Kinetics of Particles: Newton's second law

12.1 Introduction

• Newton’s first and third laws are sufficient for the study of bodies at rest (statics) or bodies in motion with no acceleration.( )

• When a body accelerates (changes in velocity magnitude or direction), Newton’s second law is required to relate the motion of the body to the Newton s second law is required to relate the motion of the body to the forces acting on it.

• Newton’s second law:

• Newton s second law:

- A particle will have an acceleration proportional to the magnitude of the resultant force acting on it and in the direction of the resultant g force.

- The resultant of the forces acting on a particle is equal to the rate of

h f li f h i l

change of linear momentum of the particle.

- The sum of the moments about O of the forces acting on a particle is equal to the rate of change of angular momentum of the particle

equal to the rate of change of angular momentum of the particle

12.2 Newton’s Second Law of Motion

• Newton’s Second Law: If the resultant force acting on a particle is not zero, the particle will have an acceleration proportional to the magnitude of resultant and in the

proportional to the magnitude of resultant and in the direction of the resultant.

• Consider a particle subjected to constant forces, a m

F a

F a

F constant mass,

3 3 2

2 1

1 = = =L= =

• When a particle of mass m is acted upon by a force the acceleration of the particle must satisfy

, Fr a

m Fr = mar F =

• Acceleration must be evaluated with respect to a Newtonian frame of reference, i.e., one that is not f f f , , accelerating or rotating.

• If force acting on particle is zero?

Particle will not accelerate i e it will remain stationary or Particle will not accelerate, i.e., it will remain stationary or continue on a straight line at constant velocity.

12.3 Linear Momentum of a Particle

• Replacing the acceleration by the derivative of the velocity yields

velocity yields

∑

=

dt v md F

r r r

( )

particle the

of momentum linear

=

=

= L

dt L v d

dt m d

r

r r

particle the

of momentum linear

= L

• 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.g

Systems of Units y

• Of the units for the four primary dimensions (force, mass, length, and time), three may be chosen arbitrarily. The

f th t b tibl ith N t ’ 2 d L

fourth must be compatible with Newton’s 2nd Law.

• International System of Units (SI Units): base units are

• International System of Units (SI Units): base units are the units of length (m), mass (kg), and time (second).

The unit of force is derived,

⎞

( )

⎛

2

2 s

m 1kg

s 1 m kg 1 N

1 ⎟ = ⋅

⎠

⎜ ⎞

⎝

= ⎛

12.5 Equations of Motion q

• Newton’s second law provides a

m Fr = r

∑

∑

• Solution for particle motion is facilitated by resolving vector equation into scalar component equations, e.g.,

f t l t

for rectangular components,

(

^F_x^{ir + F}_y ^{rj + F}_z^kr

) (

^{= m a}_x^{ir + a}_y ^{rj + a}_z^kr

)

∑

z m F

y m F

x m F

ma F

ma F

ma F

z y

x

z z

y y

x x

&&

&&

&& = =

=

=

=

=

∑

∑

∑

∑

∑

∑

• For tangential and normal components, ma

F ma

F_t = _t

∑

_n = _n

∑

ρ v2

m dt F

mdv

F_t =

∑

_n =

∑

12.6 Dynamic Equilibrium y q

• Alternate expression of Newton’s second law, a

m

F − = 0

∑

^{r r}

ector inertial v

a m ≡

−

∑

r

• With the inclusion of the inertial vector, the system

f f i h i l i i l

of forces acting on the particle is equivalent to zero.

The particle is in dynamic equilibrium.

• Methods developed for particles in staticMethods developed for particles in static

equilibrium may be applied, e.g., coplanar forces may be represented with a closed vector polygon.

• Inertia vectors are often called inertial forces as they measure the resistance that particles offer to changes in motion, i.e., changes in speed or g , , g p

direction.

• Inertial forces may be conceptually useful but are

lik h d i i l f f d

not like the contact and gravitational forces found in statics.

Sample Problem 12.3 p

APPROACH:

h i h ki i f h bl k d

• What is the kinematics of the blocks and pulley?

• What is the equation of motion for the blocks and pulley?

• Combine the kinematic relationships The two blocks shown start from rest.

The horizontal plane and the pulley are frictionless, and the pulley is assumed

p

with the equations of motion to solve for the accelerations and cord tension.

frictionless, and the pulley is assumed to be of negligible mass. Determine the acceleration of each block and the tension in the cord

tension in the cord.

Sample Problem 12.3 p

O

A B

A

B x a a

y 2

1 2

1 =

x =

y O

A :

A x m a F =

∑

y

( )

^a_A

T₁ = 100kg

B :

B

y m a

F =

∑

( ) ( ) ( )

B

B B B

a T

a m T

g m

kg 300 s

m 81 . 9 kg

300 ² ₂

2

=

−

=

−

( )

( )

^a_B

T₂ = 2940N- 300kg :

= 0

∑

Fy_y = mCaC

0 2 ₁

2 − T = T

Sample Problem 12.4 p

SOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus block as the acceleration of wedge plus the acceleration of the block relative to the wedge.

The 12-N block B starts from rest and

• Write the equations of motion for the wedge and block.

slides on the 30-N wedge A, which is supported by a horizontal surface.

l i f i i d i ( ) h

• Solve for the accelerations.

Neglecting friction, determine (a) the acceleration of the wedge, and (b) the acceleration of the block relative to the wedge.

Sample Problem 12.4 p

SOLUTION:

• The block is constrained to slide down the wedge. Therefore, their motions are dependent. Express the acceleration of block as the acceleration of wedge plus block as the acceleration of wedge plus the acceleration of the block relative to the wedge.

• Write the equations of motion for the wedge and block.

• Solve for the accelerations.

Sample Problem 12.5 p

SOLUTION:

• Resolve the equation of motion for theResolve the equation of motion for the bob into tangential and normal

components.

• Solve the component equations for the normal and tangential accelerations.

S l f th l it i t f th

The bob of a 2-m pendulum describes an arc of a circle in a vertical plane If

• Solve for the velocity in terms of the normal acceleration.

an arc of a circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position

h fi d th l it d l

shown, find the velocity and accel- eration of the bob in that position.

Sample Problem 12.5 p

SOLUTION:

• Resolve the equation of motion for theResolve the equation of motion for the bob into tangential and normal

components.

• Solve the component equations for the normal and tangential accelerations.

S l f th l it i t f th

• Solve for the velocity in terms of the normal acceleration.

Sample Problem 12.6 p

SOLUTION:

• The car travels in a horizontal circular path with a normal component of

acceleration directed toward the center of the path The forces acting on the car of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

Determine the rated speed of a highway curve of radius ρ = 400 m

• Resolve the equation of motion for the car into vertical and normal components

highway curve of radius ρ 400 m banked through an angle θ = 18^o. The rated speed of a banked highway curve is the speed at which a car should

components.

• Solve for the vehicle speed.

is the speed at which a car should travel if no lateral friction force is to be exerted at its wheels.

Sample Problem 12.6 p

SOLUTION:

• The car travels in a horizontal circular path with a normal component of

acceleration directed toward the center of the path The forces acting on the car of the path.The forces acting on the car are its weight and a normal reaction from the road surface.

• Resolve the equation of motion for the car into vertical and normal components

components.

• Solve for the vehicle speed.

12.7 Angular Momentum of a Particle g

• moment of momentum or the angular momentum of the particle about O.

=

×

= r mV Hr_O r r

k j

ir r r

• Hr_O is perpendicular to plane containing

V m rr and r φ

sin rmV H_O =

z y

x O

mv mv

mv

z y

x

k j

i Hr = θ

φ

θ

&

mr2

v rm

O

=

=

• Derivative of angular momentum with respect to time,

× +

×

=

× +

×

O = r mV r mV V mV r ma

H&r &r r r &r r r r r

z y

x

∑

∑

=

×

=

O O

M F r r

r

∑

O

• From Newton’s second law: the sum of the moments about O of the forces acting on the particle = the rate

f h f h l f h i l

of change of the angular momentum of the particle

12.8 Eqs of Motion in Radial & Transverse Components q p

(

^θ&2

)

∑

• Consider particle at r and θ, in polar coordinates,

( )

(

^{θ θ θ}

)

θ

θ && & &

&&

r r

m ma

F

r r m ma

F_{r r}

2

2

+

=

=

−

=

=

∑

∑

• This result may also be derived from conservation of angular momentum,

( )

^{θ θ}

θ &

&

d mr F

r

mr H_O

2 2

=

=

∑ ( )

( )

(

^θ

)

^θ

θ θ

&

&&

&

&

&& rr

r m dt mr F

r

2 + 2

=

∑

=

(

^{θ θ}

)

θ m r && r& &

F = + 2

∑

12.9 Conservation of Angular Momentum g

• When only force acting on particle is directed toward or away from a fixed point O, the particle

i id t b i d l f

is said to be moving under a central force.

• Since the line of action of the central force passes through O

∑

^{Mr H&r} 0 and

through O,

∑

MO ⁼ HO ⁼ 0 and constant

=

=

×mV H_O rr r r

• Position vector and motion of particle are in a plane perpendicular to Hr_O.

M it d f l t

• Magnitude of angular momentum,

0 0

0 sin

constant sin

φ φ V m r

V rm H_O

=

=

=

0 0

0mV sinφ r

momentum angular

constant

2

2 =

= H

mr H

O

O θ^&

or

momentum angular

2 = =

= r h

H_O θ^&

12.9 Conservation of Angular Momentum g

• Radius vector OP sweeps infinitesimal area

1 r2dθ

dA ²

2

= 1

D fi dA _{1 2} dθ _{1 2}θ_&

l l

• Define = θ = ₂θ =

2 2 1

2

1 r

dt r d dt

dA areal velocity

• Recall for a bod mo ing nder a central force

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

2 =

= r θ^&

h

• When a particle moves under a central force, its areal velocity is constant.

12.10 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

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,

Mm

2

n gravitatio of

constant

=

= G

r G Mm F

4 9 4

2 12 3

s lb 10 ft

4 . 34 s

kg 10 m

73 .

66 = × ⋅

× ⋅

= ^{− −}

• For particle of mass m on the earth’s surface, 2 ft m 32

81

9 =

=

=

= MG mg g m

W m 2 mg g 9.81 2 32.2 2

W

Sample Problem 12.7 p

SOLUTION:

• Write the radial and transverseWrite the radial and transverse equations of motion for the block.

• Integrate the radial equation to find an

A block B of mass m can slide freely on

expression for the radial velocity.

• Substitute known information into the transverse equation to find an A block B of mass m can slide freely on

a frictionless arm OA which rotates in a horizontal plane at a constant rateθ^&₀ .

the transverse equation to find an expression for the force on the block.

a) the component v of the velocity of B Knowing that B is released at a distance r₀ from O, express as a function of r

a) the component v_r of the velocity of B along OA, and

b) the magnitude of the horizontal force exerted on B by the arm OA.

Sample Problem 12.7 p

• Integrate the radial equation to find an expression for the radial velocity.

dv dr

dv dv

dr v dv dt

dr dr dv dt

v dv

r_&&= _&r = ^r = ^r = _r ^r

=

=

=

=

=

=

r r

r r r

r r

dr r

dr r

dv v

dr v dv dt

dr dr dv dt

v dv r

02

2 θ

θ^{& &}

&

&&

SOLUTION:

• Write the radial and transverse

equations of motion for the block

∫

^{= r}

∫

r v

r r

r r

dr r dv

v

dr r

dr r

dv v

r 2

0 0

0

θ

θ θ

&

equations of motion for the block.

: :

θ θ ma F

a m F_{r r}

=

=

∑

∑ ( )

(

^{r&&θ&& θ&r}_&^θ&

)

m F

r r m

2

0 ²

+

=

−

=

r₀

0 2

(

2 0²

)

2 0 r r

v_r =

θ

−

S b tit t k i f ti i t th

θ :

θ ma F =

∑

^{F = m}

(

^{rθ + 2rθ}

)

• Substitute known information into the transverse equation to find an expression for the force on the block.

( )

^{1 2}

Sample Problem 12.8 p

SOLUTION:

Si h lli i i d

• 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.

A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18820 km/h from y an altitude of 240 km. Determine the velocity of the satellite as it reaches it maximum altitude of 2340 km. The maximum altitude of 2340 km. The radius of the earth is 6345 km.

Sample Problem 12.8 p

SOLUTION:

Si h lli i i d

• 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.

constant sin = H_O =

v

rm φ

=

=

A A B

B B

A A

r v r v

v m r v

m r

( ) ( )

(

6345 2340

)

km km 240

km/h 6345 18820

+

= +

rB

( )

km/h 4

. 14269

B = v