13–101.
The ball of mass mis guided along the vertical circular path
using the arm OA.If the arm has a constant
angular velocity ,determine the angle at which
the ball starts to leave the surface of the semicylinder.
Neglect friction and the size of the ball.
u…45°u
#
0
r=2rccos u
SOLUTION
Since is constant,.
Ans.tan u=4rcu
#2
0
gu=tan–1
¢
4rcu
#2
0
g
≤
+Q©Fr=mar;–mg sin u=m(–4rccos uu
#2
0)
ar=r
$–ru
#2=-2rccos uu
#2
0–2rccos uu
#2
0=-4rccos uu
#2
0
u
$=0u
#
r
$=-2rccos uu
#2–2rcsin uu
$
r
#=-2rcsin uu
#
r=2rccos u
P
r
u
A
O
rc
13–102.
At
Ans.
Ans.F=3.92 N
+T©Fu=mau;–F+0.4(9.81) +0.883 sin 17.66° =0.4(0.6698)
;
+©Fr=mar;–Ncos 17.66° =0.4(–2.104)
N=0.883 N
tan c=r
dr>du=0.6u
0.6 =u=pc=72.34°
au=ru
$+2r
#u
#=0.6p(–0.2954) +2(0.6066)(1.011) =0.6698 m>s2
ar=r
$–ru
#2=-0.1772 –0.6 p(1.011)2=-2.104 m>s2
r
$=0.6(–0.2954) =-0.1772 m>s2
r=0.6(p)=0.6 pmr
#=0.6(1.011) =0.6066 m>s
u
$=-
(p)(1.011)2
1+p2=-0.2954 rad>s2
u=prad,
u
#=2
0.621+p2=1.011 rad>s
0=0.72u
#
u
$+0.36a2uu
#3+2u2u
#
u
$bu
$=- uu
#2
1+u2
22=a0.6u
#b2+a0.6uu
#b2
u
#=2
0.621+u2
v2=r
#2+arub2
vr=r
#=0.6u
#vu=ru
#=0.6uu
#
r=0.6 ur
Using a forked rod, a smooth cylinder P, having a mass of
0.4 kg, is forced to move along the vertical slotted path
where is in radians. If the cylinder has a
constant speed of determine the force of the
rod and the normal force of the slot on the cylinder at the
instant Assume the cylinder is in contact with
only one edge of the rod and slot at any instant. Hint:To
obtain the time derivatives necessary to compute the
cylinder’s acceleration components and take the first
and second time derivatives of Then, for further
information, use Eq. 12–26 to determine Also, take the
time derivative of Eq. 12–26, noting that to
determine u
$.
v
#
C=0,
u
#.
r=0.6u.
au,ar
u=prad.
vC=2m>s,
ur=10.6u2m,
r
P
r 0.6u
u p
SOLUTION
r=200(1 +cos
#
r=200(1 +cos 0°)=400 m
#
#
13–103.
The pilot of the airplane executes a vertical loop which in
part follows the path of a cardioid,
r=200(1 +cos
u
) m
,
where
u
is in radians. If his speed at A is a constant
vp
=85 m>s
, determine the vertical reaction the seat of the
plane exerts on the pilot when the plane is atA. He has a
mass of 80 kg. Hint: To determine the time derivatives
necessary to calculate the acceleration components ar and
au,
take the first and second time derivatives of
r=200(1 +cos
u
).
Then, for further information, use
Eq.12–26 to determine
u
#
.
A
fi
r 200 (1 cos ) m
fi
*13–104.
The collar has a mass of 2 kg and travels along the smooth
horizontal rod defined by the equiangular spiral
where is in radians.Determine the tangential force Fand
the normal force Nacting on the collar when if the
force Fmaintains a constant angular motion u
#=2 rad>s.
u=45°,
u
r=1eu2m,
SOLUTION
At
Ans.
Ans.F=24.8N
N=24.8N
+aaFu=mau;Fsin 45° +NCsin 45° =2(17.5462)
Q+ aFr=mar;–NCcos 45° +Fcos 45° =2(0)
c=u=45°
tan c=r
adr
du
#b=eu>eu=1
au=ru
$+2r
#u
#=0+2(4.38656)(2) =17.5462 m>s2
ar=r
$–r(u
#)2=8.7731 –2.1933(2)2=0
r
$=8.7731
r
#=4.38656
r=2.1933
u
$=0
u
#=2 rad>s
u=45°
r
$=eu(u
#)2+euu
#
r
#=euu
#
r=eu
r
F
r=e
θ
SOLUTION
u=30°
0.5
13–105.
The particle has a mass of 0.5 kg and is confined to move
along the smooth horizontal slot due to the rotation of the
arm OA. Determine the force of the rod on the particle and
the normal force of the slot on the particle when
u=30°.
The rod is rotating with a constant angular velocity
u
.
=
2
rad>s.
Assume the particle contacts only one side of
the slot at any instant.
0.5 m
O
A
r
u
·
u fi 2 rad/s
SOLUTION
$
=
3
#
=
>
$
=
>
2
0.5
r=0.5 sec 30°=0.5774 m
r
#
=0.5 sec 30° tan 30°(2) =0.6667 m>s
r
$
=
0.5
3
sec 30
°
tan
2
30
°
(2)
2+
sec
3
30
°
(2)
2+
sec 30
°
tan 30
°
(3)
4
=
4.849 m
>
s
2
ar
=
r
$–
ru
#2=
4.849
–
0.5774(2)
2=
2.5396 m
>
s
2
au
=
ru
$+
2r
#
u
#
=
0.5774(3)
+
2(0.6667)(2)
=
4.3987 m
>
s
2
Q + ΣFr=mar;
N cos 30°–0.5(9.81) cos 30°=0.5(2.5396)
N=6.3712 =6.37 N
Ans.
+ RΣFu=mau
;
F+0.5(9.81) sin 30°–6.3712 sin 30°=0.5(4.3987)
F=2.93 N
Ans.
13–106.
Solve Prob. 13–105 if the arm has an angular acceleration of
u
$
=
3 rad
>
s
2
when u
#
=
2 rad
>
s at
u=30°.
0.5 m
O
A
r
u
·
u fi 2 rad/s
13–107.
The forked rod is used to move the smooth
2-lb particle around the horizontal path in the shape of a
limaçon, . If rad, where tis in
seconds, determine the force which the rod exerts on the
particle at the instant .The fork and path contact the
particle on only one side.
t=1s
u=(0.5t2)r=(2 +cos u)ft
SOLUTION
At ,,, and
Ans.F=0.163 lb
+a©Fu=mau;F–0.2666 sin 9.46° =2
32.2 (1.9187)
+Q©Fr=mar;–Ncos 9.46° =2
32.2(–4.2346)
N=0.2666 lb
tan c=r
dr>du=2+cos u
–sin u2u=0.5rad
=-6.002 c=-80.54°
au=ru
$+2r
#u
#=2.8776(1) +2(–0.4794)(1) =1.9187 ft>s2
ar=r
$–ru
#2=-1.375 –2.8776(1)2=-4.2346 ft>s2
r
$=-cos 0.5(1)2–sin 0.5(1) =-1.357 ft>s2
r
#=-sin 0.5(1) =-0.4974 ft>s2
r=2+cos 0.5 =2.8776 ft
u
$=1 rad>s2
u=1 rad>su=0.5 radt=1s
r
$=-cos uu
#2–sin uu
$
u
$=1 rad>s2
r
#=-sin uu
u
#=t
r=2+cos uu=0.5t2
3ft
r
2ft
·
u
u
*13–108.
SOLUTION
At
Solving,
Ans.
Ans.N=4.22 lb
P=12.6 lb
;
+© Fu=mau ;
– P cos 45° –N sin 45° =3
32.2 (–128)
+c©Fr=m ar ;
P sin 45° –N cos 45° =3
32.2 (64)
c=- 45° =135°
tan c=r
(dr
du)=
4
1–cos u)
–4 sin u
(1 –cos u)22u=90°
=4
–4=- 1
dr
du=–4 sin u
(1 –cos u)2
r=4
1–cos u
au=ru
$+2 r
#u
#=0+2(–16)(4) =-128
ar=r
$–r(u)2=128 –4(4)2=64
r
$=128
r
#=-16
r=4
u=90°,
u
#=4,
u
$=0
r
$=–4 sin u u
$
(1 –cos u)2+– 4 cos u (u
#)2
(1 –cos u)2+8 sin2 u u
#2
(1 –cos u)3
r
#=–4 sin u u
#
(1 –cos u)2
r=4
1–cos u
The collar,which has a weight of 3 lb,slides along the
smooth rod lying in the horizontal plane and having the
shape of a parabola where is in radians
and ris in feet. If the collar’s angular rate is constant and
equals determine the tangential retarding
force Pneeded to cause the motion and the normal force
that the collar exerts on the rod at the instant u=90°.
u
#=4 rad>s,
ur=4>11–cos u2,
r
P
u
SOLUTION
r=1.6 cos u
r
#
=–1.6 sin uu
#
r
$
=–1.6 cos uu
#2–1.6 sin uu
$
u=45°
#
$
13–109.
Rod OA rotates counterclockwise at a constant angular
rate u
.
=
4
rad>s.
The double collar B is pin-connected
together such that one collar slides over the rotating rod
and the other collar slides over the circular rod described
by the equation
r=(1.6 cos
u
) m
. If both collars have a
mass of 0.5 kg, determine the force which the circular rod
exerts on one of the collars and the force that OA exerts on
the other collar at the instant
u=45°.
Motion is in the
horizontal plane.
A
B
0.8 m
r fi 1.6 cos u
O
u = 4 rad/s
u
13–111.
SOLUTION
Kinematic: Here, and .Applying Eqs.12–29, we have
Equation of Motion: Applying Eq.13–9, we have
(1)
(2)
laitnereffid eht fo noitulos ehT. neht, ecniS
equation (Eq.(1)) is given by
(3)
Thus,
(4)
At .From Eq.(3) (5)
At .From Eq.(4) (6)
Solving Eqs.(5) and (6) yields
Thus,
=9.81
8 (sin h 2t–sin 2t)
=9.81
8 a–e–2t+e2t
2–sin 2tb
r=-
9.81
16 e–2t+9.81
16 e2t–9.81
8 sin 2t
C1=-
9.81
16
C2=9.81
16
0=-2 C1 (1) +2C2 (1) –9.81
4
t=0, r
#=0
0=C1 (1) +C2 (1) –0t=0, r=0
r
#=-2 C1 e–2t+2C2 e2t–9.81
4 cos 2t
r=C1 e–2 t+C2 e2t–9.81
8 sin 2t
Lu
0
u
#=L1
0
2dt, u=2tu.=2 rad>s
8.0 r
#+Ns–1.962 cos u=0
(Q.E.D.)
©Fu=mau;
soc 269.1 u– Ns=0.2(4r
#)
r
$–4r–9.81 sin u=0
(Q.E.D.)
©Fr=mar ;
1.962 sin u=0.2(r
$–4r)
au=ru
$+2r
#u
#=r(0) +2r
#(2) =4r
#
ar=r
$–ru
#2=r
$–r(22)=r
$–4r
u
## =0u.=2 rad>s
A 0.2-kg spool slides down along a smooth rod.
If the rod has a constant angular rate of rotation
in the vertical plane, show that the equations of
dna era loops eht rof noitom
where is the magnitude of
the normal force of the rod on the spool. Using the
methods of differential equations, it can be shown that
the solution of the first of these equations is
If r, and are
zero when evaluate the constants and to
determine rat the instant u=p>4 rad.
C2
C1
t=0,
ur
#,r=C1e–2t+C2e2t–19.81>82 sin 2t.
N
s
0.8r
#+N
s–1.962 cos u=0,
r
$–4r–9.81 sin u=0
u
#=2 rad>s
r
u
u 2 rad/s
*13–112.
SOLUTION
At
Ans.+c©Fr=mar;
–N–130 =130
32.2(–53.33)
N=85.3 lb
ar=r
$–ru
#2=-42.67 –600(0.1333)2=-53.33 ft>s2
r
$=-2400(0.1333)2=-42.67 ft>s2
802=02+
A
600 u
#
B
2
u
#=0.1333 rad>s
y2
p=y2
r+y2
u
yr= r
#=0
yu= ru
#=600u
#
r
$=1200
A
2 cos 180°u
#2+sin 180°u
$
B
=-2400u
#2
r=-600 cos 180° =600 ft
r
#=1200 sin 180°u
#=0
u=90°
r=-600 cos 2u
r
#=1200 sin 2uu
#
r
$=1200
A
2 cos 2uu
#2+sin 2uu
$
B
T
h
e p
il
ot of an a
i
rp
l
ane executes a vert
i
ca
l
l
oop w
hi
c
h
i
n part
follows the path of a “four-leaved rose,”
where is in radians. If his speed at Ais a constant
determine the vertical reaction the seat of the
plane exerts on the pilot when the plane is at A.Heweighs
130 lb.Hint: To determine the time derivatives necessary to
compute the acceleration components and take the first
and second time derivatives of Then, for
further information, use Eq. 12–26 to determine Also,take
the time derivative of Eq. 12–26, noting that to
determine u
$.
v
#
C=0,
u
#.
r=40011+cos u2.
au,ar
vP=80 ft>s,
u
r=1–600 cos 2u2 ft,
r
r 600 cos 2
u
80 ft/s
A
u
13–114.
(1)
The velocity of the satellite orbiting around the circular orbit of radius
is given by
(2)
Solving Eqs.(1) and (2),
Ans.h=35.87(106)m =35.9Mm
yS=3072.32 m>s=3.07 km>s
yS=
C
66.73(10–12)(5.976)(1024)
h+6.378(106)
yS=
C
GMe
r0
r0=h+re=
C
h+6.378(106)
D
m
vs=2p
C
h+6.378(106)
86.4(103)
24(3600) =2p
C
h+6.378(106)
D
vs
T=2pr0
vs
A communications satellite is in a circular orbit above the
earth such that it always remains directly over a point on
the earth’s surface. As a result, the period of the satellite
must equal the rotation of the earth, which is approximately
24 hours. Determine the satellite’s altitude habove the
earth’s surface and its orbital speed.
13–115.
The speed of a satellite launched into a circular orbit
about the earth is given by Eq.13–25.Determine the
speed of a satellite launched parallel to the surface of
the earth so that it travels in a circular orbit 800 km
from the earth’s surface.
SOLUTION
360
*13–116.
The rocket is in circular orbit about the earth at an altitude
of 20 Mm. Determine the minimum increment in speed it
must have in order to escape the earth’s gravitational field.
20 Mm