Department of Mechanical Engineering
University of California at Berkeley
ME 104 Engineering Mechanics II
Spring Semester 2010
Instructor: F. Ma
Final Examination
The examination has a duration of 2 hours and 45 minutes.
Answer all questions.
All questions carry the same weight.
May 11, 2010
1. Each of the sliding bars A and B engages its respective rim of the two riveted wheels without
slipping. If, in addition to the information shown, bar A has an acceleration of 2 m/s2 to the
right and there is no acceleration of bar B, calculate the magnitude of the acceleration of
point P for the instant depicted.
2. The crank OA revolves counterclockwise with a constant angular velocity of 5 rad/s. For the
position shown, determine the angular velocity and angular acceleration of the slotted link
BC.
3. The slender 150-lb bar is supported by two identical cords AB and AC. If cord AC suddenly
breaks, determine the initial angular acceleration of the bar and the tension in cord AB.
A
g
5 ft
B
C
8 ft
4. A sphere of mass m and radius a rests on top of a larger fixed sphere of radius b. The smaller
sphere is slightly displaced so that it rolls without slipping down the larger sphere. Where
will the rolling sphere leave the fixed sphere? Is there any change in the take-off position if
two cylinders of radii a and b are used instead? The moment of inertia of a sphere of mass m
and radius r about a diameter is 2mr 2 / 5 . The moment of inertia of a cylinder of mass m and
radius r about its axis is mr 2 / 2 .
m
a
G
θ
b
g
O
5. Determine the minimum velocity v which the wheel must have to just roll over the
obstruction. The centroidal radius of gyration of the wheel is k, and it is assumed that the
wheel does not slip. What is the value of v if the wheel is a uniform disk with mass m and
radius r and h = r / 8 ?
g
Problem 1. For A and B on the riveted wheel,
( a A / B )t = rA / B α
v A / B = rA / B ω
⇒
2 = (0.26)α
⇒
⇒
α = 7.69 rad/s2
0.8 − ( −0.6) = (0.26)ω
⇒
The center O is in rectilinear motion along the x-axis and
aO = (0.16)α = 1.23 m/s2
For O and P on the wheel,
a P = aO + ( a P / O ) n + ( a P / O ) t
ω = 5.38 rad/s
= 1.23i − rP / O ω2 i + rP / O αj
= 1.23i − 0.16(5.38) 2 i + 0.16(7.69)αj
= −3.41i + 1.23j
Thus
19.8o
a P = 3.412 + 1.232 = 3.62 m/s2
vA = 0.8 m/s
100 mm
A
aA = 2 m/s2
160 mm
P
aO
ω
vB = 0.6 m/s
α
B
x
y
Problem 2. Attach a rotating xy-frame to C with the y-axis directed along CB. The pin A can only
slide along the slot on link CB, therefore both v rel and a rel act along the link CB. Then
⇒
⇒
Equating i coefficients,
v A = v C + ω × rA / C + v rel = ω × rA / C + v rel
− v A sin 30°i + v A cos 30° j = ωBC k × 8 j + vrel j
− 4(5) sin 30°i + 4(5) cos 30° j = −8ωBC i + vrel j
ωBC =
20 sin 30°
= 1.25 rad/s
8
Equating j coefficients,
v rel = 20 cos 30° = 17.32 in/sec
Since AO rotates with a constant angular velocity of 5 rad/s about fixed point O,
a A = (a A / O ) n = −4(5) 2 cos 30°i − 4(5) 2 sin 30° j = −86.60i − 50 j
In a similar manner,
 × rA / C + ω × (ω × rA / C ) + 2ω × v rel + a rel
a A = aC + ω
= α × rA / C + ω × (ω × rA / C ) + 2ω × vrel j + arel j
= αBC k × 8 j + 1.25k × (1.25k × 8 j) + 2(1.25)k × 17.32 j + arel j
= −8αBC i − 43.30i − 12.5j + arel j
It follows that
− 86.60i − 50 j = −8αBC i − 43.30i − 12.5j + arel j
Equating i coefficients,
αBC =
B
43.30
= 5.41 rad/s2
8
A
5 rad/s
8”
30°
4”
vA
x
30°
y
C
O
ω × rA / C
v rel
C
Problem 3. For the bar,
∑F
x
∑F
y
∑M
= m( aG ) x
⇒
= m( aG ) y
⇒
= IGα
⇒
G
A
4
T = m( aG ) x
5
3
mg − T sin θ = mg − T = m( aG ) y
5
12
1
T ( 4 sin θ ) = T = m82 α
5
12
T cos θ =
(1)
(2)
(3)
IGα
T
m(aG/B)t
G
B
θ
maB
mg
A
C
x
y
There are three equations containing four unknowns T, ( aG ) x , ( aG ) y and α. From kinematics,
aG = a B + aG / B
Since the bar is stationary when the cord AC breaks, ω = 0 and v B = 0 . Hence,
⇒
aG / B = ( aG / B ) t = 4α
( a G / B ) n = 4ω 2 = 0
v B2
⇒
=0
a B = (a B ) t
5
Thus a G / B is perpendicular to BG and a B is perpendicular to AB. Suppose a B makes an angle θ
with BG. Then
3
⇒
( aG ) x = ( a B ) x + ( aG / B ) x
( aG ) x = a B cos θ = a B
5
4
⇒
( aG ) y = ( a B ) y + ( aG / B ) y
( aG ) y = a B sin θ + 4α = a B + 4α
5
Eliminate a B from the above two equations,
(a B ) n =
4
( aG ) x + 4 α
3
( aG ) y =
(4)
Solve Eqs. (1) – (4) simultaneously,
27
α=
g = 4.18 rad/s2
208
20
15
T=
mα = mg = 43.27 lb
9
52
In addition, ( aG ) x = 7.43 ft/s2, ( aG ) y = 26.63 ft/s2, and a B = 12.38 ft/s2. As a consequence, ma G
has a direction as shown.
A
IGα
T
B
m(aG/B)t
G
θ
C
maG
maB
mg
In comparison,
Tst =
mg
mg
5
=
= mg = 93.75 > T = 43.27
2 cos(90° − θ ) 2 sin θ 8
Alternative Solution
Using the moment equation about B,
∑ M B = I G α ± ma G d = I G α + m(a G ) y d
1
m82 α + m( aG ) y ( 4)
12
Solve Eqs. (1), (2), (4), and (5) simultaneously,
α = 4.18 rad/s2
T = 43.27 lb
⇒
mg ( 4) =
(5)
Problem 4. Let ω and α be angular velocity and acceleration of the rolling sphere. The position
of the mass center G of the rolling sphere may be specified by the absolute coordinate θ with
respect to the vertical. From kinematics, the contact point C is the instantaneous center of zero
velocity and
vG = vC + vG /C = vG /C
⇒
( a + b)θ = aω
At any position θ before the rolling sphere leaves the fixed sphere,
∆T + ∆Vg = 0
⇒
⇒
1 2 1
mvG + I G ω2 − mg ( a + b)(1 − cos θ ) = 0
2
2
 2
12
1
2
2  ( a + b)θ 

 = mg ( a + b)(1 − cos θ )
m[( a + b)θ ] +  ma 
25
a
2


θ 2 =
⇒
For the rolling sphere,
∑F
n
10 g
(1 − cos θ )
7( a + b )
= m( aG ) n
⇒
(1)
mg cos θ − N = m( a + b)θ 2
(2)
Combine Eqs. (1) and (2),
1
mg (17 cos θ − 10)
7
When the rolling sphere leaves the fixed sphere, N = 0 and
10
= 53.97°
θ = cos −1
17
N=
(3)
The take-off position is independent of m, a, and b. For two cylinders, I G =
1
ma 2 for the rolling
2
cylinder and the same process yields
4
= 55.15°
7
If the rolling sphere is treated as a particle sliding on a smooth fixed sphere, the take-off position
is given by
2
θ = cos −1 = 48.19°
3
ω
α
n
t
θ = cos −1
G
C
θ
N
F
mg
Position 1: θ = 0
Position 2: θ > 0
Problem 5. The motion of the wheel is divided into parts: (1) impact at C and (2) rolling over the
obstruction afterwards. When contact occurs at C, the wheel pivots about the obstruction at C
and the friction F between the wheel and horizontal ground vanishes. During and immediately
after impact an unknown impulsive force R acts on the wheel over a short duration ∆t ≈ 0 . After
impact the wheel rotates about C with a velocity v2 perpendicular to GC. Since there is no
slipping,
v = rω1
v2 = rω2
Just before impact,
v
( H C )1 = I G ω1 + mv ( r − h ) = mk 2 + mv ( r − h )
r
Just after impact,
v2
+ mv2 r
r
Let s be the distance of C from the vertical line through G. Then
( H C ) 2 = I G ω2 + mv2 r = mk 2
∫∑M
C
dt = ΔH C
Δt
⇒
− ∫ mgsdt = −mgsΔt = ( H C ) 2 − ( H C )1
Δt
⇒
( H C )1 = ( H C ) 2
rh 

(1)
v 2 = v 1 − 2
2 
 k +r 
After impact, the wheel rolls on curb point C against gravity. Thus
∆T + ∆Vg = 0
For minimum v, kinetic energy after impact is totally expended in rolling over the obstruction,
1 2 1
mv 2 + I G ω22 = mgh
2
2
v2
1 2 1
⇒
(2)
mv 2 + mk 2 22 = mgh
r
2
2
Combine Eqs. (1) and (2) to eliminate v2 ,
r
v= 2
2 gh( k 2 + r 2 )
k + r 2 − rh
Suppose the wheel is a uniform disk with mass m and radius r and h = r / 8 . Since k = r / 2 ,
the minimum velocity v which the wheel must have to just roll over the obstruction is
24 gr
r
v= 2
2 gh( k 2 + r 2 ) =
= 0.4454 gr
2
k + r − rh
11
ω1
ω2
v2
v
G
G
mg
C
mg
C
h
h
F
s
s
R
Just before impact at t = 0
Just after impact at t = Δt
⇒