2016/2017 SPRING SEMESTER

AE262 DYNAMICS

Homework 1 Solutions

Prepared by: Harun Levent SAHIN

Posted: March 20, 2017

Problem 1:

1

From the figure, constraint equation of cable C1 is:

xA +xB = constant

v A + vB = 0 =⇒ v A =−vB

aA +aB = 0 =⇒ aA =−aB

Constraint equation of cable C2 is:

xB +2xD = constant

vB +2vD = 0 =⇒ vD = v A

2

aB +2aD = 0 =⇒ aD = aA

2Constraint equation of cable C3 is:

(xC −xB )+ (xB −xD ) = constant

2vC − vB − vD = 0 =⇒ vC =−v A

4

2aC −aB −aD = 0 =⇒ aC =−aA

4Constraint equation of cable portion BE is:

xB +xE = constant

vB + vE = 0 =⇒ vB =−vE

aB +aE = 0 =⇒ aB =−aE

Relative motion:v A/D = v A − vD = v A

2

aA/D = aA −aD = aA

2Note that, since block C moves downward, vC and aC are positive. Then v A , vD , aA and aD

are negative.

(a) It is given that |v A/D | = 2.5 [m/s] after 6 [s]:

|aA| = |2aA/D | = 2|v A/D |∆t

= 0.833 [m/s2]

aA =−0.833 [m/s2]

aC =−aA

4= 0.208 [m/s2] (downward)

(b) Acceleration of point E is:

aE =−aB = aA =−0.833 [m/s2] (upward)

2

Problem 2:

From the figure, constraint equation of cable C1 is:

2xA +2xB +xC = constant

2v A +2vB + vC = 0

2aA +2aB +aC = 0 (2.1)

Constraint equation of cable C2 is:

(xD −xA)+ (xD −xB ) = constant

2vD − v A − vB = 0

2aD −aA −aB = 0 (2.2)

3

Given:aC /B = aC −aB =−140 [mm/s2] =⇒ aC = aB −140 [mm/s2] (2.3)

aD/A = aD −aA = 240 [mm/s2] =⇒ aD = aA +240 [mm/s2] (2.4)

Substituting (2.3) and (2.4) int (2.1) and (2.2) gives 2 equations and 2 unknowns:

2aA +2aB + (aB −140) = 0 =⇒ 2aA +3aB = 140 [mm/s2] (2.5)

2(aA +240)−aA −aB = 0 =⇒ aA −aB =−480 [mm/s2] (2.6)

Solving (2.5) and (2.6) simultaneously gives:

aA =−260 [mm/s2] and aB = 220 [mm/s2]

aC = 80 [mm/s2] and aD =−20 [mm/s2]

(a) Velocity of block C after 7 [s] is:

vC = (vC )0 +ac t = 0+80 · (7)

vC = 560 [mm/s] (downward)

(b) Change in position of D after 10 [s] is:

∆xD = (vD )0t + aD t 2

2= 0+ −20 · (12)2

2

∆xD =−1440 [mm] (upward)

4

Problem 3:

From the figure, constraint equation of entire cable is:

xB + (xB −xA)+2(d −xA) = constant

2vB −3v A = 0

2aB −3aA = 0 =⇒ aA = 2aB

3= 0.333 [m/s2]

Constraint of point C:2(d −xA)+ yC /A = constant

−2v A + vC /A = 0

−2aA +aC /A = 0

(a) The relative acceleration of portion C of the cable with respect to slider block A is:

aC /A = 2aA = 0.667 [m/s2] (upward)

(b) The velocity of portion C of the cable after 3 [s] is:

v A = (v A)0 +aA t = 1 [m/s](→)

vC /A = (vC /A)0 +aC /A t = 2 [m/s](↑)

From the vector addition:

|vC | =√

v2A + v2

C /A = 2.236 [m/s](∠θ)

where θ is:

tanθ = vC /A

v A= 2

1= 2 =⇒ θ = 63.4o

5

Problem 4:

Given: r = (Rt cosωn t )i+ ct j+ (Rt sinωn t )k. Differentiating to obtain v and a:

v = dr

d t= R(cosωn t −ωn t sinωn t )i+ cj+R(sinωn t +ωn t cosωn t )k

a = dv

d t= R(−2ωn sinωn t −ω2

n t cosωn t )i+R(2ωn cosωn t −ω2n t sinωn t )k

(a) The magnitudes of the velocity and acceleration of the particle are:

v2 = v2x + v2

y + v2z

v2 =R2(cos2ωn t −2ωn t cosωn t sinωn t +ω2n t 2 sin2ωn t )+ c2

+R2(sin2ωn t +2ωn t sinωn t cosωn t +ω2n t 2 cos2ωn t )

v2 = R2(1+ω2n t 2)+ c2

v =√

R2(1+ω2n t 2)+ c2

a2 = a2x +a2

y +a2z

a2 =R2(4ω2n sin2ωn t +4ω3

n t sinωn t cosωn t +ω4n t 2 cos2ωn t )

+R2(4ω2n cos2ωn t −4ω3

n t cosωn t sinωn t +ω4n t 2 sin2ωn t )

a2 = R2(4ω2n +ω4

n t 2)

a = Rωn

√4+ω2

n t 2

(b) The radius of curvature of the path described by the particle when t = 0 can be foundas:Tangential component of acceleration when t = 0 is:

at = dv

d t= R2ω2

n t√R2(1+ω2

n t 2)+ c2

at |t=0 = 0

Normal component of acceleration when t = 0 is:

an =√

a2 −a2t

an |t=0 = a|t=0 = 2Rωn

Note that: an |t=0 = v2|t=0ρ , where ρ is the radius of curvature of particle when t = 0, then:

ρ = R2 + c2

2Rωn

6

Problem 5: This document is prepared in Mathcad Prime 3.0. I strongly recommend you to

learn and use any version of Mathcad. You can download METU licensed version fromhttps://ftpstu.cc.metu.edu.tr Given:

≔θ ((t)) +t3 4 t

≔r ((t)) +t3 2 t2

In order to find time ellapsed for one revolution of the rod solve the cubic equation as:

≔t 0

＝+t3 4 t ⋅2 π

≔tfinal =find ((t)) 1.17

Radial and transverse components of the velocity are:

≔vr ((t)) →――d

dtr ((t)) +⋅3 t2 ⋅4 t

≔vθ ((t)) ―――→⋅r ((t)) ――d

dtθ ((t))

expand+++⋅3 t

5 ⋅6 t4 ⋅4 t

3 ⋅8 t2

[s]

Radial and transverse components of the acceleration are:

≔ar ((t)) −――d

d

2

t2r ((t)) ⋅r ((t))

⎛⎜⎜⎝――d

dtθ ((t))

⎞⎟⎟⎠

2

≔aθ ((t)) +⋅r ((t)) ――d

d

2

t2θ ((t)) ⋅2 ――

d

dt⋅r ((t)) ――

d

dtθ ((t))

―――→ar ((t))expand

+−−−−−−⋅6 t ⋅18 t6 ⋅24 t5 ⋅48 t4 ⋅16 t3 ⋅32 t2 ⋅9 t7 4

―――→aθ ((t))expand

+++⋅36 t4 ⋅60 t

3 ⋅24 t2 ⋅32 t

Note that: ≔t , ‥0 0.01 tfinal Magnitudes of velocity and accelerations are:

≔x ((t)) ⋅r ((t)) cos ((θ ((t))))

≔y ((t)) ⋅r ((t)) sin((θ ((t))))

≔v ((t))‾‾‾‾‾‾‾‾‾‾‾‾

+vr ((t))2

vθ ((t))2

≔a ((t))‾‾‾‾‾‾‾‾‾‾‾‾

+ar ((t))2

aθ ((t))2

-2.45

-2.1

-1.75

-1.4

-1.05

-0.7

-0.35

0

0.35

-3.15

-2.8

0.7

-0.6 0 0.6 1.2 1.8 2.4 3 3.6 4.2-1.8 -1.2 4.8

x ((t))

y ((t))

(a)

7

10.5

14

17.5

21

24.5

28

31.5

35

0

3.5

38.5

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.20 0.1 1.3

t

vr ((t))

vθ ((t))

v ((t))

(b)

-140

-70

0

70

140

210

280

350

-280

-210

420

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.20 0.1 1.3

t

ar ((t))

aθ ((t))

a ((t))

(c)