A
B
α =12 rad/s 2
0.3 m
ω =4 rad/s
C
0.4 m
D
aC=5 m/s 2
Figure 1: Problem 1
1. For the instant represented, corner C of the rectangular plate has an acceleration of 5 m/s2 in the negative y-direction, and the plate has a clockwise
angular velocity of 4 rad/s which is decreasing by 12 rad/s each second.
Determine the acceleration of A at this instant. Solve by scalar-geometric
and by vector-algebraic methods.
Solution. The acceleration of A is written in term of the given acceleration of C. Thus
aA = aC + (aA/C )n + (aA/C )t
(1)
where we may write
aC = −5j
(aA/C )n
(2)
= ω × (ω × r)
= −4k × {−4k × (0.4i + 0.3j)}
= −4k × (−1.6j + 1.2i)
= −6.4i − 4.8j
(aA/C )t
=
=
ω̇ × r
12k × (0.4i + 0.3j)
=
−3.6i + 4.8j
(3)
(4)
We now substitute these results into the relative-acceleration equation.
aA
= (−6.4 − 3.6)i + (−5 − 4.8 + 4.8)j
= −10i + 4.6j
The magnitude of aA is
p
(−10)2 + 52 = 11.18 m/s2
1
(5)
(6)
R
θ
O
vO
aO
P
Figure 2: Problem 2
2. The wheel of radius R rolls without slipping, and its center O has an
acceleration aO . A point P on the wheel is a distance r from O. For given
values of aO , R, and r, determine the angle θ and the velocity vO of the
wheel for which P has no acceleration in this position.
Solution. Unit vectors i, j, k are attached to the wheel. The ki-plane is
horizontal, j with the downward direction is vertical and the wheel moves
along the i-axis. The acceleration equation of P can be expressed as
aP
= aO + aP/O
= aO + ω × (ω × r P/O ) + α × r P/O
= aO i + ωk × [ωk × r(cos θi + sin θj)] + αi × r(cos θi + sin θj)
= (a0 − rω 2 cos θ − rα sin θ)i + (−rω 2 sin θ + rα cos θ)j
(7)
Substituting ω = v0 /R and α = a0 /R into Eq. (7) gives
r
r
r
r
aP = a0 − 2 v02 cos θ − a0 sin θ i + − 2 v02 sin θ + a0 cos θ j
R
R
R
R
(8)
When aP = 0, we have
r 2
r
v cos θ − a0 sin θ
R2 0
R
r
r
− 2 v02 sin θ + a0 cos θ
R
R
a0 −
=
0
(9)
=
0
(10)
Simultaneous solution of Eqs. (9) and (10) yields
θ
=
v0
=
r
sin−1
R
r
Ra0 p
4
R2 − r 2
r
(11)
(12)
3. The disk rolls without slipping on the horizontal surface, and at the instant
represented, the center O has the velocity and acceleration shown in the
figure. For this instant, the particle A has the indicated speed u and
time-rate-of-change of speed u̇, both relative to the disk. Determine the
absolute velocity and acceleration of particle A.
2
y
A
u=7 m/s 2
vO =3 m/s
O
aO =5 m/s2
u=2 m/s
0.24 m
x
0.30 m
Figure 3: Problem 3
Solution. Unit vectors, I, J , K, are attached to the ground. The IJplane is vertical and K and I are in the horizotal. The wheel is in the
IJ -plane. Unit vectors, i, j, k, are attached to x-y-z axes. Unit vectors,
iA , j A , kA , are also attached to the system with the origin at A. The
k-vector coincides to the k-vector, the iA j A -plane is in the ij-plane and
j A is in the line from O to A. The angle from the I-vector to i is θ, and θA
form i to iA . The ij-plane has the angular velocity of ω, and the angular
acceleration of α with respect to the IJ -plane. The iA j A -plane has the
angular velocity of ωA , and the angular acceleration of αA with respect to
the ij-plane. The time derivatives of i and j are
di
= ωj,
dt
dj
= −ωi
dt
(13)
The iA and j A are
iA
jA
=
=
cos (θ + θA )I + sin (θ + θA )J
− sin (θ + θA )I + cos (θ + θA )J
(14)
(15)
The time derivatives of iA and j A are
diA
= (ω + ωA )j A ,
dt
dj A
= −(ω + ωA )iA
dt
(16)
The position vector to A is
pA = pO + pA/O = pO + rj A
(17)
The velocity vector v A can be calculated as
vA = vO + r
dj A
= v O − r(ω + ωA )iA
dt
(18)
For θ = θA = 0 and the given conditions, we have v = −3i, r = 0.24,
ω = v0 /R = 10, ωA = −u/r = −8.333, and iA = i.
v A = −3i − 0.24(10 − 8.333)i = −3.4i m/s
3
(19)
The acceleration vector aA is
aA
diA
dt
= a0 − r(α + αA )iA − r(ω + ωA )2 j A
= a0 − r(ω̇ + ω̇A )iA − r(ω + ωA )
(20)
For θ = θA = 0 and the given conditions, we have a0 = 5I = 5i, iA = i,
j A = j, α = −5/0.3 = −16.67, αA = 7/0.24 = 29.17. Thus
aA = 5i − 0.24(−16.67 + 29.17)i − 0.24(10 − 8.333)2j = 2i − 0.667j (21)
4