Engineering Dynamics Homework, Exercises of Mechanical Engineering

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Homework Assignment 3 Problem 3 - 1

Given:

1. Coefficient of kinetic friction =Mr
2. Rough Area distance ( 1 ) = R
3. Radius of loop = R
4. Block mass = M
5. Spring stiffness = k To find: S^ (minimum^ spring^ compression) for which particle does not lose contact with the loop! Solution: Step 1 : Reference frames

oh

÷

m

Rough Area (Mn)

Rough Area (Mn)

i

£ 2 : Free Body Diagrams

'N'

÷

S

oh 7 m

S ÷

oh (^) m

POSITION 1

POSITION 2

OB

Fgv^ "Ñ my 7 i o ↑ Concentrate on position 2 in particular-

'B'frame

ñ

E

;^ i

O

Steps- Governing Equations of Motion IF = MBE block OR, Fg→ + FN = m BE^ block. % mgñ + NI = m (Viet + LEG)

But it = ◦ because the loop is perfectly circular.

OR, (^) mg + N = mr¥

(scalar equation along ni )

OR, (^) VE = V 2 = (mg+N)R m Consider the limiting case i.e. when N → o (jcuons^ tta ctlos)tD witI^ h loop Them; mg≥> N Therefore, I^ =^ Vgr ☆^ Time^ to^ invoke^ the^ next^ Governing^ Equation {ÑW = AKE

LHS: Includes work done by all forces acting on the system Contributing forces: ① Gravity 2 ③^ Spring^ force.

② Friction

W gravity = f Fg. ds→ I = (^) f- mgj. [dy;] I = - mg(Yi-Yi)=- 2 mg R

[F. as

[Msmgi. dei

(since dot product only consider motion along 5 )

A

B (^) Wtriction =

mg?Rough Sfce.

N

IFFrll.MN' ⇒ N.mg

= - Msmg ( 22 - 21 ) = - Msmg R

c Wspring^ force^ = i

• mm^.^ ke

kxi.de? = Skadn I = * ¥!

V

S

= KE

Eg. l

Revisiting Eq. =^1282 - 2 mg R-Ms Mg R = ±

l mvi-Emre

(start from ◦ rest)

OR, (^) KEI-^2 mg^ R-Ms^ mgr^ =^2 -^ gr

OR, *^ =^5 mg_R+^ Us^ mgr

OR, 8 =

min

Mgr ( 5 + 2 Mr) k

An rs

Note: This is the minimum value of ± because this is just enough that block loses contact with loop.

If S^ initial^78 min^ :^ Loop^ is^ completed S initial < 8 min :^ Block^ falls^ of^ at^ top. LIMITING CASE!! S initial = 8 min:

Problem 3 - 2

Given:# 11 = 3 m/s To find: IN Solution:

3 m/s

n^ B

c

1. 8 m (^0) = 600 £^1.^2 m

r

Slept: Reference frames

StepI: FREE BODY DIAGRAMS

ie

3 m/s

_ B

1. 8 m e"^1.^2 m

r

ñ

A

O

O

3 m/s

B

1. 8 m

E

e" 1. 2 m A

t

A

⇒I

A and C are the positions 1 and 2.

Steps: Governing Equations of motion

r

if^ £ I^ Wiz^ - -^ KE N: No. of contributing forces towards total work done. Here N = 1. "Tension is an internal force".

"d W 9 Ifg. ds {^1 OR WE = - mg ↑. (daitdy ↑) I = - my/ dy § = - mg ( 42 - 41 )

c

Ñ

Ñ

Now; FOA = rai and af=-sino it cos 09 04 FOA = - l sinoi - loos 09 So, (^) TY = - loos 0 = Yi Aside... OR, WE = - mg C- 0. 8 + Icoso) = - mg (- 0. 8 + 1. 2. ±) = - my (- 0 - 2 ) = 0. 2 mg

Now, back to RHS in Eq. ①i 0 - 2 mg = ± my (V 27 4 2 )

⇒ Eg.①

OR, 0.^4 g^ =^ VE-(^3 )^2 ☑ = ③. 92 + 9 ) m/s

(Vi = V 1 = 3 m/ 5 )

OR, OR, V.^ =^3.^594 m/s^ And