Задачи по инженерной динамике: подготовка к экзамену, Study Guides, Projects, Research of Mechanical Engineering

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Rennselaer Polytechnic Institute School of Engineering

ENGR 2090 Practice Problems: Exam 1: Engineering Dynamics

Section 03, Instructor: Professor Singh Due: NA Semester: Note the following: Fall 2022 Max points: NA

• Use the blueprint provided on LMS for solving dynamics problems. Writing gov-erning equations and solving using vector equations will result in a significantpercentage of the assigned credit being awarded.
• Although these problems are only representative, your crib sheet cannot includethese problems and their solutions.
• The exam will primarily include the following concepts: Kinematics-Kinetics, WorkEnergy principle, 1D Kinematics, Mass-pulley Systems, Friction, Relative Velocity,Relative Acceleration.

1

1. For the following expressions for acceleration, find the corresponding expressions forvelocity. Assume that the initial velocity at (t = 0) in the ‘N’ basis is ~v(t = 0) = 7.5 ˆi. (a) (b) ~~aa((tt) =) = ⇢lne( !⇢t^ eˆi! (^) t+ 3 (^) ) ˆi + (5cos 2 (sin!t)ˆ(2j!t) + 3cos 2 (!t))ˆj

2

3. The spring is unstretched when x = 0. If the body moves from the initial position xspring on the body and (b) determine the work done on the body by its weight. If 1 = 100 mm to the final position x 2 = 200 mm, (a) determine the work done by the the velocity of the block reduces to half of its initial value during its motion between x 1 and x 2 , find the initial velocity of the block.

4

4. The system is released from rest with the spring initially stretched 25 mm. First,compute the acceleration of the cart in this initial condition.velocity v of the 60-kg cart after it has moved 100 mm down the incline. Next, calculate the Also determine the maximum distance d traveled by the cart before it stops momentarily.Neglect the mass and friction of the pulleys and wheels.

5

Pr ob le m 2

B

p = 500 ' 30 ° ( i

i O

'N '

ñ

e n A

R-o T A = ie r + ro d o Now, s i nce path i s CIRCULAR,' II So , R-^ Of^ A^ =^ r o to / RO JA/ = 30 m i/ he r OR, (^) r 8 = 3 0 m i/ hr = 4 474 s OR, (^8) = 41 r ad /s 500 OR, (^8) = 0 - 088 rad/ s

No w , at th i s i nst ant i er = e t =

• e^ ^ i No w,

ñ

ñ

i i 3 0 °

Q ab = (^) - 5 m i/ ur /s ay

• (^) - 7. 33 f t / s" q, Wh at ar e t he com po nents o f-t his accel er atio n alo ng ↑ and j? NIB =

^

( i B. E ) i (^) -^ @^ a-^ B.^5 )^5

-^ - [^7.^33 95.^ Cos^3099 +^ s^ in^3 022 ]^ it^1 -^7 -^33 ai.^1 -^ sizogi^ +^ cos^ so^ %^ )]^ ;

1 "= 9 , co s 3 0 ° + 92 £ 3 0 ° ^ co s 3 0 °

÷

j^ ^^ =-^ ay,^ s^ in^30 °^ +^92

NE B = - 7. 33 cos 3 0 ° ↑ + 7. 33 s i n 30 ° j^ n O R;

But at^ t^ h^ is^ i nstant ;^ R-^02 A^ =^ N^ IA^ (^ si^ nce^ t^ he uni t ve ct o rs ar e coi ncid ent) So,

Naw AIB_ NIA - N IB

No w, R-^02 A^ = (i^ -^ ro^ "^ Je^ r^ +^ Cr^ o^ +^ ai^ o^ l^ io

Becaus e i t i^ s^ a^ cir^ cular^ ar^ c; F - o , É^ =^0

So, R-^ o^ @^ A - - ri te r +^ Oe^ i

And i N^ IA^ =^ - rot C- E ) = r ot i

Re visi ti ng eq. ① i N & AIB = NIA_Na^ →^ B

• -^ ro^ t^ +^7.^33 m^30 )^ i^ +^7.^3 3 sui^30 j^ +^4 s^ "

N IA/ B = 10 - 2199 % + 3. 665 ↑ f t/ 5

And , AN IA/ 13 1 /= 10 - 86 ft /s u

Y as

P ro b le m 3 -

PAY

ATTE N T IO N T O

ALL STE PS.

St ep 1 : Re fe r ence fr ame s

/ m

K

g-

k

Tr ans for mat i o ns : i = ai cos o (^) i

j

ñ

• ai s i nce^10 I = - ai si e + ai coso s t ep 2 : f re e Bo dy Diagr am

k

u^ m

K

g- :

• Asi / m

Ñ

E - •

Fg St eph: Go ver ni ng E quati o n W 1 - 2 D KE

Spri ng , gravi t y

Wipi ng = f f spri ng' d i

1 N u

• ka"g. d rag = - KI 2 /" "= 0 - 1

- 4 × 10. 2

I

v

• o- F) ✗ 1000

= - 60 J

W gr avit y = f F gravity - di

• f - m yi - de al =

• my C- ai sio + ai co s o f. d a ai

21

N

an l

Mys i 0 (^4 -^2 1 )^ =^ m^ g.^ s^ i^ n^2 0 °^ (^0 -^2 -^0 -^1 ) = 7 × 9 - 8 ✗ Sui 20 ° ✗ 0. 1 = 2. 34 62 J

O R, 2 -^3462 -^60 = £ m ust - ≤m oi

O R, ( Give n i n t he pr o bl em t h at Vp = Vi / 2 )

OR, We^2 1.^963

OR, (^) D= 4. 6865 m /s (^) As