X2 t06 07 miscellaneous dynamics questions (2013)

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