solucionario riley Dynamics.pdf

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10 February 2005

Dear Recipient,

You are privileged to obtain a partial collection of answers for MEG 207 Dynamics. Thisdocument was compiled from PDF files collected during Fall of 2002 and Spring 2003. I havecomplied them into one document with Bookmarks to aid you in quickly finding the solutions youneed. Unfortunately this isn’t a complete compilation of all of the answers, but comprehensiveenough to help complement your studies.

Engineering Mechanics, Dynamics, Second Edition.ISBN: 0-471-05339-2

Happy Studies!


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Name:___________________Last First

UNLV, DEPARTMENT OF MECHANICAL ENGINEERING

MEG 207, SPRING 2002, FIRST TEST

Closed Book, one page of handwritten notes allowed. Enter the answer for each questioninto the space provided. Enter SI units in all answer spaces with brackets ( ).

1. (15 points) A vehicle traveling at 108 km/h suddenly decelerates at a rate of 3 m/s2

.a) Determine the time needed for the vehicle to come to rest.b) Determine the distance traveled between the beginning of braking, and the full stop.

Answers: a) Tstop = 10 ( s units)

b) dStop = 150 ( m units)

2. (20 points) A ball of mass m is thrown horizontally from a bridge 30 m above ground.

It touches ground at distance d = 15 m. Determine the ball's initial velocity. No friction.

Answer v0 = 6.065 ( m/s )

A (x0,y0)

B

v0 g

horiz. distance

d = 15 m

h = 30 m

x

y

1(a) v0 1081000

3600⋅:= in m/s a 3−:=

v t( ) v0 a t⋅+ General equation. Solving for t when v=0

gives: t v0

a−:= and t 10=

1 (b) v*dv = a*dx or a distance⋅1

20 v0−( )

2⋅

distancev0

2−

2 a⋅:= distance 150=

Problem 2 d 15:= h 30:= Angle is zero.

Horizontal: d = v0*t. Vertical: y(t) = y0 -1/2*g*t^2. At point B, y = 0.

g 9.81:=

Given

d v0 t⋅

0 h 0.5 g⋅ t2

⋅−

res Find t v0,( ):= res2.473

6.065

=


220/253

3. (20 points) Pin P moves at constant speed of 3 m/s ina counterclockwise sense around the circular slot withradius r = 2 m. Determine (a) the angular velocity of P.

(b) the total acceleration vector ( Use polar coordinates:er and eθ directions) of pin P when θ = 30 degrees.(c) the magnitude and angle of the resultant accelerationvector in Cartesian x-y coordinates.

(b) no radial or angular accel. In neg. er direction we

have r*ω2

= - 2*2.25 m/s2

(c) Resultant Acceleration is purely inward, thus:ax = -4.5*cos 30

o

and ay = -4.5*sin 30o

Answer

ω = v/r = 3/2 = 1.5 rad/s

a = -4.5 e 0 eθ m/s2)

a = 4.5 m/s2 at -150 de rees (ma nitude and an le in x- coordinates)

4. (25 points) Wind-driven rain is falling

with a speed of 30 m/s at an angle of 200 to the vertical as shown at left.

Determine the angle ψ at which the rainis seen by passengers inside the bus. Thebus is traveling at 72 km/h.

V rain = V Bus + V Rain/bus V bus = 20m/s

x-and y-components of V Rain/bus:

v R/B,x = -V bu ---V R*sin(20deg) = -30.2 m/s

v R/B,y =---V R*cos(20deg) = -28.19 m/s V Rain/bus

= tan-1

(29.19/30.2) = 43 deg.

An le seen b movin assen ers = 43 de . ( )

θθθθ

R = 2 m

P

x

y

v = 72 km/h

yy

20 0

R a i n

ψ ψψ ψ

4.5

ψ


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5. (20 points) In the pulley system shown at left, the cable is attached at C. Mass Bmoves to the left at vB = 3 m/s, and acceleratesalso to the left at aB = 0.5 m/s

2

. Using the x-y

frame with origin at C, determine:(a) the velocity of A(b) the acceleration of A

L = 2xB(t) +(Y0 --- yA(t))

Differentiation gives:2 vB = vA

and

2 aB = aA

Answer

v A = 6 m/s ( )

a A = 1 m/s2 ( )

A

Bx

y

C


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Na m e:___________________La st F ir st

U NLV, D E P AR TMENT OF MEC H ANIC AL ENG INEE R ING

MEG 207, Spring 2002, Third Test

Closed B ook, one pa ge of ha ndw ritten n otes a llowed. E nter t he a nswer for each question

into the space provided. Ent er SI units in all answer spaces with brackets ( ).

1. (25 points) The thin, uniform rod (Length L, mass m, I n er t i a : I G = m L2

/ 12 .) rotates

about fixed point A as it is released from rest in the horizontal position shown. At the

instant immediately following the release, determine

(a ) th e a ngula r a ccelerat ion of th e rod.

(b) the a ccelera tion, a B , of th e point B .

Sum of moments a bout A:

mg *L/2 = I A*α

I A = I G + m*(L/2)2 = (1/12+ 1/4)*m L 2 == m L

2

/3

I n ser t in g in t o t h e fir st eq ua t ion

gives:

mg *L/2 = mL2

/3 *α

after simplification and regrouping we get:

αRod = 3g/(2L)

Part (b):

A is center of rotation. Thus a B = L*α = 3g/2

Answe r s: a ) αRod = 3g/(2L)

b) a B = - 3g/2 in ---j-di re cti on

i

J

Unit vectors

B

L

g

A

m


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2. (30 points) A uniform plat e of steel of length L = 1 m a nd m a ss m = 200 kg is placedo nt o a t r u c k b e d a s sho wn. The p la t e c a nno t sl ide a t p o int A, b u t c a n r o t a t e a b o u t A.

Determine the maximum allowable acceleration of the truck so that the plate does not

rotate.

B x At ma x. a ccel, cont a ct force B x = 0.

From free-body diagram:

Ax = m *a xAy = m *g

Ax*L /2*sin (α) --- Ay*L /2*cos (α) = I g*α = 0

Inserting the top two equations into the third :

Ax Ay

m*a x = m*g* cos(α)/sin(α)

a ma x = g* cot(α) = 5.664 m /s2

Answer aTruck,max = 5.664 ( m /s2

Unit s)

3. (20 points) The s prin g (k = 1200 N/m) is in it ia llycompressed by 2 m. The 3-kg block shown is not

attached to the spring. After release from rest , the

block tr a vels along th e rough surfa ce (µk = 0.2).

Determine t he position x final a t w hich th e block comes

to rest.

Forces doing work:

Spring: F srping ds

Friction F frict ds

T1 + U 1-2 = T2 w i t h T1 = T2 = 0. Thus U 1-2 = ½ *k*(2meters)2

---µk*m*g*x = 0

x = 2*1200/(0.2*3kg&9.81m /s2

) = 407.7 m

An sw er x final = 407.7 ( m U nit s)

i

J

L

60 0

Truck

B

Aa

truck

Mass = 3 kg

gk = 1.2 kN/m

2 meters

Spring is compressedat time of release

µk

= 0.2

x

Uncompressedspring position

m g


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4. (25 Points) A parcel of ma ss m is relea sed at a n initia l velocity v 0 = a s shown.(a ) D e t er m ine t he m inim u m velocit y a t B (im m edia t e ly b efor e r ea c hing t he u pp er

conveyor) so tha t the pa rcel does not drop at B .

(b) Determine the minimum velocity v 0 at point A so that the parcel reaches point B at

th e minimum velocity computed in part (a ).

G iven: Radius r , ma ss m, g. No friction. Neglect the pa ckage t hickness!

(Use Energy method)

Answer (a)

v2/r

g

P a r t (b )

T1 + U 1-2 = T2 w i t h T1 = ½ *m *v02

a n d T2 = ½ *m*g*r

O n ly g r a v it y d oes w o rk : m g d s

Thus

½ *m*v02

- 2r*m*g = ½ *m*g*r

After simplification, we get:

v02 = 5gr

An s w er : (a ) v B,min = (gr)1/2

(b ) v0,min = (5gr)1/2

v0

g

r

CB

A

m

In order not to falloff a t B , g r a vit y

must be offset by the

centripetal

acceleration:

v2

/r = g

or

v2

,min = g*r


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Na m e:_____KE Y ________L a st F ir st

UNIVERSITY OF NEVADA, LAS VEGAS

D EP AR TMENTOF MEC H ANIC AL ENGINEER ING

MEG 207, Spring 2002, Final ExaminationClosed Book, tw o pages of han dw ritten notes al low ed. Enter t he an swer for each question into the

space provided. Enter correct dimensional SI units where applicable.

1. (10 points) A point ma ss of 10 kg is tossed horizont a lly from 30 m a bove ground. Themass land s on th e ground a t a distan ce of 40 m from a point on the ground directly below

the tossing point. Determine the initial velocity of the

mass.

v0*t = x1 = 40 (1)

y 1 = 0 = y 0 ----½ *g*t2

(2)

fr om (1): t = sq rt (2/g* y 0) = 2.47 s

inserting into (1) gives: v0 = 40/t = 16.17 m/s

Answ er: v 0 = 16.17 m/s

2. (15 points) A point mass m is suspended from two wires AB an d CD. D etermine thetension in the other wire CD (a) before AB is cut

(b) immediately after AB is cut.

(a ) La w of sines: A/sin (40o

) = mg /sin(120o

)

Thus: A = 0.742*mg

(b) After AB is cut: Reaction B disa ppea rs. U sing pola r coordina tes:

Summing in radial dir: A-mg*cos(20o

) = m*r_ddot = 0, thus A = mg*cos(20o

) = 0.94*mg

(a ) Tension in Ca ble CD before cut = 0.742*mg

(b) Tension in Cable CD after cut = 0.94*mg

mg

A

B A, eR

mg eθθθθ


229/253

3. (20 points) A collar with ma ss 5 kg slides in th e vertica l plane a long t he curved rodshow n. It is a tta ched to a n elastic spring w ith un deformed length of 150 mm, and k = 600

N/m. The collar is released from rest at A.

Determine the collar's speed at B.

T1 + V1 = T2 + V2 T1 = 0

½ k (x -x0)2

+ mgh = ½ mv22

+ ½ k(x2-x0)2

0.1meters mg*0.2 0.05 met ers

inserting the da ta a nd solving for v2 gives:

v22

= k(0.12 ---- 0.052) + 2mg*0.2

v22

= 0.9 + 3.924 = 4.824 m2

/s2

vB = 2.196 ( m/s unit s)

4. (15 points) The 12-kg mass B is dropped with a horizontal velocity v 0 = 2.5 m/s ont o th e30-kg luggage cart (µk = 0.5), which is initially at rest and can roll freely. Determine the

velocity of the luggage cart after mass B has reached

the sa me velocity a s th e car t.

An approximate solution can be found using the energy

principle. This neglects the energy loss due to friction

a s th e package slides on the cart.

T1 + V1 = T2 + V2 T1 = ½ mBv12

U se dat um line at A. there is no cha nge in potential: V1 = V2 = 0

So we get: T1 = ½ mBv12

= ½ (mB + m cart)*v22

v22

= (12*6.25)/42 = 1.78 m2

/s2

Answ er (a ) v = 1.336 ( m/s unit s)

g


230/253

5. (20 points) Center B of t he double pulley ha s a velocity of 0.6 m/s, a nd a n a ccelera tionof 2.4 m/s

2

, both directed downwa rd. R= 2m, r = 0.8 m. Determine the total acceration

vector of point D.Point C is an instantaneous center.

ω = vB/r = -0.75 ra d/s clockw ise

α = a B/r = -3 r a d/s2

clockwise

a D = (0 ---- 2.8* ω2)∗i + 2.8*α*j

a D = ---- 1.575∗i -8.4*j

a D = ---- 1.575∗i -8.4*j ( m/s unit s)

6. (20 points) The uniform rod AB with mass m an d length L is relased from rest at ana ng le of Θ0 a s shown. Assu m ing t ha t no sliding occu r s, det er m ine (a ) t he a ng u la r

acceleration of the rod just after release. (b) the normal reaction and friction force at point

A just a fter release.

Pure rotation about A.

Summing moments about A:

-1/2*L *sin (θ0)*mg = I B*α where IB = mL2 /3

thus α = -3/2*g /L* sin (θ0)

(B ) Rea ctions a t A:

Forces in x-dir: F = m*a x = m*1/2*L*cos(θ0)*α

Forces i n y-dir : N ---- mg = -m*1/2*L*sin (θ0)*α

or N = m g[1- ¾ * sin2

(θ0) ]

ÿRod = α = -3/2*g /L* sin (θ0)

N = m g[1- ¾ * sin2

(θ0) ] F = ¾ *mg* cos(θ0)* sin(θ0)

i

J

L

g

A

Bθθθθ0000

mg

½ *L *α


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Note: Exam results are up significantly in comparison to exam 2.

Name:_____KEY________Last First

UNLV, DEPARTMENT OF MECHANICAL ENGINEERING

MEG 207, Spring 2003, Third Test

Closed Book, one page of handwritten notes allowed. Enter the answer for each question into the space

provided. Enter SI units in all answer spaces with brackets ( ).

1. (25 points) A disk(mass m= 2kg, r= 0.1m Inertia: I G = mr 2 /2.) is rigidly attached to massless rod AB

(Length R = 0.5m) and released from rest in the horizontal position. At the instant immediately following

the release, determine

(a) the angular acceleration of the rod.

(b) the acceleration, aB, of the point B.

ANSWER

Newton: Sum of moments about A: using L = R. m B = m a ss a t B

mB *g*R= I A*α

I A = I G + m B*(R)2 = (0.01/2+ 0.25)* mB = 0.255*m B

Inserting into th e first equa tion gives:

mBg*R = 0.255*mB *α

after simplification and regrouping we get:

αRod = R*g /(0.255) = 0.5*9.81/0.255 = 19.2 ra d/s2

Part (b):

A is center of rotation. Thus a B = R*α = 0.5∗α = 9.62 m/s2

Answers: a) αRod = 19.2 ra d/s2

b) aB = R*α = 0.5∗α = 9.62 m/s2

g

A (fixed)

B V B

r = 0 . 1

m

R=0.5m


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2. (25 points) A uniform rectangular crate of height = 0.5 m, width 0.2m and mass m = 200 kg is placed

onto a truck bed as shown. The crate cannot slide at point B, but can rotate about B. Determine the

maximum allowable acceleration of the truck so that the crate does not rotate.

[

N = mg

Moments about G: 0.25*FB - 0.1*mcrate*g = 0 (1)

Forces on crate in x-direction: FB = mcrate*atruck (2)

Insert (2) into (1): 0.25* mcrate*atruck = 0.1*mcrate*g

atruck = g/2.5

Answer aTruck,max = 3.924 ( m/s2

Units)

F Bi

J

Crateg

Ba

truck

0.2m

0 . 5 m

N

m g


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3. (30 points) The spring (k = 1500 N/m) is initially uncompressed. The 3-kg block shown is released from

rest 4m above the spring and falls onto the spring. Using the energy method, determine

(a) the velocity at which the block hits the spring at y = 0 m.

(b) the amount of maximum spring

compression, δ, caused by the

impacting block.

Answer

T2 – T1 = - INT( [-mg]*dy = ½*m*v22

T1 = 0 v22

= 2gy =8g =78.48 m2 /s

2

(c)

T3 – T2 = ½*k*δ

2

+ m*g*δ

(d) T3 = 0

T2 =½*m*v22

= ½*3*kg*78.48 m

2

/s

2

= ½*1500* δ

2

+ m*g*δ

We now solve for δ

(substitute d = δ a t r i gh t ) T2 750 d

2⋅ 3 9.81⋅ d⋅+

Answer vimpact (y=0) = 8.86 ( m/s Units)

δmax. compression = 0.377 ( m Units)

k = 1.5 kN/m

y

3 k g

4 m

g

δδδδSpring is initiallyuncompressed


237/253

4. (20 Points) A point mass pendulum (mass m, Length L) is released from rest at an initial angle θ0 = 450

as shown.

(a) Determine the maximum velocity of the pendulum

(b) Determine the pendulum tension at θ = 0.

Given: Length L , mass m, g. No friction. (Use Energy method)

(a)

T2 – T1 = ½*m*v22

= mgL(1- cos(θ0)

T1 = 0 (release from rest)

v22

= 2g L(1- cos(θ0)

(b) Max. tension Force exists at (2): Sum of forces in radial dir. (inward towards center B):

T –mg = m* v22 /L

T = m(g+ v22 /L)

Answer: (a) vmax = [ 2g L(1- cos(θ0)]1/2

(b) Max.Tension Force at θ = 0 = m(g+ v22 /L)

m

gRope length L

B

fixed

θ0 = 450

θ = 0θ = 0θ = 0θ = 0

1

2

v2=vmax

h =L(1- cos(θθθθ0)


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Name:________KEY______Last First

UNIVERSITY OF NEVADA, LAS VEGAS

DEPARTMENT OF MECHANICAL ENGINEERING

MEG 207, SPRING 2003, Final Examination

Closed Book, three pages of handwritten notes allowed. Enter the answer for each question into the space provided.Enter correct dimensional SI units where applicable.

1. (15 points) A player throws a ball horizontally from point A at an elevation 2m above ground.The ball lands on the flat ground at point B, at a distance of 20 m from the origin.

Determine(a) the initial velocity at which the ball was thrown

(b) the time elapsed between points A and B

B

g

horiz. distance

d = 20 m

yo

= 2 m

y

A (x0,y

0)

vo

x

The only acceleration is g in –j-direction. vo = vo*i. After integrating twice in i-direction:

xB = 20 = vo*t + 0 (xo = 0) (1)

After integrating in j-direction: yB = 0 = 0 – ½*g*t2

+ 2 (yo = 2) (2)

From (2): t2

= 4/g ÿ inserting into (1) gives: vo = 20/t = 20*(g/4)1/2

= 31.3 m/s

Answer: vo = 31.3 m/s

t = 0.64 s


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2. (15 points) A constant force F = 50 N is applied to massm=10 kg, which is initially at rest with the attached

spring (k = 200 N/m) unstretched. Determine(a) the maximum velocity of the mass.

(b) the maximum stretch of the spring.

(a) T2 – T1 = F*x - ½*k* x2

= ½*m*v22

F x⋅1

2k ⋅ x

2⋅−

1

2m⋅ v x( )( )

2⋅

The maximum velocity occurs

when the derivative of the velocity is zero:F k x⋅− m v x( )⋅

xv x( )

d

d⋅

= 0, x = F/k = 50/200 = 0.25,

thus v2

max = 1.25 (m/s)2

(b) Both start and final velocities must be zero: T2 – T1 = F*x - ½*k* x2

= 0

Two solutions: x1 = 0 (start point) x2 : 2F - k* x = 0

x2 = 2F/k = ½ meters

(a) vmax = 1.12 m/s(b) xmax = 0.5 m

3. (20 points) Knowing that vΑ= 5 m/s is constant to the left,determine for the instant shown :

(a) the angular velocity of rodBC

(b) the angular acceleration of rod AB.

(a) Vector equation:VB = VA + VB/A

Thus: RB/C*ωBC*j = -vA*i + RB/A*ωAB*(-sin30o

i + -cos30o j) Components:

i: vA = RB/A*ωAB*sin30oÿ ωωωωAB = vA /(RB/A*sin30

o) = - 5/(10*0.5) = -1 rad/s

j: RB/C*ωBC = RB/A*ωAB *cos30oÿ ωωωωBC = -RB/A*ωAB cos30

o /RB/C = +10*0.866/4 = 2.165 rad/s;

(b) aA = 0. only centripetal and angular accel terms exist. Vector eq. AB = AA + AB/A-RB/C*ωBC

2*i + RB/C*αBC j = RB/A*ωAB

2*(-sin30

o j + cos30

oi) + RB/A*αAB*(-sin30

oi - cos30

o j)

j: RB/C*αBC = -RB/A*ωAB2

*sin30o

- RB/A*αAB* cos30o

i: -RB/C*ωBC2

= RB/A*ωAB2

* cos30o

- RB/A*αAB*sin30o

αAB = -(RB/C*ωBC2

+ RB/A*ωAB2

* cos30o

)/(RB/A*sin30o) = - (4*4.69 + 10*1*0.866)/5 = - 5.48 rad/s

2

ωBC = 2.165 rad/s

αAB = - 5.48 rad/s2

cw

i

J

Unit vectors vA = 5 m/s

B10 m

4 m

30 o

A

CV

B/A

Mass =

10 kg

k = 200 N/m

Spring is unstretched at time of release

no friction

F= 50 N

unstretched

spring position

x

g


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4. (15 points) The frictionless system of masses A (40 kg) and B (10 kg) is released from rest.Determine: (a) the acceleration of mass A.

(b) the tension in the cable

Mass A moves in i-direction, and B in j-directiononly.

Mass A moves in I-direction, and B in j-direction only.

Newton: A: xmT F A x ÿÿ==ÿ Constraint eq.: x = -yB: xm ymgmT F B B B y ÿÿÿÿ −==−=ÿ *

Ba

B

mm

gm x

+=ÿÿ = 98.1/50 = 1.962 m/s

2

(b) Using first eq.: T = 40kg*1.962 m/s2

= 78.5 N

(a) aA = 1.962 m/s2

(b) T = 78.5 N

5. (15 points) Gear D is stationary. Gear C has radius r C = 100 mm. Bar AB has a length of 200

mm. As Bar AB rotates counterclockwise at ωAB = 5 rad/s, determine

(a) The angular velocity of gear C(b) The total acceleration vector (in Cartesian i-j

coordinates) of point P on gear C. Point P islocated 100 mm below B, normal to the line AB.

i

J

Unit vectors(a) The Contact of gear D with gear B is an inst. Center. We have: vB = 200*ωΑB = 100*ωCω

C = 2*5 = 10 rad/s(b) AP = AB + AP/B aP = - 200*52*i + 100*10

2*j

ωC = 10 rad/s

Vector aP = - 5 i + 10 j m/s2

Am = 40 kg

Bm = 10 kg

g

i

J

Unit vectors

P

Ctr


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6. (20 points) The uniform rod AB with mass m is released from rest at an angle of 60 degrees asshown. Assuming that no sliding occurs, determine (a) the angular acceleration of the rod just after

release. (b) the normal reaction and friction force at point B. Moment of Inertia for the rod, at centerof mass: IG = mL

2 /12.

We have a pure rotation about

fixed point B.

Parallel axis Theorem: IB = IG+ m(L/2)

2=mL

2(1/12+1/4) =

mL2 /3

Newton about B:

αα *3

*2

1*

2

2 Lm

I mg L

M B B

==−=ÿ

α = -3g/(4L) (oriented in k-direction normal to the i-j plane)

(b) Reactions: Newton in x- and y- directions:

16 / 34

g L

y −=−= α ÿÿ )0(433.0*866.0*2

>+=−= ct Crossprodug L

x α ÿÿ

ymmg N F y ÿÿ=−=ÿ N = m(g – 3g/16) = 13mg/16

433.0*mg xmF F x ===ÿ ÿÿ

α

L/2

0.866*L

L/4

αRod = 3g/(4L)

Ν = 13mg/16 F = 0.433*mg

i

J

L

g

B

60 0

A

mg


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Documents

solucionario riley Dynamics.pdf