Newton’s Second Law - Purdue University

Introduction to Dynamics (N. Zabaras)

Newton’s Second Law

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

April 10, 2016

mailto:[email protected]

http://www.zabaras.com/


Equations of Motion

• Newton’s second law provides

amF

• Solution for particle motion is facilitated by

resolving vector equation into scalar component

equations, e.g., for rectangular components,

zmFymFxmF

maFmaFmaF

kajaiamkFjFiF

zyx

zzyyxx

zyxzyx

• For tangential and normal components,

2vmF

dt

dvmF

maFmaF

nt

nntt

2


Linear Momentum of a Particle

• Replacing the acceleration by the derivative of

the velocity yields

particle theof momentumlinear

L

dt

Ldvm

dt

d

dt

vdmF

• Linear Momentum Conservation Principle:

If the resultant force on a particle is zero, the

linear momentum of the particle remains

constant in both magnitude and direction.

3


Sample Problem

The two blocks shown start from

rest. The horizontal plane and the

pulley are frictionless, and the

pulley is assumed to be of

negligible mass.

Determine the acceleration of each

block and the tension in the cord.

4


Sample Problem

• Write equations of motion for blocks and pulley.

:AAx amF

AaT kg1001

:BBy amF

B

B

BBB

aT

aT

amTgm

kg300-N2940

kg300sm81.9kg300

2

22

2

:0 CCy amF

02 12 TT

SOLUTION:

• Write the kinematic relationships for the dependent

motions and accelerations of the blocks.

ABAB aaxy21

21

x

y

O

5


Sample Problem

N16802

N840kg100

sm20.4

sm40.8

12

1

221

2

TT

aT

aa

a

A

AB

A

• Combine kinematic relationships with equations of

motion to solve for accelerations and cord tension.

ABAB aaxy21

21

AaT kg1001

A

B

a

aT

21

2

kg300-N2940

kg300-N2940

0kg1002kg150N2940

02 12

AA aa

TT

x

y

O

6


Angular Momentum of a Particle• moment of momentum or the

angular momentum of the particle about O. VmrHO

• Derivative of angular momentum with respect to time,

O

O

M

Fr

amrVmVVmrVmrH

• It follows from Newton’s second law that the sum

of the moments about O of the forces acting on

the particle is equal to the rate of change of the

angular momentum of the particle about O.

zyx

O

mvmvmv

zyx

kji

H

• is perpendicular to plane containingOH

Vmr

and

2

sin

mr

vrm

rmVHO

7

rr r v e e(in polar coordinates)


Eqs of Motion in Radial & Transverse Components

rrmmaF

rrmmaF rr

2

2

• Consider particle at r and , in polar coordinates,

rrmF

rrrm

mrdt

dFr

mrHO

2

22

2

2

• The 2nd Eq. above may also be derived from

conservation of angular momentum,

8

Rate of change

of angular momentum

Moments of external

forces around O


Conservation of Angular Momentum

• When only force acting on particle is directed

toward or away from a fixed point O, the particle

is said to be moving under a central force.

• Since the line of action of the central force

passes through O, and 0 OO HM

constant OHVmr

• Position vector and motion of particle are in

a plane perpendicular to .OH

• Magnitude of angular momentum,

000 sin

constantsin

Vmr

VrmHO

2

2

constant

angular momentum

unit mass

O

O

H mrv mr

Hr h

m

or

9

Central force

rr r

v r

v e e


Conservation of Angular Momentum

• Radius vector OP sweeps infinitesimal area

drdA 221

• Define 2

212

21 r

dt

dr

dt

dAareal velocity

• Recall, for a body moving under a central force,

constant2 rh

• When a particle moves under a central force,

its areal velocity is constant.

10


Newton’s Law of Gravitation• Gravitational force exerted by the sun on a planet

or by the earth on a satellite is an important

example of gravitational force.

• Newton’s law of universal gravitation - two particles

of mass M and m attract each other with equal and

opposite force directed along the line connecting

the particles,

2

3 412 9

2 4

m ft66.73 10 34.4 10

kg s lb s

MmF G

r

G

constant of gravitation

• For particle of mass m on the earth’s surface,

222 s

ft2.32

s

m81.9 gmg

R

MGmW

11


Sample Problem

A satellite is launched in a

direction parallel to the surface of

the earth with a velocity of 18820

mi/h from an altitude of 240 mi.

Determine the velocity of the

satellite as it reaches it maximum

altitude of 2340 mi. The radius of

the earth is 3960 mi.

SOLUTION:

• Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular

momentum at A and B and solve for

the velocity at B.

12


Sample Problem

SOLUTION:

• Since the satellite is moving under a

central force, its angular momentum is

constant. Equate the angular

momentum at A and B and solve for

the velocity at B.

sin constant

3960 240 mi18820mi h

3960 2340 mi

O

A A B B

AB A

B

rmv H

r mv r mv

rv v

r

hmi12550Bv

13


Trajectory of a Particle Under a Central Force

• For particle moving under central force directed towards the force center,

022 FrrmFFrrm r

• Second expression is equivalent to from which,,constant 2 hr

rd

d

r

hr

r

h 1and

2

2

2

2

2

• After substituting into the radial equation of motion and simplifying,

ru

umh

Fu

d

ud 1where

222

2

• If F is a known function of r or u, then particle trajectory may

be found by integrating for u = f(), with constants of

integration determined from initial conditions.14

2

2 2 2 2

2 2 2 2 2

1

1 1 1

dr dr d h dr dh

dt d dt r d d r

d r d dr d d d d d h d hh h

dt d dt dt d d r dt d r r d r r

Use:


Trajectory of a Satellite

constant

1where

22

2

2

2222

2

h

GMu

d

ud

GMmur

GMmF

ru

umh

Fu

d

ud

• Consider earth satellites subjected to only gravitational

pull of the earth,

• Solution is equation of conic section,

tyeccentricicos11 2

2

GM

hC

h

GM

ru

• Origin, located at earth’s center, is a focus of the conic

section.

• Trajectory may be ellipse, parabola, or hyperbola

depending on value of eccentricity.

15


Trajectory of a Satellite

tyeccentricicos11 2

2

GM

hC

h

GM

r

• Trajectory of earth satellite is defined by

• hyperbola, > 1 or C > GM/h2. The radius vector

becomes infinite for

2

1111 cos

1cos0cos1

hC

GM

• parabola, = 1 or C = GM/h2. The radius vector

becomes infinite for

1800cos1 22

• ellipse, < 1 or C < GM/h2. The radius vector is finite

for all . For =C=0, we obtain a circle.

16


Trajectory of a Satellite• Integration constant C is determined by

conditions at beginning of free flight, =0, r =

r0 ,

2

2

0

0 022

0 0 0 0

11 cos 0

1 1,

GM Ch

r h GM

GM GMC where h r v

r h r r v

00

200

2

2

or 1

r

GMvv

vrGMhGMC

esc

• Satellite escapes earth orbit for

• Trajectory is elliptic for v0 < vesc and

becomes circular for = 0 or C = 0,

0r

GMvcirc

17


Time to Complete an Orbit

• Recall that for a particle moving under a

central force, the areal velocity is constant,

i.e.,

constant212

21 hr

dt

dA

• The Periodic time or time required for a satellite

to complete an orbit is equal to area within the

orbit divided by areal velocity,

h

ab

h

ab

2

2

where

10

1021

rrb

rra

18

Introduction to Dynamics (N. Zabaras) 19

Newton’s Second Law - Purdue University

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