AAST/AEDT. Center of mass

Transcription

1 AAST/AEDT AP PHYSICS C: Center of mass Let us run an experiment: We take an object of a random shape and pull it by applying a force as shown on a diagram. The object experiences translational and rotational motions. (Translational motion is the motion when all the points of the object have the same path) Among all the directions of the applied force only one direction will create a purely translational motion (Diag. ). Such direction of force that produces translational motion exists for each point of the object. Some of them are presented on a diagram 3. If we extend the directions along which those forces are applied we will discover that all those directions intersect at a single point. Such point is defined as object s center of mass. Center of mass can be also defined as point of application of the net of all gravity forces that are applied to the different parts of the object. Let us derive a formula that allows us to estimate the position of the center of mass in a case of a linear object. Let the system consist of two masses, m and m, located at points X and X and connected by a massless rod as shown on a diagram. Our goal is to define the position of the center of mass C. Criteria to determine C is that a torques produced by the gravity forces exerted on both masses relatively to the center of mass must be equal. Thus, we can write m g (X c -X ) = m g (X -X c ) g can be canceled and after simple algebraic manipulations the formula to determine position of the center of mass is: m X m X X c m m Although we derived our formula in case of two masses located along one line, the result may be extended for a random number of masses. In general position of the center of mass for random number of masses can be determined as.

2 X c m X m X... m X m m... m n M n n n i m X i i In a dimensional space Y-position of the center of mass can be determined in a similar way. n Yc miyi M i Center of the mass position can be also determined through the vector r (r - is a vector directed from the origin to the point with the mass), See example on the diagram) r com M n mi r i i If we take a time derivative from the equation above, the result would be a formula for the velocity of the center of mass. V com M n mi V i i For the solid bodies with a continuous distribution of mass the sum () should be replaced by integration. The formulas will be: X c xdm Yc ydm Zc zdm M M M Second Newton s law for a system of particles In many cases we have to deal with a motion of an object that starts as a single mass and then explodes in numerous fragments. The simplest examples are fireworks, shell, grenade, etc. In those three cases the objects were moving in a filed of a gravity force. The gravity force in this case is defined as an external force. The forces that exerted on the particles as the result of an explosion of the above mentioned objects we define as an internal. The internal forces on each of the particles are non zero. However the net of all internal forces exerted on all particles is zero (III N Law). Thus, the system continues to move as a whole under the influence on the external force. It can be proven that the point of application of the external force is a center of mass. Thus, the second N Law for a system of particles can be written as F=ma com, Impulse and Momentum

3 3 Let us run an experiment: The ball is moving with a velocity of V o and a force of F is applied on it for the time interval of t. As the result the ball s velocity changes from V o to V. From the second Newton s law we know We also know that F m a If we substitute expression for a into the equation for F, we obtain V V F m o t The next step is to transfer t from the left side of the equation to the right one. Ft m(v Vo ) mv mvo mv Where mv - means the change of this product. At the left side of the equation above the expression is Ft. The product of the force and the time interval when the force is applied on the object is defined as an impulse. IMPULSE = Ft At the right side of the equation we have mv The product of an object s mass and velocity is defined as momentum MOMENTUM = mv Usually momentum is represented by the symbol P Using that terminology, we can read the equation () as APPLIED IMPULSE IS EQUAL TO THE MOMENTUM CHANGE Both impulse and momentum are vectors V V a o t MOMENTUM CONSERVATION LAW Let us run another experiment. Two balls with the masses of m and m are moving toward one another with the velocities of V o and V o. Double subscript (o) means - initial velocity of the first object. The balls collide. After the collision their velocities become V and V. ()

4 4 According to the 3 Newton s Law F =-F () We can express the forces as ( Newton s law) F = ma and F = ma, Where a and a - are the accelerations of the balls. If we plug in expressions for accelerations into () the result is ma = - ma () We also know, the acceleration is the ratio of the velocity change over the time V a Vo t V a Vo t After substitution of those expressions into () we have V m Vo V m Vo t t We can cancel t and multiply mass times velocities. The result is m V m V o m V m V o Expressions with the subscript o (that means that the expression relates to the events before collision) we move to the right side of the equation, and all others we move to the left. The result is: m V m V m V o m V o Now, on the left side we have total momentum after and on the right side - total momentum before the collision. We have derived the momentum conservation law. It states. The total momentum of an isolated system of bodies remains constant or total momentum before any interaction equals to the total momentum after them.. Write the data. Problems solving strategy. Define which objects had momentum before the interaction and write an expression for the total momentum of all the objects before the interaction.

5 5 3. Define which objects have momentum after the interaction and write an expression for the total momentum of all the objects after the interaction. 4. Write the momentum conservation law - total momentum before the interaction has to be equal to the total momentum after the interaction 5. From the obtained equation express the unknown variable and substitute numbers. Example: A kg hockey puck moving at 48 m/s is caught by a 75-kg goalie at rest. With what speed does the goalie slide on ice? m = 0.05 kg V o = 48 m/s m =75 kg V o =0 V=? In the problem only the puck had a momentum before the collision (interaction). Thus, the total momentum before is m V o After the collision both objects are moving together. The total momentum after collision is (m +m )V According to the momentum conservation law m V o = (m +m )V In this equation the only unknown variable is V. We can isolate it and plug in the numbers. V = m V o /(m +m )= 0.05 kg *48 m/s/(75 kg kg) = m/s JET PROPULSION Let us run an experiment. A test-tube partially filled with water and a stopper is installed on a cart. Initially the stopper and the cart are at rest. So, the total initial momentum m cart V cart +m stopper V stopper =0 If we heat the water, the stopper shots and the cart starts to move backward. According to the momentum conservation law the total momentum does not change. Thus, after the shot the total momentum also remains zero. If we express the V cart from the equation above we have: V Cart m stopperv stopper m Cart Sign - shows us, that the direction of the stopper s motion is opposite to the direction of the cart s motion.

6 6 That motion with the total momentum that is equal to zero is defined as jet propulsion. The same principle is used for the rocket operation. Initially the velocity of the fuel gases and of the rocket is zero. Thus, the total initial momentum. When the fuel starts to burn, the gases start to eject, but the total momentum has to remain zero. Thus, V rocket m fuel V fuel m rocket The formula provides us with two possible ways of increasing the rocket s velocity. One way is to increase the velocity of the ejected gases. The second is to decrease the rocket s mass Problem: A man wit a mass of 70 kg is standing at the rear of a boat of mass 80-kg in the middle of the placid lake. The boat is 5 m long. The man walks to the front of the boat How far does the boat move relative to the lake bottom? Neglect water resistance. Solution. As you can observe from the diagram on the right, when the man moves in one direction according to the jet propulsion principle the boat starts to move in an opposite one. Because the net momentum should me zero, the equation for the momentum conservation law is M man V man - M boat V boat =0, or M man V man = M boat V boat () Assuming that the motion is uniform we can express the velocities in a scalar shape as V man = d/t = (L-x)/t, and V boat = x/t, where L is the length of the boat and x is the distance that boat has traveled. When we substitute those expressions into (), we can cancel time and isolate x. L x x M man M boat or M manl M manx M boatx t t M manl x M M man m rocket V rocket+ m fuel V fuel = 0 boat Collisions There are two types of collisions - elastic and inelastic

7 7. Elastic collision is the collision when both mechanical energy and momentum are conserved. A good example of an elastic collision is the collision of the two steel balls. When we have elastic collision only conservative forces are acting inside the physical system Let us assume that the ball with a mass of m and with a velocity of V o collides with the ball m at rest. After the collision the ball s velocities are V and V respectively. Our goal is to evaluate those velocities. For the observed collision the mechanical energy conservation law is mv m V o mv KE o KE KE or After cancellation of in the denominator we have V mv m V m o Momentum conservation law for the same collision is V mv m m o V We obtained a set of two equations with two unknown variables V and V. The set can be solved

8 8 m m m The final result is V Vo and V Vo m m m m The formulas allows us to make several very interesting conclusions about the elastic collision. If m >m then V is positive. That means that both balls after the collision are moving in one and the same direction.. If m <m then V is negative. That means that both balls after the collision are moving in an opposite directions. 3. If m =m then V is 0 and V =V o. That means that the first ball stops after the collision and the second one is moving forward with the same speed. A good example of this is the collision of the two billiard balls. 4. If m =(infinitive mass). A good example of this is the collision with the wall, then V = -V o and V =0. That means that the first ball renounced with the same speed. Inelastic collision Only momentum conservation law is valid for the collision. Mechanical energy conservation law does not take place, because partially mechanical energy transfers into the heat form of energy. A good example of this collision is the collision of the two clay balls. The momentum conservation law for the collision is mv o mv o ( m m ) V or V m m We can evaluate the amount of mechanical energy dissipated into heat. H = KE 0 - KE +,

9 9 where KE0- is the initial kinetic energy of the first ball before the collision and the KE + - is the total energy of both balls after the collision. In a real life we do not have perfectly elastic or totally inelastic collisions. To observe the quality of collision physicists use an elasticity coefficient. If it is - the collision is elastic. If it is 0 it is completely inelastic. Two Dimensional Collisions Let us imagine that you are playing pool and you hit the target ball in such a way that after a collision, two balls are moving in different directions as shown on a diagram. We can also assume that balls masses are different. In this case we can apply the energy conservation law mv mv f mv f and two projections of the momentum conservation law. Along axis X momentum conservation law can be presented as Along axis Y m V = m V cos + m V cos 0 = - m V cos + m V cos There are 7 quantities in the equation. Depending on the problem, if 4 of them are given, the other 3 can be determined. Problem: A bullet of mass m flying horizontally at a speed V o strikes a block of mass M suspended on a string with a length of L and gets stuck in it. To what angle to the vertical does the block deviate? The problem should be divided into three separate problems. What velocity block obtained as the result of inelastic collision with the bullet? We apply the momentum conservation law and obtain mv o = (m+m)v block, or V block m Vo m M. What was the final height of the deviated block?

10 0 As the result of the collision block obtained KE and it was finally transformed into the potential one. We apply the energy conservation law and obtain KE PE or ( m M ) Vblock ( m M ) gh Vblock h g 3. What is the angle of deviation? From triangle ABC it is obvious that cos L - h L, thus cos L h L Home Assignment: Momentum Chapter 9 Problems page 9 #33, 4, 5, 7, 0, 04, 9.5, 7, 0(WA), 0, 35, 36, 0, 37, Collisions; Chapter 9: Page 09.. #3, 5, 8, 9, 33, 0, 99, 48, 50, 54, 56, 58, 6, 66,67

11 a m b a m 3b.6 m 3c x-position does n't change. y-position shifts up 5a -0.5 m E- m down from N 8a,b x = 0cm, y = 0 cm 8c 6 cm from the m 7 km/h 3a m 3b m/s m show two solutions m 0a km/h 0b km/h kgm/s 5b #NAME? 5 c 0 force m/s m/s 35a kgm/s 35b degrees

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