Newton’s Second Law of motion can be used to set up a ...assets.openstudy.com/updates/...1315780551917-108newtonssecon… · down. Newton’s Second Law can be used to set up a

© Dr. Honeycutt Newton’s Second Law

Newton’s Second Law of motion can be used to set up a

differential equation to describe the object’s motion.

The acceleration of an object is directly proportional to

and in the direction of the net force acting on an object

and is inversely proportional to the object’s mass.

This assumes that the mass of the object is not changing.

More correctly the net force is equal to the change in the

object’s momentum, which is its mass times velocity.

In working a problem a direction for + needs to be set. When an object

is moving in only one direction + is usually chosen to be in the direction

of motion. If the object changes direction during its motion + may be

chosen to be upward, north, east, right, the direction of the initial

motion, or the direction of the motion in the time frame of interest.

Regardless of this choice, the interpretation of the results are the same.

Acceleration is the derivative of velocity and second derivative of position,

resulting in either a first or second order differential equation depending

on whether the dependent variable is velocity or position. i.e. which is the

physical quantity of interest.

ma F =

dt

mv d F

) ( =

2

2

dt

x d m =

dt

dv m = ma F =


Consider an object with constant mass moving vertically up or

down. Newton’s Second Law can be used to set up a diff. eq.

which describes the object’s motion over time. This equation

depends on the force of air resistance, R, acting on the object.

v

v

R

R

W = mg

W = mg As a resistive force, R will act opposite

the direction of motion and when entered

into the sum its sign depends on its

direction and the direction taken to be +.

W, weight, is the force of gravity acting

on the object and always acts downward.

When entered into the sum its sign

depends on the direction taken to be +.

downward motion

positive up

positive down

upward motion

positive up

positive down

ma = – W + R

ma = W – R

v

v +

v +

v

ma = – W – R

ma = W + R

An object speeds up when v and a are in

the same direction and slows down when

they are in opposite directions. g = 9.8 m/s2

= 32 ft/s2

For an object moving through the air at

a slow speed, R is proportional to the

object’s speed. For very fast objects it

is proportional to speed squared.

ma F =


Write a differential equation for the velocity of a ball falling

vertically downward if the force of air resistance is proportional

to the ball’s speed.

v

R = bv

W = mg

only downward motion

positive down

ma = W – R

v +

v

R = bv

W = mg

v

R = bv

W = mg

as the ball falls through the air it speeds up

as it speeds up the force of air resistance increases

as the force of air resistance increases the downward

acting net force, and consequently the magnitude of

the acceleration, decreases so it’s speed increases

less each second

the speed reaches a limiting value as t ∞ (terminal

velocity) where the force of air resistance balances with

the weight and the acceleration becomes zero

mg bv dt

dv m =

v m

b g

dt

dv =


A 5-kg ball is dropped and encounters air resistance which is

proportional to its speed. Write the velocity as a function of time

if the limiting velocity is 39.2 m/s.

limiting velocity occurs as t ∞

0

dropped v(0) = 0

b

mg v

=

m

b = v

m

b g

dt

dv

dt m

b =

b

mg v

dv

c t m

b + =

b

mg v ln

c t m

b

e +

m bt ce

= b

mg v =

m bt m bt ce ce

= b

mg v =

b

mg ce v

m bt + =

0 = + b

mg c ) 0 ( = v

b

mg c =

) ( b

mg 1

m bt e ) ( t v =

9 . 32 = = L v

b

mg

25 . 1 = = L v

mg b

4

1 = =

L v

g

m

b

) 1 ( 2 . 39 ) ( 4 t

e t v =


A 5-kg ball is dropped and encounters air resistance which is

proportional to its speed. Write the velocity as a function of time

if the limiting velocity is 39.2 m/s.

How long does it take for the ball to reach 95% of vL?

How far does the ball fall in this time? let y(0) = 0

4( 39.2)

= dt

dy ) 1 (

4 t

L e v = ) 1 ( 2 . 39 ) (

4 t e t v =

95 . 0 1 4 = t

e 05 . 0 4 = t

e 05 . 0 ln 4 =

t s t 98 . 11 05 . 0 ln 4 = =

c + e t + 4

8 . 156 t = 2 . 39 dt e t y t =

4 ) 1 ( 2 . 39 ) ( c y + = = 8 . 156 0 ) 0 (

8 . 156 8 . 156 2 . 39 ) ( 4 + = t

e t t y

m 66 . 320 8 . 156 ) 05 . 0 ( 8 . 156 ) 98 . 11 ( 2 . 39 = + y ) 98 . 11 ( =


Write a differential equation for the velocity of a ball thrown

vertically upward if R = 0, if R is proportional to the ball’s speed,

and if R is proportional to the square of its speed.

+ up v +

R W = mg

v R

W = mg

R = 0

ma = – W + R ma = – W – R

R = b|v|

R = bv2

same up and down:

no need to break the

problem into parts

different up and down:

work the problem in parts

same up and down:

no need to break the

problem into parts

recall R

represents the

magnitude of

the force

sub. a pos. v gives sub. a neg. v gives +

R mg m dt

dv m =

m

R g

dt

dv m =

g dt

dv = g

dt

dv =

m

bv g

dt

dv =

m

bv g

dt

dv =

m

bv 2

g dt

dv =

m

bv 2

+ g dt

dv =


A 0.2-kg ball is thrown vertically upward from the top of a 60-m

building with an initial velocity of 20 m/s. Write an equation for

the ball’s position assuming no air resistance and g = 10 m/s2.

g dt

dv = dt 10 = t v ) (

c t t v + = 10 ) (

t t v 10 20 ) ( = 20 ) 0 ( = = c v

) ( t v dt

dy = dt t ) 10 20 ( = dt t v ) ( = t y ) (

c t t t y + = 2 5 20 ) (

2 5 20 60 ) ( t t t y + = 60 ) 0 ( = = c x

+ up

W = mg m = 0.2 kg

y(0) = 60 m

v(0) = 20 m/s

g = 10 m/s2


How high does the ball go and how long is it in the air?

t t v 10 20 ) ( =

2 5 20 60 ) ( t t t y + =

find the max. height

1st set v = 0 and solve for t

2nd find y for that t

find the time to ground

set y = 0 and solve for t

20 m/s

10 m/s

0 m/s

40 m/s

30 m/s

20 m/s

10 m/s

0s

1s

2s

6s

5s

4s

3s

35 m/s

25 m/s

15 m/s

5 m/s

35 m

25 m

15 m

5 m

t t v 10 20 0 ) ( = =

s t top 2 =

m 80 = y h max ) 2 ( =

2 5 20 60 0 ) ( t t t y + = =

0 12 4 2 = t t

0 ) 2 )( 6 ( = + t t

s t 6 = s t 2 =


Write an equation for the position assuming the air resistance is

proportional to its speed and the same initial conditions.

+ up

m = 0.2 kg

y(0) = 60 m

v(0) = 20 m/s

g = 10 m/s2

v +

R W = mg

v R

W = mg

all the steps with

variations on c

20

v R =

20

1 = b

4 10

v =

) 2 . 0 )( 20 (

v 10 =

m

bv g

dt

dv =

) ( 4

1 40 + v =

dt

dv 40 = c 20 ) 0 ( = v 60 = c

dt 4

1 =

+ v

dv

40

c t + 4

1 v = + 40 ln

4 40

t ce v

= +

40 4 = t

ce v

40 60 ) ( 4 = t

e t v

c t + 40 e t =

240 4

dt e t y t =

) 40 60 ( ) ( 4

c + = 240 y = 60 ) 0 (

300 = c

4 240 40 300 ) (

t e t t y =

= dt

dy


How high does the ball go and how long is it in the air?

find the max. height

1st set v = 0 and solve for t

2nd find y for that t

find the time to ground

set y = 0 and solve for t

this can’t be solved

algebraically so we’ll use

trial and error here

40 60 ) ( 4 = t

e t v 4

240 40 300 ) ( t

e t t y =

0 40 60 4 = t

e

3

2

60

40 = 4 = t

e

3

2 ln =

4

t

s 622 . 1 t top 3

2 ln 4

=

3

2 240 ) 622 . 1 ( 40 300 =

max h

m h 126 . 75 max =

0 240 40 300 4 = t

e t

45 . 6 ) 000 . 6 ( y

306 . 0 ) 250 . 6 ( y

033 . 0 ) 240 . 6 ( y

0056 . 0 ) 239 . 6 ( y

s t air in 239 . 6


Write an equation for the position assuming the air resistance is

proportional to its speed squared and the same initial

conditions. How high does the ball go and how long is it in the air?

+ up

m = 0.2 kg

y(0) = 60 m

v(0) = 20 m/s

g = 10 m/s2

v +

R W = mg

v R

W = mg

different up and down:

work the problem in parts

y(0) = hmax

v(0) = 0 m/s

new initial conditions for the

downward motion where t = 0

corresponds to ttop of upward

initial conditions of the

upward motion are the

given initial conditions

800

2 v

R = 800

1 = b

m

bv g

dt

dv

up

2

= m

bv g

dt

dv

down

2

+ =

) ( 160

1 = 10 1600

2 + v 160

10 2

= v

) 2 . 0 )( 800 (

2

v

= dt

dv

up

) ( 160

1 = 1600

2 v 160

10 2

+ = v

) 2 . 0 )( 800 ( 10

2

+ = v

dt

dv

down


Upward motion m = 0.2 kg

y(0) = 60 m

v(0) = 20 m/s

g = 10 m/s2

integration table or

use trig substitution

radians

) ( 160

1 = 1600

2 + v dt

dv

up

dt 160

1 =

+ v

dv

40 2 2

c t + 160

1 v =

40 arctan

40

1

c + t 4

1 = v

40 arctan

+ c t

4

1 tan =

v

40

+ = c t v

4

1 tan 40

( ) c tan 40 = v 20 ) 0 ( =

( ) 5 . 0 tan = c 4636 . 0 ) 5 . 0 ( tan 1 =

c

( ) t t v 25 . 0 4636 . 0 tan 40 ) ( =

( ) = dt t y 25 . 0 4636 . 0 tan 40

= dt

dy

c t y + = ) 25 . 0 4636 . 0 sec( ln 160

c + = ) 4636 . 0 sec( ln 160 y = 60 ) 0 (

c + = 85 . 17 60 85 . 77 = c

) 25 . 0 4636 . 0 sec( ln 160 85 . 77 ) ( t t y =


Upward motion m = 0.2 kg

y(0) = 60 m

v(0) = 20 m/s

g = 10 m/s2

new initial conditions

y(0) = 77.85 m

v(0) = 0 m/s

for any complete time on the

way down 1.8546s needs to be

added to t results

( ) t t v up 25 . 0 4636 . 0 tan 40 ) ( =

) 25 . 0 4636 . 0 sec( ln 160 85 . 77 ) ( t t y up =

0 = ( ) 25 . 0 4636 . 0 tan 40 ) ( = t t v top

0 25 . 0 4636 . 0 = t

s 8546 . 1 = t top ) 4636 . 0 ( 4 =

m 85 . 77 = 1 ln 160 85 . 77 = h max ) 0 sec( ln 160 85 . 77 =


Downward motion m = 0.2 kg

y(0) = 77.85 m

v(0) = 0 m/s

g = 10 m/s2

integration table or

use partial fractions

v(0) = 0 m/s

) 40 ( 160

1 2 2 = v dt

dv

down

dt 160

1 =

v

dv

40 2 2

c t + 160

1

c + t 5 . 0 v

v =

+

40

40 ln

t ce

v

v 5 . 0

40

40 =

+

1 = c

t 1e

v

v 5 . 0

40

40 =

+

t e v

5 . 0 ) 40 ( + v 40 =

v t

e 5 . 0

40 40 = t ve

5 . 0 +

) 1 ( 40 5 . 0 t

e ) 1 ( 5 . 0 t

e v = +

2 1

2 1

) 25 . 0 cosh( t

) 25 . 0 sinh( t

25 . 0 e

t ) (

25 . 0 e

t

25 . 0 e

t ) (

25 . 0 e

t

40 =

40 =

1 5 . 0

e t +

1 5 . 0

e t

40 = 1

5 . 0 e

t +

1 5 . 0

e t

40 v =

25 . 0 e

t +

25 . 0 e

t

) 25 . 0 tanh( 40 ) ( t t v =

v

v +

40

40 ln

80

1 =


Downward motion m = 0.2 kg

y(0) = 77.85 m

v(0) = 0 m/s

g = 10 m/s2

cosh is always + so no

abs. val. is needed

ln(1) = 0

tup = 1.8546s needs to

be added for t total

c t + = )] 25 . 0 ln[cosh( 160 25 . 0

4 dt t

t

) 25 . 0 cosh(

) 25 . 0 sinh( 40 t y = ) (

) 25 . 0 tanh( 40 ) 25 . 0 cosh(

) 25 . 0 sinh( 40

1

1 40 5 . 0

5 . 0

t t

t

e

e v t

t

= = +

= =

dt

dy

c + = )] 0 ln[cosh( 160 y = 85 . 77 ) 0 ( 85 . 77 = c

)] 25 . 0 ln[cosh( 160 85 . 77 ) ( t t y =

)] 25 . 0 ln[cosh( 160 85 . 77 t = 0 y bottom =

4866 . 0 160

85 . 77 = )] 25 . 0 ln[cosh( = t

626 . 1 4866 . 0 = = e ) 25 . 0 cosh( t

4 ) 626 . 1 ( cosh 1 = t

s t down 2723 . 4

s t air in 127 . 6

u

du


e

e t

t

t 5 . 0

5 . 0 lim 40

5 . 0

5 . 0 =

e

e t

t

t 1

1 lim 40

5 . 0

5 . 0

+

What is the ball’s speed when it hits the ground, what percent of

the limiting velocity is this?

the – indicates

downward motion

limiting velocity

occurs as t ∞

∞ L’Hopital’s Rule

∞

this velocity is only for

downward motion so

only use the time for

downward motion, not

the total time

speed of impact is ~79% of the

limiting velocity

) 25 . 0 tanh( 40 ) 25 . 0 cosh(

) 25 . 0 sinh( 40

1

1 40 5 . 0

5 . 0

t t

t

e

e v t

t

= = +

=

s m / 40 = = t v t

) ( lim

s m v L / 40 =

s m / 55 . 31 = v bottom )) 2723 . 4 ( 25 . 0 tanh( 40 =

7887 . 0 / 40

/ 55 . 31 =

s m

s m


Equations for velocity and position in two part problems.

upward t = 0 to 1.8546s

downward t = 0 to 4.2723s

( ) t t v 25 . 0 4636 . 0 tan 40 ) ( =

) 25 . 0 4636 . 0 sec( ln 160 85 . 77 ) ( t t y =

) 25 . 0 tanh( 40 ) ( t t v =

)] 25 . 0 ln[cosh( 160 85 . 77 ) ( t t y =

< s t s 127 . 6 855 . 1 t 855 . 1 )) ( 25 . 0 tanh( 40

( ) s t t 855 . 1 25 . 0 4636 . 0 tan 40

= t v ) (

< s t s 127 . 6 855 . 1 t 855 . 1

s t t 855 . 1 ) 25 . 0 4636 . 0 sec( ln 160 85 . 77 t )

= y ( )) 25 . 0 cosh( ln 160 85 . 77 (


Compare the three problems:

t t v 10 20 ) ( =

2 5 20 60 ) ( t t t y + =

R = 0

< s t s 127 . 6 855 . 1 t 855 . 1 )) ( 25 . 0 tanh( 40

( ) s t t 855 . 1 25 . 0 4636 . 0 tan 40

= t v ) (

< s t s 127 . 6 855 . 1 t 855 . 1

s t t 855 . 1 ) 25 . 0 4636 . 0 sec( ln 160 85 . 77 t )

= y ( )) 25 . 0 cosh( ln 160 85 . 77 (

800

2 v

R =

40 60 ) ( 4 = t

e t v 4

240 40 300 ) ( t

e t t y =

20

v R =

m h max 80 = s t top 2 = s t air in 6 =

m h 13 . 75 max = s t top 622 . 1 s t air in 239 . 6

m h max 85 . 77 = s t top 855 . 1 = s t air in 127 . 6


____

____

____

R = 0 y(t)

t

2 4 6

20

40

60

80

t

v(t)

2 4 6

40

20

20

800

2 v

R =

20

v R =


t

v(t)

4 8 12

40

20

20

If thrown from a greater height the velocity curves continue as

shown. In the cases with air resistance both approach their

limiting velocity of 40 m/s.


Related Homework Problems:

1.1: 25 a, b, c

1.2: 9, 10

2.3: 20 – 24, 28

note: |v| is speed just v is needed in equations if attaching

appropriate signs – don’t forget the hidden sign in v

#23: in physics freefall means no force other than gravity in

this problem he means without the parachute

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Newton’s Second Law of motion can be used to set up a ...assets.openstudy.com/updates/...1315780551917-108newtonssecon… · down. Newton’s Second Law can be used to set up a