net force on the apple-Earth system is zero, and momentum is conserved for it. During the fall the Earth attains a very small speed v. So, by conservation of momentum: apple m V F v Earth M F m V = M v

Sample Problem 4 A crate of raspberry donut filling collides with a tub of lime Kool Aid on a frictionless surface. Which way on how fast does the Kool Aid rebound? answer: Let’s draw v to the right in the after picture. 3 (10) - 6 (15) = -3 (4.5) + 15 v v = -3.1 m/s Since v came out negative, we guessed wrong in drawing v to the right, but that’s OK as long as we interpret our answer correctly. After the collision the lime Kool Aid is moving 3.1 m/s to the left. before 6 m/s 10 m/s 3 kg 15 kg after 4.5 m/s v 3 kg 15 kg

Conservation of Momentum in 2-D To handle a collision in 2-D, we conserve momentum in each dimension separately. Choosing down & right as positive: before: px = m1 v1 cos1 - m2 v2 cos2 py = m1 v1 sin1 + m2 v2 sin2 m2 m1 2 1 v2 v1 after: px = -m1 va cosa + m2 vb cos b py = m1 va sina + m2 vb sin b m1 m2  b a vb va Conservation of momentum equations: m1 v1 cos1 - m2 v2 cos2 = -m1 va cosa + m2 vb cos b m1 v1 sin1 + m2 v2 sin 2 = m1 va sina + m2 vb sin b

Conserving Momentum w/ Vectors BEFORE p1 m2 m1 2 1 p 2 p before p1 p 2 p a m1 m2 AFTER p after  b a p a p b p b This diagram shows momentum vectors, which are parallel to their respective velocity vectors. Note p1 + p 2 = p a + p b and p before = p after as conservation of momentum demands.

Exploding Bomb A e c m A c m e after before A bomb, which was originally at rest, explodes and shrapnel flies every which way, each piece with a different mass and speed. The momentum vectors are shown in the after picture. continued on next slide

Exploding Bomb (cont.) Since the momentum of the bomb was zero before the explosion, it must be zero after it as well. Each piece does have momentum, but the total momentum of the exploded bomb must be zero afterwards. This means that it must be possible to place the momentum vectors tip to tail and form a closed polygon, which means the vector sum is zero. If the original momentum of the bomb were not zero, these vectors would add up to the original momentum vector.

2-D Sample Problem 152 g A mean, old dart strikes an innocent mango that was just passing by minding its own business. Which way and how fast do they move off together? before 40 34 m/s 0.3 kg 5 m/s Working in grams and taking left & down as + : 152 (34) sin 40 = 452 v sin 152 (34) cos 40 - 300 (5) = 452 v cos after Dividing equations : 1.35097 = tan 452 g  = 53.4908  Substituting into either of the first two equations : v v = 9.14 m/s

Alternate Solution 40 Shown are momentum vectors (in g m/s). The black vector is the total momentum before the collision. Because of conservation of momentum, it is also the total momentum after the collisions. We can use trig to find its magnitude and direction.  5168 p 40 1500 Law of Cosines : p2 = 5168 2 + 1500 2 - 2  5168  1500 cos 40 p = 4132.9736 g m/s Dividing by total mass : v = (4132.9736 g m/s) / (452 g) = 9.14 m/s sin  sin 40 Law of Sines : =  = 13.4908 1500 4132.9736 Angle w/ resp. to horiz. = 40 + 13. 4908   53.49

Comments on Alternate Method Note that the alternate method gave us the exact same solution. This method can only be used when two objects collide and stick, or when one object breaks into two. Otherwise, we’d be dealing with a polygon with more sides than a triangle. In using the Law of Sines (last step), the angle involved (ß) is the angle inside the triangle. A little geometry gives us the angle with respect to the horizontal.

Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are  then the magnitude of angular momentum w/ resp. to point Q is given by L = r p = m v r. In this case L points out of the page. If the mass were moving in the opposite direction, L would point into the page. The SI unit for angular momentum is the kg  m2 / s. (It has no special name.) Angular momentum is a conserved quantity. A torque is needed to change L, just a force is needed to change p. Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning. v r m Q

Angular Momentum: General Definition If r and v are not  then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product: L = r  p This formula works regardless of the angle. As you know from our study of cross products, the magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. In this case, by the right-hand rule, L points out of the page. If the mass were moving in the opposite direction, L would point into the page. v  r m Q

Moment of Inertia I = m r 2 I =  mi ri 2 = m1 r12 + m2 r22 Any moving body has inertia. (It wants to keep moving at constant v.) The more inertia a body has, the harder it is to change its linear motion. Rotating bodies possess a rotational inertial called the moment of inertial, I. The more rotational inertia a body has, the harder it is change its rotation. For a single point-like mass w/ respect to a given point Q, I = m r 2. For a system, I = the sum of each mass times its respective distance from the point of interest. m r m2 Q m1 r1 I = m r 2 r2 Q I =  mi ri 2 = m1 r12 + m2 r22

Moment of Inertia Example Two merry-go-rounds have the same mass and are spinning with the same angular velocity. One is solid wood (a disc), and the other is a metal ring. Which has a bigger moment of inertia relative to its center of mass? r r   m m answer: I is independent of the angular speed. Since their masses and radii are the same, the ring has a greater moment of inertia. This is because more of its mass is farther from the axis of rotation. Since I is bigger for the ring, it would more difficult to increase or decrease its angular speed.

Angular Acceleration  =   / t As you know, acceleration is when an object speeds up, slows down, or changes directions. Angular acceleration occurs when a spinning object spins faster or slower. Its symbol is , and it’s defined as:  =   / t Note how this is very similar to a =  v / t for linear acceleration. Ex: If a wind turbine spinning at 21 rpm speeds up to 30 rpm over 10 s due to a gust of wind, its average angular acceleration is 9 rpm / 10 s. This means every second it’s spinning 9 revolutions per minute faster than the second before. Let’s convert the units: 9 rpm 9 rev / min 9 rev 9  (2  rad) = = = = 0.094 rad / s2 10 s 10 s min  10 s (60 s)  10 s Since a radian is really dimensionless (a length divided by a length), the SI unit for angular acceleration is the “per second squared” (s-2).

Torque & Angular Acceleration Newton’s 2nd Law, as you know, is Fnet = m a The 2nd Law has a rotational analog:  net = I  A force is required for a body to undergo acceleration. A “turning force” (a torque) is required for a body to undergo angular acceleration. The bigger a body’s mass, the more force is required to accelerate it. Similarly, the bigger a body’s rotational inertia, the more torque is required to accelerate it angularly. Both m and I are measures of a body’s inertia (resistance to change in motion).

Linear Momentum & Angular Momentum If a net force acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular acceleration, which means its angular momentum changes. Recall, angular momentum’s magnitude is given by L = m v r (if v and r are perpendicular) v So, if a net torque is applied, angular velocity must change, which changes angular momentum. r m proof:  net = r Fnet = r m a = r m  v / t =  L / t So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum. continued on next slide

Linear & Angular Momentum (cont.) Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r , we get L = m v r = m r (r ) = m r 2  = I  This is very much like p = m v, and this is one reason I is defined the way it is. In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times angular speed. L = I  p = m v

Spinning Ice Skater I  = L = I  Why does a spinning ice skater speed up when she pulls her arms in? Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the weights far from that axis, his moment of inertia is large. When he pulls his arms in as he’s spinning, the weights are closer to the axis, so his moment of inertia gets much smaller. Since L = I  and L is conserved, the product of I and  is a constant. So, when he pulls his arms in, I goes down,  goes up, and he starts spinning much faster. I  = L = I 

Comparison: Linear & Angular Momentum Linear Momentum, p Tendency for a mass to continue moving in a straight line. Parallel to v. A conserved, vector quantity. Magnitude is inertia (mass) times speed. Net force required to change it. The greater the mass, the greater the force needed to change momentum. Angular Momentum, L Tendency for a mass to continue rotating. Perpendicular to both v and r. A conserved, vector quantity. Magnitude is rotational inertia times angular speed. Net torque required to change it. The greater the moment of inertia, the greater the torque needed to change angular momentum.