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 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
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
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
∫ 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
−
∴