Exam01 Soln

ME 274 – Spring 2009 Examination No. 1 PROBLEM NO. 1

SOLUTION y

Given: The sliding collar moves vertically along the shaft producing an oscillation in OA. At the instant shown, AB and OA are horizontal and vertical, respectively, and the velocity of B is not changing. Find:

x B

vB

For the instant shown, determine

A

L

O

a) the velocity of A.

r

b) the angular acceleration of OA.

Velocity AB: v A = v B + ! AB " r A/ B = #vB j + (! AB k ) " ( L i ) = ( #vB + L! AB ) j OA:

( )

v A = vO + ! OA " r A/O = 0 + (! OA k ) " r j = ( #r! OA ) i

Equating and balancing coefficients: i : ! OA = 0 " j : #vB + L! AB = 0 "

vA = 0 ! AB = vB / L

Acceleration 2 2 2 a A = a B + ! AB " r A/ B # $ AB r A/ B = 0 + (! AB k ) " ( L i ) # $ AB i + L! AB j AB: ( L i ) = #L$ AB OA:

( )

2 a A = aO + ! OA " r A/O # $ OA r A/O = 0 + (! OA k ) " r j # 0 = ( #r! OA ) i

Equating and balancing coefficients: j : L! AB = 0 "

! AB = 0 2 L ( vB / L ) L$ AB vB2 = = = r r Lr 2

i:

2 #L$ AB

In summary, vA = 0

" v2 % ! OA = $ B ' k # Lr &

= #r! OA

"

! OA

ME 274 – Spring 2009 Examination No. 1 PROBLEM NO. 2

SOLUTION

Given:

The mechanism shown below is driven by a motor attached at B. The angular velocity of ωAB is a constant 2 rad/s in the direction shown.

Find:

For the position shown below, determine a) the angular velocity of OC. b) the velocity of A relative to an observer positioned at O and rotating with OC. c) the angular acceleration of OC.

Express your answers as vectors in terms of either their xyz or their XYZ components. Y Velocity AB v A = v B + ! AB " r A/ B B

(

)

= 0 + ( #2k ) " #0.17 j = ( #0.34 i ) m / sec

Also, v A = vO + ( v A/O )rel + ! OC " r A/O

!AB 30°

y

0.17 m

= 0 + vrel i + (! OC k ) " ( 0.1i )

x

= vrel i + 0.1! OC j C

Equating and balancing coefficients: j : 0.1 ! OC = 0 " ! OC = 0

i : vrel = #0.34 m / sec

"

60°

( v A/O )rel = ( #0.34 i ) m / sec

Acceleration AB

(

) (

O

)

2 a A = a B + ! AB " r A/ B # $ AB r A/ B = # ( #2 ) #0.17 j = 0.68 j m / sec 2 2

Also, a A = aO + ( a A/O )rel + ! OC " r A/O + 2 # OC " ( v A/O )rel + # OC " (# OC " r A/O ) = arel i + (! OC k ) " ( 0.1i ) = arel i + 0.1! OC j

Equating and balancing coefficients:

j : 0.68 = 0.1! OC

"

i : arel = 0

"

! OC = ( 6.8k ) rad / sec 2

( a A/O )rel = 0

A 0.1 m

X

ME 274 – Spring 2009 Examination No. 1 PROBLEM NO. 3

SOLUTION

Part (a) – 4 points

A wheel rolls without slipping as its center O moves to the right with a constant speed of vO. vA R Point A is on the outer circumference of the vO = constant A wheel. At the instant shown, A is located O aA directly to the left of O. At this instant (circle the correct answer), no slip a) the speed of A is decreasing. b) the speed of A is constant. C c) the speed of A is increasing. Projection of aA onto vA is POSITIVE. d) numerical values for the speed of O and the radius R of the wheel are needed in order to determine if the speed of A is decreasing, constant or increasing.

Particle P travels on a path whose coordinates are given in polar coordinates as r = 2 cos! , where r is in feet and θ is in radians. When " ! = radians , it is known that !! = "3 rad / sec and !!! = 0 . 3

Part (b) – 2 points

At this instant (circle the correct answer), a) r! < 0 . r! = !2"! sin" b) r! = 0 . = !2 ( !3) 3 / 2 = +3 3 ft / sec c) r! > 0 .

(

)

Part (c) – 2 points

At this instant (circle the correct answer), r <0. a) !! !! r = !2"!! sin" ! 2"! 2 cos" r = 0. b) !!

r >0. c) !!

= !2 ( !3) (1 / 2 ) = !9 ft / sec 2 2

PROBLEM NO. 3 Part (d) – 4 points

Particle P moves in the x-y plane with an acceleration of a P = aP i with aP = 10 m / sec 2 = constant . At the instant shown below, P has a speed vP = 5 m / sec with the velocity of P given by v P = vP cos! i + vP sin! j . Circle the angle θ below that

gives the SMALLEST radius of curvature ρ for the path of P: a) ! = 0 b) ! = 30°

y

vP2

c) ! = 60°

an = aP sin! =

d) ! = 90°

vP2 "= = minimum when ! = 90° aP sin!

e) ! = 120°

"

et

#

vP

θ aP

P

f) ! = 150° g) ! = 180° h) The radius of curvature ρ of the path of P is independent of the angle θ. i) More information is needed about the path of P in order to assess the dependence of ρ on θ.

en

x

2 ft

PROBLEM NO. 3

2 ft

Part (e) – 4 points

E

Blocks A and B are connected by an inextensible cable, with this cable being wrapped around pulleys at A and E. Assume the radii of the pulleys to be small. Block A moves downward with a speed of vA . Let vB represent the speed of B for s A > 0 . Circle the answer below that most accurately describes the speed of B as compared to the speed of A:

sA sB

a) 0 < vB < 2vA

A B

b) vB = vA c) vB = 2vA

vA

d) vB > 2vA

e) More information is needed about s A and vA to assess the magnitude of vB as compared to vA .

L = s B + 2 s A2 + 4 = constant

!

" s % A vB = 2 $ ' vA < 2vA for ALL s A > 0 $# s 2 + 4 '& A

Part (f) – 4 points

Collar B, shown below, is free to move vertically and the rigid disk is free to rotate about point O. At the instant shown, point A on the disk is on the same horizontal line as point O. At this instant, the angular velocity of AB is: a) counterclockwise A

b) clockwise

r

c) zero d) indeterminate without explicit values for L, r, and ωOA The instant center for AB is at infinity. Therefore, the angular velocity of AB is zero.

ωOA

O

L

vA

B

vB