Ch 3 Momentum

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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 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 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.



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 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 CampusExample 3.15 (page139)Rocket in a Gravitational Field



On the receiving end as a stream of water from a hose, a push is felt.

ConsiderMomentum Transport




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


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


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


Ch 3 Momentum

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