Upload
amaranath-govindolla
View
225
Download
1
Tags:
Embed Size (px)
Citation preview
Slide 1
Contents:Centre of MassConservation of MomentumImpulseMomentum and flow of massMomentum transportProblems
Chapter 3 : MOMENTUMDr. Amaranath, BITS Pilani, Dubai CampusMomentum
Dr. Amaranath, BITS Pilani, Dubai CampusThe momentum of a particle isdefined as the product of its massand its velocity:
Momentum is a vector, in thesame direction as ( is ascalar) and its units are kg.m/s.Newtons second law can be writtenin terms of momentum as
i.e.,
Introduction
Dr. Amaranath, BITS Pilani, Dubai CampusIn 2nd chapter (Newtons Laws of Motion), the concept of point mass is introduced and used.In reality a) extended bodiesb) flow of mass (variable mass)
In this Chapter, we generalise the Newtons Laws so as to suit to the real situations.Momentum (P): Quantity of Motion. Vector. SI Unit: Ns is preferred to
Introduction
Dr. Amaranath, BITS Pilani, Dubai CampusThere are two classes of forces that act on and within systems. INTERNAL FORCES are forces between an object within the system and another object within the system. EXTERNAL FORCES are forces between an object within the system and an object outside the system.
Internal and External ForcesDr. Amaranath, BITS Pilani, Dubai CampusWe can classify each force as external or internal.The sum of all forces on all forces on all the particles is then
Because of Newton's third law, the internal forces all cancel in pairs, and
So,
OrExternal and internal forces
Dr. Amaranath, BITS Pilani, Dubai CampusIt is easy to find the center of mass of a homogeneous symmetric object, as shown in figure at the left.
Dr. Amaranath, BITS Pilani, Dubai CampusThe effective position of a system of particles is the point that moves as though all of the systems mass were concentrated there and all external forces were applied there.
The effective position is called as the center of mass of a system. It represents the average location for the total mass of a system
Centre of Mass
Dr. Amaranath, BITS Pilani, Dubai CampusFour masses 1 kg, 2 kg, 3 kg and 4 kg are placed at the four vertices of a square of side 1m as shown in the figure. Find CM of the arrangement.4kg3kg2kg1kgYX1m1mCentre of MassDr. Amaranath, BITS Pilani, Dubai CampusConsider the 4 masses in the coordinate system are placed as shown
Center of mass
Solution4kg3kg2kg1kgYX1m1m
Dr. Amaranath, BITS Pilani, Dubai Campus. Three objects are located in the x-y plane as shown in the figure. Determine the x coordinate of the center of mass for this system of three objects. Note the masses of the objects: mA = 6.0 kg, mB = 2.0 kg, and mC = 4.0 kg.
Dr. Amaranath, BITS Pilani, Dubai CampusThe total momentum of a system is equal to the total mass times the velocity of the center of mass. The center of mass of the wrench in the figure at the right moves as though all the mass were concentrated there.
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusSolid Bodies
If objects have uniform density,
For objects such as a golf club, the mass is distributed symmetrically and the center-of-mass point is located at the geometric center of the objects.Dr. Amaranath, BITS Pilani, Dubai CampusCentre of mass formulae:
Dr. Amaranath, BITS Pilani, Dubai CampusEx 3.3 (page 119)Centre of mass of a Non-uniform rod:
Dr. Amaranath, BITS Pilani, Dubai CampusExample 3.4 (page120) : Centre of mass of triangular sheet
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusBecause of uniform density and uniform mass,
Therefore
From similar triangles,
Substituting
By similar method, Alternate method
Dr. Amaranath, BITS Pilani, Dubai CampusFor any system, the forces that the particles of the system exert on each other are called internal forces.Forces exerted on any part of the system by some object outside it are called external forces.Conservation of momentum: If the vector sum of the external forces on a system is zero, the total momentum of the system is constant. Mathematically, it can be written as when F = 0 then, P = constant.
Conservation of Momentum
Dr. Amaranath, BITS Pilani, Dubai CampusWe know that F = dP/dt , where F is the external force acting on the system and P is the total momentum of the system.
For an isolated system of particles, the total momentum of the system remains constant.
Mathematically, it can be written as when F = 0 then, P = constant.
Conservation of MomentumDr. Amaranath, BITS Pilani, Dubai CampusEx. 3.6 Spring Gun Recoil :
Dr. Amaranath, BITS Pilani, Dubai CampusBefore firing of the gun,As there are no horizontal external forces, Px. Initial is conservedSince the system is initially at rest, Px. Initial = 0 But according to the LCM, Px. Initial = Px,final After firing of the gun,The gun recoils the some speed Vf and its final horizontal momentum is MVf , to the left.Guns final velocity is also Vf .
Dr. Amaranath, BITS Pilani, Dubai CampusAt the same instant the marbles speed relative to the gun is v0 Hence the final speed relative to the table is v0 cos - Vf Using conservation of momentum, we have0 = m(v0 cos - Vf ) = MVf
Or
The same problem can also be solved by using Newtons Laws directly.
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusSolution:
Dr. Amaranath, BITS Pilani, Dubai CampusThe impulse of a force is the product of the force and the time interval during which it acts. On a graph of Fx versus time, the impulse is equal to the area under the curve, as shown in figure to the right.
Impulse
Dr. Amaranath, BITS Pilani, Dubai CampusImpulse-momentum theorem: The change in momentum of a particle during a time interval is equal to the impulse of the net force acting on the particle during that interval.
Impulse
Dr. Amaranath, BITS Pilani, Dubai CampusImpulseThe relation between force and momentum is F = dP/dtThe integral form of force-momentum relationship is
The change in momentum is the time integral of force. is called IMPULSE.
Dr. Amaranath, BITS Pilani, Dubai CampusTypically, a tennis ball is in contact with the racket forapproximately 0.01 s. The ball flattens noticeably due to thetremendous force exerted by the racket.
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Solution:Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusIn the following examples, the Newtons laws can not be applied blindily, as the mass is flowing or mass is a variable RocketCalculation of reaction force on a fire hoseCalculation of acceleration of a snow ball which grows larger as it rolls downhill.
When we apply the integral form, it is essential to deal with the same set of particles throughout the time interval ta to tb. Consequently, the mass of the system cannot change during the time of interest.Momentum and Flow of Mass
Dr. Amaranath, BITS Pilani, Dubai CampusExample 3.11 (page 134), Mass flow and momentum
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusSand falls from a stationary hopper onto a freight car which is movingwith uniform velocity v. The sand falls at the rate dm/dt. How muchforce is needed to keep the freightcar moving at the speed v?
(because u = 0)
The required force is F =
Example 3.12, (page135) Freight Car and Hopper
Dr. Amaranath, BITS Pilani, Dubai CampusExample 3.13 (page 136)Leaky Freight car
Dr. Amaranath, BITS Pilani, Dubai CampusAs a rocket burns fuel, its mass decreases, as shown in Figure below.
Rocket propulsion Rocket propulsionDr. Amaranath, BITS Pilani, Dubai Campus
The x-velocity of the burned fuel relative to our coordinate system is
and the x-component of momentum of the ejected mass
Thus the total x-component of momentum, P2, of the rocket plus ejected fuel at time t+dt is
Or
And is simplified to
neglecting ( ), because it is a product of two small quantities
Dr. Amaranath, BITS Pilani, Dubai CampusThe net force or thrust is
The x-component of acceleration of the rocket is
If the exhaust ex, speed is constant, we can integrate above equation to find a relationship between the velocity at any time and the remaining mass m. At time =0, let the mass be o and the velocity o. Then we rewrite the equation as
Dr. Amaranath, BITS Pilani, Dubai CampusWe change the integration variables to and so we can use and m as the upper limits (the final speed and mass). Then we integrate both sides, using limits o to and o to m, and take the constant outside the integral:
Dr. Amaranath, BITS Pilani, Dubai CampusExample 3.14(page 138)Rocket in free space
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusExample 3.15 (page139)Rocket in a Gravitational Field
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
On the receiving end as a stream of water from a hose, a push is felt.
ConsiderMomentum Transport
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai Campus
Dr. Amaranath, BITS Pilani, Dubai CampusSolution Velocity vo
Where is the density and vol is the volume
Thus
DvotEx. No. 3.16Dr. Amaranath, BITS Pilani, Dubai Campus3.1 The density of a thin rod of length l varies with the distance x
from one end as . Find the position of the center of mass. Solution
Problems
Dr. Amaranath, BITS Pilani, Dubai Campus
Suppose a cannon shell traveling in a parabolic trajectory (neglecting air resistance) explodes in flight, splitting into two fragments with equal mass (figure below). The fragments follow new parabolic paths, but the center of mass continues on the original parabolic trajectory, just as though all the mass were still concentrated at that point. Dr. Amaranath, BITS Pilani, Dubai CampusEx. No. 3.4
Dr. Amaranath, BITS Pilani, Dubai CampusA object of mass 4m breaks into masses m and. The mass m is ejected backwards at x=0The mass 3m moves forwards at x = 2LTherefore, the center-of-mass of the two fragments must be
Or
So the landing point of 3m isSolutionm3mYX02L
Dr. Amaranath, BITS Pilani, Dubai Campus3.5 A circus acrobat of mass M leaps straight up with initial velocity vo from a trampoline. As he rises up, he takes a trained monkey of mass m off a perch at a height h above the trampoline. What is the maximum height attained by the pair?Ex. No. 3.5Dr. Amaranath, BITS Pilani, Dubai CampusTo find speed of acrobat before he grabs the monkey
Using momentum conservation to find speed of monkey and acrobat just after he grabs the monkey
Also . ThereforeSubstituting for v,
Therefore total height = h + x
SolutionhMM+mvov=0x
v
Dr. Amaranath, BITS Pilani, Dubai Campus3.9 A freight car of mass M contains a mass of sand m. At t=0 a constant horizontal force 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 constant rate dm/dt. Find the speed of the freight car when all the sand is gone. Assume the freight car is at rest at t=0.SolutionChoose the positive-direction to point in the direction that the car is moving.Lets the amount of sand of mass ms 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 .
Ex. No. 3.9Dr. Amaranath, BITS Pilani, Dubai CampusThe momentum at time t
where mc(t) is the mass of the car and sand in it at time t . Denote by mc,0 = mc + ms where the mc is the mass of the car and ms is the mass of the sand in the car at t = 0, and ms(t)= bt is the mass of the sand that has left the car at time t since
mc(t) +ms vtmc(t)v +vt + t
v +vms
Dr. Amaranath, BITS Pilani, Dubai CampusThus mc(t) = mc,0 (t) bt = mc + ms bt. momentum at time t + t is given byTherefore force,
Substituting
Or simplifying and converting to differentiation
Dr. Amaranath, BITS Pilani, Dubai CampusRewriting
Integrating both sides
Thus the velocity of the car as a function of time is
Dr. Amaranath, BITS Pilani, Dubai Campus
Ex. No. 3.20Dr. Amaranath, BITS Pilani, Dubai CampusThe rocket equation is
Or
The mass cancels and
C can be found from
The solution is Solution
Dr. Amaranath, BITS Pilani, Dubai CampusAn inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up from the ground with a speed 0, at a constant rate /.The problem is to find the maximum heightat which the garbage can rides. What assumption must be fulfilled for the maximum height to be reached?Ex. No. 3.17
Dr. Amaranath, BITS Pilani, Dubai CampusAssumptions: 1. All water hits the can2. If water hits the can travelling upwards at speed v, it reflects and travels downward of speed v (by conservation of energy)Thus the rate of change of momentum at the top of the can is
At height h, o2 2 =2, OrChange in momentum = W = mgThus,
Solution
Dr. Amaranath, BITS Pilani, Dubai Campus