X2 T07 07 miscellaneous dynamics questions
116
Miscellaneous Dynamics Questions e.g. (i) (1992) – variable angular velocity The diagram shows a model train T that is moving around a circular track, centre O and radius a metres. The train is travelling at a constant speed of u m/s. The point N is in the same plane as the track and is x metres from the nearest point on the track. The line NO produced meets the track at S. diagram in the as and Let TOS TNS
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Transcript of X2 T07 07 miscellaneous dynamics questions
Miscellaneous Dynamics Questions
e.g. (i) (1992) – variable angular velocityThe diagram shows a model train T
that is
moving around a circular track, centre O and radius a
metres.
The train is travelling at a constant speed of u
m/s. The point N
is in the same plane as
the track and is x
metres from the nearest point on the track. The line NO
produced
meets the track at S.diagramin theasand Let TOSTNS
uadtdθ and of in terms Express a)
uadtdθ and of in terms Express a) al
uadtdθ and of in terms Express a) al
dtda
dtdl
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
coscoscos
that;deduceand 0sinsinthat Show b)
aaxu
dtd
ax-a
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
coscoscos
that;deduceand 0sinsinthat Show b)
aaxu
dtd
ax-a
),(exterior OTNTOSTNONTO
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
coscoscos
that;deduceand 0sinsinthat Show b)
aaxu
dtd
ax-a
),(exterior OTNTOSTNONTO
NTO
NTO
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
coscoscos
that;deduceand 0sinsinthat Show b)
aaxu
dtd
ax-a
),(exterior OTNTOSTNONTO
NTO
NTO
sinsin
;In xaaNTO
uadtdθ and of in terms Express a) al
dtda
dtdl
au
dtd
dtdau
coscoscos
that;deduceand 0sinsinthat Show b)
aaxu
dtd
ax-a
),(exterior OTNTOSTNONTO
NTO
NTO
sinsin
;In xaaNTO
0sinsin
sinsin
xaa
xaa
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
when NT
is a tangent;
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
when NT
is a tangent;
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
when NT
is a tangent;radius)(tangent 90 NTO
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
when NT
is a tangent;radius)(tangent 90 NTO
90
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
when NT
is a tangent;radius)(tangent 90 NTO
90
cos90cos90cos
xaau
dtd
0coscos
respect to with atedifferenti
dtdxa
dtd
dtda
t
0coscoscos dtdxaa
dtda
aua
dtdxaa coscoscos
coscos
cosxaa
udtd
track. the tol tangentiais when 0 that Show c) NTdtd
N
T
O
when NT
is a tangent;radius)(tangent 90 NTO
90
cos90cos90cos
xaau
dtd
0dtd
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
53
0
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
0when
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
0when
0cos20cos0cosaa
udtd
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
0when
0cos20cos0cosaa
udtd
au3
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
0when
0cos20cos0cosaa
udtd
au3
au53
5
d) Suppose that x
= aShow that the train’s angular velocity about N
when is
times the angular velocity about N
when 2
2when
53
0
T
ON
a
2a
a5 52cos
51
2cos
cos22
cos
2cos
aa
u
dtd
522
51
51
aa
u
au5
0when
0cos20cos0cosaa
udtd
au3
au53
5
0n locity wheangular ve the
times53 is
2n locity wheangular ve theThus
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
ls
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
ls
dtdl
dtdsv
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
dtd
ddv
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
dtd
ddv
lv
ddv
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
dtd
ddv
lv
ddv
vddv
l
1
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
dtd
ddv
lv
ddv
vddv
l
1
2
211 v
dvd
ddv
l
(ii) (2000) A string of length l
is initially vertical and has a mass P
of m
kg attached to it. The mass P
is given a
horizontal velocity of magnitude V and begins to move along the arc of a circle in a counterclockwise direction.Let O
be the centre of this circle and A
the initial
position of P. Let s
denote the arc length AP, , dtdsv
AOP and let the tension in the string be T. The acceleration due to gravity is g
and there are no frictional forces acting on P.
For parts a) to d), assume the mass is moving along the circle.a) Show that the tangential acceleration of P
is given by
2
2
2
211 v
dd
ldtsd
dtdv
dtsd2
2ls
dtdl
dtdsv
dtd
ddv
lv
ddv
vddv
l
1
2
211 v
dvd
ddv
l
2
211 v
dd
l
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sinmg
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
sin211 2 gv
dd
l
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
sin211 2 gv
dd
l
sin21 2 glv
dd
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
sin211 2 gv
dd
l
sin21 2 glv
dd
cglv
cglv
cos2
cos21
2
2
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
sin211 2 gv
dd
l
sin21 2 glv
dd
cglv
cglv
cos2
cos21
2
2
glVccglV
Vv
22
0,when
2
2
b) Show that the equation of motion of P
is
sin211 2 gv
dd
l
T
mg
sm
sinmg
sinmgsm sings
sin211 2 gv
dd
l
cos12 that Deduce c) 22 glvV
sin211 2 gv
dd
l
sin21 2 glv
dd
cglv
cglv
cos2
cos21
2
2
glVccglV
Vv
22
0,when
2
2
cos12cos22
2cos2
22
22
22
glvVglglvV
glVglv
21cosy Explain wh d) mvl
θT-mg
21cosy Explain wh d) mvl
θT-mg
21cosy Explain wh d) mvl
θT-mg T
21cosy Explain wh d) mvl
θT-mg T cosmg
21cosy Explain wh d) mvl
θT-mg T cosmgxm
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm But, the resultant force towards the centre is centripetal force.
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
cos1231cos0 glglml
mg
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
cos1231cos0 glglml
mg
cos2cos ggmmg
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
cos1231cos0 glglml
mg
cos2cos ggmmg
mgmg cos3
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
cos1231cos0 glglml
mg
cos2cos ggmmg
mgmg cos3
31cos
21cosy Explain wh d) mvl
θT-mg T cosmgxm
cosmgTxm
2
2
1cos
cos
mvl
mgT
mgTl
mv
But, the resultant force towards the centre is centripetal force.
0at which of value theFind .3 that Suppose e) 2 TglV
cos121cos 2 glVml
mgT
cos1231cos0 glglml
mg
cos2cos ggmmg
mgmg cos3
31cos
radians911.1
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
f) Consider the situation in part e). Briefly describe, in words, the path of P
after the tension T
becomes zero.
When T = 0, the particle would undergo projectile motion, i.e. it would follow a parabolic arc.Its initial velocity would be tangential to the circle with magnitude;
21cos mvl
mgT
2131 mv
lmg
3
32
glv
glv
(iii) (2003)A particle of mass m
is thrown from the top, O, of a very tall building
with an initial velocity u
at an angle of to the horizontal. The particle experiences the effect of gravity, and a resistance proportional to its velocity in both directions.
The equations of motion in the horizontal and vertical directions are given respectively by
gykyxkx and where k is a constant and the acceleration due to gravity is g.
(You are NOT required to show these)
cosresult theDerive a) ktuex
cosresult theDerive a) ktuex
xkdtxd
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
yugyk
kt sinlog1
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
yugyk
kt sinlog1
gkugykkt sinloglog
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
yugyk
kt sinlog1
gkugykkt sinloglog
gkugykkt
sinlog
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
yugyk
kt sinlog1
gkugykkt sinloglog
gkugykkt
sinlog
ktegku
gyk
sin
cosresult theDerive a) ktuex
xkdtxd
x
u xxd
kt
cos
1
xuxk
t coslog1
cosloglog1 uxk
t
coslog1
ux
kt
coscos
coslog
kt
kt
uex
eu
xu
xkt
condition initial andmotion ofequation
eappropriat thesatisfies sin1t Verify tha b) gegkuk
y kt
gykdtyd
y
u gykydt
sin
yugyk
kt sinlog1
gkugykkt sinloglog
gkugykkt
sinlog
ktegku
gyk
sin
gegkuk
y kt sin1
c) Find the value of t
when the particle reaches its maximum height
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
cos
cos
kt
kt
uedtdx
uex
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
cos
cos
kt
kt
uedtdx
uex
0
coslim dteux kt
t
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
cos
cos
kt
kt
uedtdx
uex
0
coslim dteux kt
t
tkt
te
kux
0
1coslim
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
cos
cos
kt
kt
uedtdx
uex
0
coslim dteux kt
t
tkt
te
kux
0
1coslim
1coslim
kt
te
kux
c) Find the value of t
when the particle reaches its maximum height0when occursheight Maximum y
0sinlog1ugyk
kt
gkugk
t sinloglog1
ggku
kt sinlog1
d) What is the limiting value of the horizontal displacement of the particle?
cos
cos
kt
kt
uedtdx
uex
0
coslim dteux kt
t
tkt
te
kux
0
1coslim
1coslim
kt
te
kux
kux cos
Exercise 9E; 1 to 4, 7
Exercise 9F; 1, 2, 4, 7, 9, 12, 14, 16, 20, 22, 25
