Dynamics

DynamicsNewton’s Laws Of Motion

DynamicsNewton’s Laws Of Motion

Law 1“ Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter, nisi quatenus a viribus impressis cogitur statum illum mutare.”

DynamicsNewton’s Laws Of Motion

Law 1“ Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter, nisi quatenus a viribus impressis cogitur statum illum mutare.”

Everybody continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.

DynamicsNewton’s Laws Of Motion

Law 1“ Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter, nisi quatenus a viribus impressis cogitur statum illum mutare.”

Everybody continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.

Law 2“ Mutationem motus proportionalem esse vi motrici impressae, et

fieri secundum rectam qua vis illa impritmitur.”

DynamicsNewton’s Laws Of Motion

Law 1“ Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter, nisi quatenus a viribus impressis cogitur statum illum mutare.”

Everybody continues in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed upon it.

Law 2“ Mutationem motus proportionalem esse vi motrici impressae, et

fieri secundum rectam qua vis illa impritmitur.”

The change of motion is proportional to the motive force impressed, and it is made in the direction of the right line in which that force is impressed.

motionin change force Motive α

motionin change force Motive αonaccelerati αF

motionin change force Motive αonaccelerati αF

onaccelerati constant ×=F

motionin change force Motive αonaccelerati αF

maF =∴

onaccelerati constant ×=F

(constant = mass)

motionin change force Motive αonaccelerati αF

maF =∴

onaccelerati constant ×=F

(constant = mass)

Law 3“ Actioni contrariam semper et aequalem esse reactionem: sive

corporum duorum actions in se mutuo semper esse aequales et in partes contrarias dirigi.”

motionin change force Motive αonaccelerati αF

maF =∴

onaccelerati constant ×=F

(constant = mass)

Law 3“ Actioni contrariam semper et aequalem esse reactionem: sive

corporum duorum actions in se mutuo semper esse aequales et in partes contrarias dirigi.”

To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed in contrary parts.

Acceleration

Acceleration

dtdvdt

xdx

=

= 2

2

Acceleration

dtdvdt

xdx

=

= 2

2

(acceleration is a function of t)

Acceleration

dtdvdt

xdx

=

= 2

2

(acceleration is a function of t)

= 2

21

vdxd

(acceleration is a function of x)

Acceleration

dtdvdt

xdx

=

= 2

2

(acceleration is a function of t)

= 2

21

vdxd

(acceleration is a function of x)

dxdv

v= (acceleration is a function of v)

Acceleration

dtdvdt

xdx

=

= 2

2

(acceleration is a function of t)

= 2

21

vdxd

(acceleration is a function of x)

dxdv

v= (acceleration is a function of v)e.g. (1981) Assume that the earth is a sphere of radius R and that, at a point r from the centre of the earth, the acceleration due to gravity is proportional to and is directed towards the earth’s centre.

A body is projected vertically upwards from the surface of the earth with initial speed V.

( )R≥

2

1r

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=vrm

resultantforce

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

2rr

µ−=

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

2rr

µ−=

grRr −== ,when

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

2rr

µ−=

grRr −== ,when

2

2i.e.

gR

Rg

=

−=−

µ

µ

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

2rr

µ−=

grRr −== ,when

2

2i.e.

gR

Rg

=

−=−

µ

µ

2

2

rgR

r−=

(i) Prove that it will escape the earth if and only if where g is the magnitude of the acceleration due to gravity at the earth’s surface.

gRV 2≥

O

grVvRr −=== ,,

0=v

2rmµ

rm resultant

force2r

mrm

µ−=

2rr

µ−=

grRr −== ,when

2

2i.e.

gR

Rg

=

−=−

µ

µ

2

2

rgR

r−=

2

22

21

rgR

vdrd −=

cr

gRv

cr

gRv

+=

+=

22

22

2

21

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

gRV

gRVr

gRv

rr

2

22

limlim

2

22

2

−=

−+=

∞→∞→

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

gRV

gRVr

gRv

rr

2

22

limlim

2

22

2

−=

−+=

∞→∞→

0 Now 2 ≥v

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

gRV

gRVr

gRv

rr

2

22

limlim

2

22

2

−=

−+=

∞→∞→

0 Now 2 ≥v

022 ≥−∴ gRV

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

gRV

gRVr

gRv

rr

2

22

limlim

2

22

2

−=

−+=

∞→∞→

0 Now 2 ≥v

022 ≥−∴ gRV

gRV 22 ≥

cr

gRv

cr

gRv

+=

+=

22

22

2

21

VvRr == ,when

gRVc

cRgR

V

2

2 i.e.

2

22

−=

+=

gRVr

gRv 2

2 22

2 −+=

If particle never stops then ∞→r

gRV

gRVr

gRv

rr

2

22

limlim

2

22

2

−=

−+=

∞→∞→

earth. theescapes particle thei.e.

stops,never particle the2 if Thus gRV ≥

0 Now 2 ≥v

022 ≥−∴ gRV

gRV 22 ≥

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

,2 If gRV =

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

rgR

v

gRgRr

gRv

22

22

2

222

=

−+=,2 If gRV =

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

rgR

v

gRgRr

gRv

22

22

2

222

=

−+=,2 If gRV =

rgR

v22= (cannot be –ve, as it does not return)

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

rgR

v

gRgRr

gRv

22

22

2

222

=

−+=,2 If gRV =

rgR

v22= (cannot be –ve, as it does not return)

rgR

dtdr 22=

(ii) If ,prove that the time taken to rise to a height R above the earth’s surface is;

gRV 2=

( )gR

2431 −

gRVr

gRv 2

2 22

2 −+=

rgR

v

gRgRr

gRv

22

22

2

222

=

−+=,2 If gRV =

rgR

v22= (cannot be –ve, as it does not return)

rgR

dtdr 22=

22gRr

drdt =

∫=R

R

drrgR

t2

22

1

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

( )RRRRgR

−= 223

22

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

( )RRRRgR

−= 223

22

( ) RRgR

1223

2 −=

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

( )RRRRgR

−= 223

22

( ) RRgR

1223

2 −=

( )243

−=g

R

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

( )RRRRgR

−= 223

22

( ) RRgR

1223

2 −=

( )243

−=g

R

( )gR

2431 −=

∫=R

R

drrgR

t2

22

1

∫ RRR

R

2 to from g travellinas2

R

R

rgR

2

2

3

2 32

2

1

=

( )RRRRgR

−= 223

22

( ) RRgR

1223

2 −=

( )243

−=g

R

( )gR

2431 −= ( ) seconds 24

31 is surface searth'

above height a torise taken totime

gR

R

−

∴

Exercise 8C; 3, 15, 19