Momentum and the Flow of Mass 1 Introduction: So far we have restricted ourselves to considering systems consisting of discrete objects or point-like objects that have fixed amounts of mass. We shall now consider systems in which material flows between the objects in the system, for example we shall consider coal falling from a hopper into a moving railroad car, sand leaking from railroad car fuel, grain moving forward into a railroad car, and fuel ejected from the back of a rocket, In each of these examples material is continuously flows into or out of an object. We have already shown that the total external force causes the momentum of a system to change, !Fexttotal=d!psystemdt. (1.1.1) We shall analyze how the momentum of the constituent elements our system change over a time interval [t,t + !t], and then consider the limit as !t " 0. We can then explicit calculate the derivative on the right hand side of Eq. (1.1.1) and Eq. (1.1.1) becomes !Fexttotal=d!psystemdt= lim!t"0!!psystem!t= lim!t"0!psystem(t + !t) #!psystem(t)!t. (1.1.2) We need to be very careful how we apply this generalized version of Newton’s Second Law to systems in which mass flows between constituent objects. In particular, when we isolate elements as part of our system we must be careful to identify the mass !m of the material that continuous flows in or out of an object that is part of our system during the time interval !t under consideration. We shall consider four categories of mass flow problems that are characterized by the momentum transfer of the material of mass !m. Category 1: There is a transfer of material into the object but no transfer of momentum in the direction of motion of the object. Consider for example rain falling vertically downward into a moving cart. A small amount of rain !m has no component of momentum in the direction of motion of the cart.Category 2: The material continually leaves the object but it does not transport any momentum away from the object in the direction of motion of the object. For example, consider an ice skater gliding on ice holding a bag of sand that is leaking straight down with respect to the moving skater. Category 3: The material continually hits the object providing an impulse resulting in a transfer of momentum to the object in the direction of motion. For example, suppose a fire hose is used to put out a fire on a boat. The incoming water continually hits the boat impulsing it forward. Category 4: The material continually is ejected from the object, resulting in a recoil of the object. For example when fuel is ejected from the back of a rocket, the rocket recoils forward.We must carefully identify the momentum of the object and the material transferred !m at time t in order to determine !psystem(t ). We must also identify the momentum of the object and the material transferred !m at time t + !t in order to determine !psystem(t + !t)as well. Recall that when we defined the momentum of a system, we assumed that the mass of the system remain constant. Therefore we cannot ignore the momentum of the transferred material at time t + !t even though it may have left the object, it is still part of our system (or at time t even though it has not flowed into the object yet). 2 Worked Examples 2.1 Worked Example Coal Car An empty coal car of mass 0m starts from rest under an applied force of magnitude F. At the same time coal begins to run into the car at a steady rate b from a coal hopper at rest along the track. Find the speed when a mass cm of coal has been transferred. Solution: We shall analyze the momentum changes in the horizontal direction which we call the x-direction. Because the falling coal does not have any horizontal velocity, the falling coal is not transferring any momentum in the x-direction to the coal car. So we shall take as our system the empty coal car and a mass cm of coal that has been transferred. Our initial state at 0t = is when the coal car is empty and at rest before any coal has been transferred. The x-component of the momentum of this initial state is zero, px(0) = 0. (1.2.1) Our final state at ft t= is when all the coal of mass c fm bt= has been transferred into the car that is now moving at speed fv. The x-component of the momentum of this final state is px(tf) = (m0+ mc)vf= (m0+ btf)vf. (1.2.2)There is an external constant force xF F= applied through the transfer. The momentum principle applied to the x-direction is Fxdt0tf!= "px= px(tf) # px(0). (1.2.3) Since the force is constant, the integral is simple and the momentum principle becomes Ftf= (m0+ btf)vf. (1.2.4) So the final speed is vf=Ftf(m0+ btf). (1.2.5) 2.2 Worked Example: Emptying a Freight Car A freight car of mass cm contains a mass of sand sm. At 0t = a constant horizontal force of magnitude F is applied in the direction of rolling and at the same time a port in the bottom is opened to let the sand flow out at the constant rate /sb dm dt=. Find the speed of the freight car when all the sand is gone. Assume that the freight car is at rest at 0t =. Solution: Choose the positive x-direction to point in the direction that the car is moving. Let’s take as our system the amount of sand of mass sm! that leaves the freight car during the time interval [ , ]t t t+ !, and the freight car and whatever sand is in it at time t.At the beginning of the interval the car and sand is moving with speed v so the x-component of the momentum at time t is given by px(t) = (!ms+ mc(t))v), (1.2.6) where ( )cm t is the mass of the car and sand in it at time t. Denote by ,0c c sm m m= + where cm is the mass of the car and sm is the mass of the sand in the car at 0t =, and ( )sm t bt= is the mass of the sand that has left the car at time t since ms(t) =dmsdtdt0t!= bdt0t!= bt. (1.2.7) Thus mc(t) = mc,0! bt = mc+ ms! bt. (1.2.8) During the interval [ , ]t t t+ !, the small amount of sand of mass sm! leaves the car with the speed of the car at the end of the interval v v+ !. So the x-component of the momentum at time t t+ ! is given by px(t + !t) = (!ms+ mc(t))(v + !v). (1.2.9) Throughout the interval a constant force F is applied to the car so 0( ) ( )limx xtp t t p tFt! "+ ! #=!. (1.2.10) From our analysis above Eq. Error! Reference source not found. becomes 0( ( ) )( ) ( ( ) )limc s c stm t m v v m t m vFt! "+ ! + ! # + !=!. (1.2.11) Eq. …
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