me3133: d exam #3- s 2012 100 points closed book...

ME3133: DYNAMICS

EXAM #3- SPRING 2012 100 POINTS - CLOSED BOOKFORMULA SHEET ALLOWED

CLASS ID: VERBOSE KEY

Instructions:

1. SET CELL PHONES OFF/SILENT -- THEY ARE NOT TO BE USED DURING EXAM

2. FREE BODY DIAGRAMS or ACTIVE FORCE DIAGRAMS are required in order to receive ANY

credit for KINETICS problems (i.e. those involving forces in the analysis).

3. DO NOT OPEN this exam until you are told to do so. The exam pages will be separated for

grading; therefore, your CLASS ID # must appear at the top of every page you want graded.

You will have exactly 105 minutes (1 hour 45 minutes) to complete this examination. Use

only your CLASS ID on exam pages as this preserves anonymity in grading.

4. Work only on the front side of any page. Extra paper will be available from the proctor if

necessary. Each extra sheet of paper should relate to one and only one exam problem. Be

certain you indicate on the original question sheet that your solution is continued on another

page and clearly label each sheet at the top with your Class ID # and the problem number.

5. Define all acronyms & abbreviations the first time you use them in every problem. Use

illustrations, figures and other pictorial representations where appropriate to assist with your

explanation. This alone, however, does not constitute a complete solution unless explicitly

requested as such in the exam question. Be certain you concisely explain your work if you

hope to receive full or partial credit. You are expected and encouraged to utilize your technical

insight and intuition in the formulation and resolution of these questions. Remember, however,

that others may not share your insights; therefore, briefly explain any unusual shortcuts or

assumptions you use in solving the problems.

6. If you get stuck on any problem, move on to the next one and return to it when you have

completed the remainder of the exam.

7. You may use pencils, erasers, a straight edge, protractor, and/or compass to complete the

examination.

8. Calculators: You may use a calculator ONLY for basic algebraic operations and

trigonometric functions. NO OTHER PRE-PROGRAMMED routines are to be used in the

generating the final numbers for a solution. Your instructor is more interested in your ability to

devise a solution strategy than in punching keys on a calculator. Therefore, formulate the

problem as far as you can using variables or symbols. Then only your last step is plug and

chug or number crunching. It also localizes numerical errors to the last step.

9. Portions of the questions contained in this examination may come directly from your

textbooks.

1. (9)

2. (30)

3. (30)

4. (30)

Total (100=99 Points + 1 Test Eval Below)

Test Evaluation (Circle One In Each Row) (NOTE: WORTH ONE POINT ON EXAMINATION):

CONTENT: Too Easy 1 2 3 4 5 Too Difficult

LENGTH: Too Short 1 2 3 4 5 Too Long

Class ID #: VERBOSE KEY

ME 3133: Dynamics Spring 2012 - Exam #3

(1) CONCEPT QUESTIONS (9 PTS)

(A) (8 pts) Each rigid body is shown below with velocity vectors at two points. For each VALID

velocity configuration :

(i) Indicate the location of the INSTANT CENTER of zero velocity;

(ii) Indicate its direction of rotation.

Figure A Figure B

Figure C Figure D

(B) (1 pt) A bowling ball is thrown down the alley with a strong backspin as shown (i.e. not rolling).

Is its INSTANT CENTER above or below the center at A? (Circle one) ABOVE BELOW



(2) GIVEN: (30 PTS) The mass D is raised and lowered by the electric powered lift shown. It

uses a belt drive connecting pulley A (r=50 mm) to a second pulley B (r=75mm) fixed to spool

C which carries the lift cord. Spool C (thus pulley B) has an initial clockwise angular velocity

of ωC0= -6 rad/s when the motor is engaged to begin braking the decent and gives pulley A a

counter-clockwise angular acceleration defined by

αΑ = 0.6 t2 + 0.75 (rad/s

2)

where t is in seconds and α is in radians/second2. For the instant where t=2.0 seconds:

FIND:

(A) The ACCELERATION (AD) of the block D,

(B) The VELOCITY (VD) of the block D,

(C) The ACCELERATION (AP) of the point P where the cord (un)winds from the rim of spool C.

(Bonus- 2Pts) The net vertical DISPLACEMENT (∆YD) of the block D during this time interval.

SOLUTION:

* The belt speed matches and equalizes that of the rim points (E&F) on pulleys A & B

respectively. It also matches the tangential accelerations of those rim points.

VE = ωA k × rE = −0.050ωA e t = −0.075ωBe t = ωB k × rF = VF (m /s)

AEt= αA k × rE = −0.050αAet = −0.075αBet = αB k × rF = AFt

(m / s2)

∴ωB =−0.050

−0.075ωA =

2

3ωA & αB =

−0.050

−0.075αA =

2

3αA

* Spool C is fixed to pulley B and has a common angular velocity & acceleration. The rim point P

has the same velocity as the block D and the acceleration of D is equal to the tangential

acceleration of P.

AD = APt= αB k × rP/B = 0.150αBetP

= (0.150)2

3

αA etP

=1

10αA j (m / s2 ) (ettP

= j )



EXTRA WORKSHEET PROBLEM # 2

* Knowing the angular acceleration αΑ, the acceleration of D as a function of time becomes

AD =1

10αA j=

1

100.6t

2 + 0.75( )j=3

50t2 +

3

40

j (m /s

2)

and in particular at t=2.0 seconds

AD t= 2.0=

3

502

2 +3

40

j = 0.315(m /s

2)j Ans A

* Integrate AD with time to find velocity VD using ωC0= -6 rad/s to get the initial speed of D.

VD = ADdt +∫ V0 =3

50t 2 +

3

40

0

t f

∫ dt + 0.15ωC 0

j (m /s)

=1

50t f

3 +3

40t f + 0.15(−6)

j =1

50t f

3 +3

40t f −

9

10

j (m /s)

and again at t=2.0 seconds

VD t f = 2.0=

1

5023 +

3

402 −

9

10

j = −59

100 j (m /s) Ans B

* Point P also has a normal acceleration in addition to the tangential component equal to AD.

AP =APn +APt = −vP

2

ρi+ AD j= −

−0.59( )2

0.15i+ 0.315j (m / s2 )

≅ −2.32i+ 0.32 j (m / s2 ) Ans C

* For the Bonus, continue integrating the expression for VD to get the net displacement.

and lastly at t=2.0 seconds



(3) GIVEN: (30 PTS) The two disks (1 & 2) roll without slipping on the plane surface

shown. Link AB is attached to the respective disk rims with pin joints. At the instant shown,

the disk 1 has a clockwise angular velocity of 2 radians per second, pin A is positioned

between disk centers and B is directly above O2. Use INSTANT CENTERS to estimate:

FIND:

(A) The VELOCITY (VO1) of the center of disk 1.

(B) The ANGULAR VELOCITY (ωAB) of the bar AB,

(C) The ANGULAR VELOCITY (ω2) of disk 2 on the right.

SOLUTION:

* Rolling contact for both disks thus P & R are the instant centers for disks 1 & 2 respectively.

* Since A is both on disk one and link AB, the line through AP must also contain M=ICAB, the

instant center for AB.

* Similarly, B is both on disk two and link AB, the line through RB must also contain M ==> M

lies at the intersection of these two lines.

* Point P & ω1 resolve VA from which point M can be used to resolve ωAB.

VA = ω1 rA /P = −2 2 = 2 2 = ωAB 3 2 = ωAB rA /M ( ft /s)

=⇒ ωAB = −1

3ω1 = −

1

3(−2) =

2

3r /s CCW Ans B




* Then, point M & ωAB resolve VB from which point R can be used to resolve ω2.

VB = ωAB rB /M =2

3 2 =

4

3= ω2 2 = ω2 rB /R ( ft /s)

=⇒ ω2 = −ωAB = −2

3r /s or CW Ans C



(4) GIVEN: (30 PTS)

In the mechanism shown, the rotating rod OB drives a compound slider consisting of two pin-

connected (C) collars (or sleeves) which are constrained to move along the fixed circular track

and the rod OB respectively. Rod OB rotates counterclockwise with a constant angular

velocity ω=3 rad/s. For the instant shown where θ=45°

FIND:

(A) Determine the VELOCITY of pin C connecting the two collars,

(B) What are the ANGULAR VELOCITIES of the two collars connected by pin C,

(C) Determine the ACCELERATION of pin C connecting the two collars.

(Bonus: 2 pts) Determine the ANGULAR ACCELERATIONS of the two collars connected by C.

SOLUTION:

* Place rotating axes at O with x axis along OB. The straight collar slides on OB and rotates with

it. For convenience of describing the circular collar motion, choose a fixed/inertial axes at the

center of the circle (D) with horizontal and vertical

VC = VO + ωOBk × rC /O + VRel = / O + ωOBk × 0.4 2i + vRel i (m /s)

vc I = 0.4 2ωOB j+ vRel i (m /s) ⇒ vc

2

2i −

2

2j

= 0.4 2ωOB j+ vRel i (m /s)




* Note that the j axes only involves unkown vC and can be resolved directly, then used to find vrel..

j's ⇒ −2

2vc = 0.4 2ωOB (m /s)

⇒ vc = −4

5* (3) = −2.4 or VC = −2.4I (m /s) Ans A

* For angular velocities.

ωStraightCollar = ωOB = 3 (r /s) CCW Ans B1

VC = vC I = ωCircularCollark × rE / D = ωCircularCollark × 0.4J = −0.4ωCircularCollarI

⇒ ωCircularCollar =vC

−0.4=

−2.4

−0.4 (r /s) ⇒ ωCircularCollar = 6 (r /s) CCW Ans B2

* For acceleration of pin C, vrel will have to be determined before proceeding => resolve the i’s.

i' s ⇒ vRe l =

2

2vc (m /s) ⇒ vRe l =

2

2(−2.4) (m /s) =

−6 2

5 (m /s) & VRel ≅ −1.7i (m /s)

Now looking to the rotating axes acceleration formula. C travels a circle so it has both normal

and tangential acceleration. The motion relative to the rotating axes is rectilinear along x (i).

AC =AC n+AC t

= −vc

2

rC /D

J + actI = −(

2.4 2

0.4)

2

2i +

2

2j

+ ac t

2

2i −

2

2j

=AO + αOBk × rC /O −ωOB

2 rC /O + 2ωOBk ×VRel +A Rel

= / O + / O − rC /OωOB

2i + 2ωOBvRel j + aRel i (m /s2) = −0.4 2(32)i + 2(3)(−

6 2

5)j + aRel i (m /s2)

Note that the j axes only involves aCl and can be resolved directly, then used to find arel.

j's ⇒ −(2.4 2

0.4)

2

2−

2

2act

= 2(3)(−6 2

5)(m /s2)

⇒ act=

−2

22(3)(−

6 2

5) + (

2.4 2

0.4)

2

2

=

72

5− (

2.4 2

0.4) ≅ 0 !

thus AC =AC n = −(2.4 2

0.4)J = −14.4J (m /s2) Ans C

* Bonus: Angular accelerations follow similar logic to angular velocity relations, .

ωOB ⇒ constant ⇒ αStraightCollar = αOB = 0 (r /s) Ans Bonus1

and

AC t= aC t

I = αCircularCollark × rE /D = αCircularCollark × 0.4J = −0.4αCircularCollarI

⇒ αCircularCollar =aC t

−0.4=

0

−0.4 (r /s) ⇒ αCircularCollar = 0 Ans Bonus2

me3133: d exam #3- s 2012 100 points closed book...

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