Download Midterm examination: Dynamics

Midterm examination: Dynamics
Hiroki Okubo
June 1 2011
y
θ
A
path
ρ
r
eθ
er
A
r
θ
Figure 2: Sliding block.
x
O
Figure 1: Polar coordinates.
(c) Differentiate the position vector with respect to the time and show the vector expression for the velocity. (5)
Solution. We differentiate the position
vector.
1. Figure 1 shows the polar coordinates r and θ
which locate a particle traveling on a curved
path. Unit vectors er and eθ are established
in the positive r- and θ-directions, respectively.
dr
dt
=
(a) Determine the position vector r to the
particle at A. (5)
Solution. The position vector is r =
rer .
(b) Find the time derivatives of er and eθ using er and eθ , respectively. (5)
Solution. Unit vectors i and j are attached to the xy axes. The er and eθ can
be described as
er
eθ
= cos θi + sin θj
= − sin θi + cos θj
deθ
dt
=
θ̇(− sin θi + cos θj)
=
θ̇eθ
=
θ̇(cos θi + sin θj)
=
−θ̇er
der
dt
ṙer + rθ̇eθ
ṙer + r
(5)
(d) Determine the acceleration. (5)
Solution. We differentiate the velocity.
d2 r
dt2
=
=
der
deθ
+ ṙ θ̇eθ + rθ̈eθ + rθ̇
dt
dt
(r̈ − rθ̇2 )er + (rθ̈ + 2ṙθ̇)eθ
(6)
r̈er + ṙ
2. Determine the maximum speed v which the
sliding block may have as it passes point A
without losing contact with the surface as
shown in Fig. 2. The radius of curvature at A
is ρ. (10)
Solution. The equation of motion in the normal direction is
(1)
(2)
The time derivatives of them can be calculated by
der
dt
=
(3)
mg − N = m
v2
ρ
(7)
where N is the reaction force from the surface.
The force is zero, N = 0, when the block loses
(4)
1
A
B
where m is the mass, N is the reaction force
from the slide and θ = 30◦ . Eliminating N
from Eqs. (12) and (13) yields
a
30
9.81
a = g tan θ = √ = 5.66 m/s2
3
Figure 3: Sliding collar.
5. The acceleration of a particle, a, is assumed
to be written a = −C1 − C2 v 2 where C1 and
C2 are constants, and v is the velocity. If the
particle has an initial velocity v0 , derive an
expression for the distance D required for it to
a stop. (10)
Solution. Integration of the acceleration gives
the distance
contact with the surface. Substituting N = 0
into Eq. (7) gives
v=
√
gρ
(8)
√
If the speed at A were less than gρ, an upward normal force exerted by the surface on
the block would exist. In order for the block
to have a speed at A which is greater than
√
gρ, some type of constraint would have to
be introduced to provide additional downward
force.
dv
dt
dv
v
dt
3. A particle moves in a circular path of 0.4-m
radius. Calculate the magnitude a of the acceleration of the particle if its speed is 0.6 m/s
but is increasing at the rate of 1.2 m/s each
second. (10)
Solution. With the constant velocity given,
we can compute the acceleration from
a=
0.62
= 0.9 m/s2
0.4
Z
where the magnitude of a is
p
a = 0.92 + 1.22 = 1.5 m/s2
=
=
0
ma
−C1 − C2 v 2
=
−(C1 + C2 v 2 )
ds
=
ds
=
D
=
D
=
(9)
ds
dt
vdv
1
2 2
C1 1 + C
C1 v
Z 0
1
vdv
−
C2 2
C1 v0 1 + C
v
1
0
C2 2
1
ln | 1 +
v |
−
2C2
C1
v0
−
1
C2 2
ln | 1 +
v |
2C2
C1 0
(15)
where s is the displacement of the particle.
(10)
(11)
4. The collar A is free to slide along the smooth
shaft B mounted in the frame in Fig. 3. The
plane of the frame is vertical. Determine the
horizontal acceleration a of the frame necessary to maintain the collar in a fixed position
on the shaft. (10)
Solution. The equations of motion of the collar give
−mg + N cos θ
N sin θ
=
0
With the unit vectors en and et , the total acceleration becomes
a = 0.9en + 1.2et m/s2
(14)
(12)
(13)
2