Problem 1

The rocket is traveling in free flight along an elliptical trajectory The planet has amass k times that of the earth's. If the rocket has an apoapsis and periapsis as shownin the figure, determine the speed of the rocket when it is at point A.

Units Used:

Mm 103 km=

Given:

k 0.60=

a 6.40 Mm=

b 16 Mm=

r 3.20 Mm=

G 6.673 10 11−× Nm2

kg2⋅=

Me 5.976 1024 kg×=

Solution: Central - Force Motion: Substitute Eq

rar0

2G M

r0 v02

1−= with r0 rp= a=

M k Me=

ba

2G M

a v02

1−=

ab

2G M

a v02

1−⎛⎜⎝

⎞⎟⎠

= 1ab

+⎛⎜⎝

⎞⎟⎠

2G k Me

a vp2

=

vp2G k Me b

a b+( )a= vp 7.308

kms

=

2 2 20 0 0 0 0

1 1 1 cose eGM GMr r r v r v

θ

= − +

For θ = Π

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0 0h r v=

0

0c

GMv

r=

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0

0c

GMv

r=

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300 N

50 N

1,5 m 9,81 m/s2

2,15 m/s

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25 kg

2,4 m/s 3 m

4 m

9,81 m/s2

2,55 kg

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9,81 m/s2

75 kN

15 kN

1,5 m/s

1,5 m/s

1,05 m/s

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Problem 10

The box of weight Wb slides on the surface for which the coefficient of friction is μk. The box hasvelocity v when it is a distance d from the plate. If it strikes the plate, which has weight Wp and isheld in position by an unstretched spring of stiffness k, determine the maximum compressionimparted to the spring. The coefficient of restitution between the box and the plate is e. Assume thatthe plate slides smoothly

Given:

Wb 10 N= Wp 50 N=

μk 0.3= k = 6500 N/m

v = 4.5 m/s e 0.8=

Solution:

Given12

Wbg

⎛⎜⎝

⎞⎟⎠

v2 μk Wb d−12

Wbg

⎛⎜⎝

⎞⎟⎠

vb12=

Wbg

⎛⎜⎝

⎞⎟⎠

vb1Wbg

⎛⎜⎝

⎞⎟⎠

vb2Wpg

⎛⎜⎝

⎞⎟⎠

vp2+=

e vb1 vp2 vb2−=12

Wpg

⎛⎜⎝

⎞⎟⎠

vp22 1

2kδ2

=

vb1

vb2

vp2

δ

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

Find vb1 vb2, vp2, δ,( )=

vb1

vb2

vp2

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

4.09

-2.32

1.28

⎛⎜⎜⎝

⎞⎟⎟⎠

ms

=δ 0.036 m=

d = 0.6 m g = 9.81 m/s 2

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Problem 11

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Problem 12

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Problem 13

Determine the differential equation of motion for the dampedvibratory system shown.What type of motion occurs?

Given:

M 25 kg= k 100Nm

= c 200N s⋅m

=

Solution:

M g k y yst+( )− 2c y'− M y''=

M y'' k y+ 2c y'+ k yst+ M g− 0=

Equilibrium k yst M g− 0=

M y'' 2c y'+ k y+ 0= (1)

y''2cM

y'+kM

y+ 0=

By comparing Eq.(1) to Eq. 22-27

pkM

= p 2.00rads

=

cc 2M p= cc 100.00 Nsm

⋅=

Since c 200.00N s⋅m

= > cc 100.00N s⋅m

= , the system is overdamped

and will not oscillate. The motion is an exponential decay.

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Problem 14

A weight W is suspended from a spring having a stiffness k. If the weight is given an upwardvelocity of v when it is distance d above its equilibrium position, determine the equation whichdescribes the motion and the maximum upward displacement of the weight, measured from theequilibrium position. Assume positive displacement is downward.

Given:

ωnk gW

= ωn 13.90rads

= y A cos ωn t( ) B sin ωn t( )+=

A d−= Bv−

ωn=

y A cos ωn t( ) B sin ωn t( )+=

A = -0.05 m B = -0.22 m

ωn 27.12rads

=

C A2 B2+=

Solution:

W 10 N= k = 0.75 N/mmv = 6 m/s

d = 50 mm g = 9.81 m/s 2

C = -0.22 m

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Problem 15

A weight W is attached to a spring having a stiffness k. The weight is drawn downward a distanced and released from rest. If the support moves with a vertical displacement δ = δ0 sinωt, determinethe equation which describes the position of the weight as a function of time.

Given:

δ0 13 mm=

ω = 4 rad/s

Solution:

For Static Equilibrium W kδst=

Equation of Motion is then

k y δst+ δ−( ) W−W−g

⎛⎜⎝

⎞⎟⎠

y''=

Wg

⎛⎜⎝

⎞⎟⎠

y'' k y+ kδ0 sin ωt( )=

y''k gW

⎛⎜⎝

⎞⎟⎠

y+k gW

⎛⎜⎝

⎞⎟⎠

δ0 sin ωt( )=

The solution consists of a homogeneous part and aparticular part

y t( ) A cosk gW

t⎛⎜⎝

⎞⎟⎠

B sink gW

t⎛⎜⎝

⎞⎟⎠

+δ0

1Wω2

k g−

sin ωt( )+=

The constants A and B are determined from the initial conditions.

A d= Bδ0− ω

1Wω2

k g−

⎛⎜⎝

⎞⎟⎠

k gW

= Cδ0

1Wω2

k g−

= pk gW

=

y A cos p t( ) B sin p t( )+ Csin ωt( )+=

where A 0.33 ft= B 0.0232− ft=

C 0.05 ft= p 8.97rads

= ω 4.00rads

=

W 20 N=

k = 160 N/m

d = 100 mm

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