978-0133915426 Chapter 21 Part 4

21–58.

SOLUTION

Hence

Ans.

Solving,

Ans.

Ans.Az=Bz=7.5 lb

By=-3.90 lb

Ay=-1.69 lb

©Fz=m(aG)z;Az+Bz–15 =0

©Fy=m(aG)y;Ay+By=-a15

32.2 b(1)(12)

©Fx=m(aG)x;Bx=0

©Mz=Izv

#

z–(Ix–Iy)vxvy,Ay(1) –By(1) =c1

12 a15

32.2 b

A

3(0.5)2+(2)2

B

d(12) –0

©My=Iyv

#

y–(Iz–Ix)vzvx,Bz(1) –Az(1) =0

©Mx=Ixv

#

x–(Iy–Iz)vyvz,0=0

v

#

x=v

#

y=0, v

#

z=12 rad>s2

vx=-4, vy=vz=0,

v

xyz +Æ*v=12k+0*(–4i)={12k} rad>s2

The 15-lb cylinder is rotating about shaft AB with a constant

angular speed .If the supporting shaft at C,

initially at rest, is given an angular acceleration

,determine the components of reaction at the

bearings Aand B. The bearing at Acannot support a force

component along the xaxis,whereas the bearing at Bdoes.

aC=12 rad>s2

v=4 rad>s

x

z

y

1ft

0.5 ft

1ft

A

B

G

ω

21–59.

SOLUTION

Using Eqs. 21–25:

Ans.

Ans.©M

z

=0.006 sin ucos u=(0.003 sin 2 u)N#m

©Mz=0–[0 –1.5(10–3)](2 sin u)(2 cos u)

©My=(–0.036 sin u)N#m

©My=1.5(10–3)(–12 sin u)–[1.5(10–3)–0](6)(2 sin u)

©Mx=0–0=0

Iy=Iz=1

12(0.8)(0.15)2=1.5(10–3)

Ix=0

v

#

z=0

v

#

y=-2u

#

sin u=-12 sin u

v

#

x=2u

#

cos u=12 cos u

vz=u

#

=6

vy=2 cos u

u.

2 rad/s

z

x

y

Auu

.

= 6 rad>s,

u

.

The thin rod has a mass of 0.8 kg and a total length of 150 mm.

It is rotating about its midpoint at a constant rate

while the table to which its axle A is fastened is rotating at

fastened is rotating at Determine the x, y, z moment

components which the axle exerts on the rod when the rod is

in any position

. 2 rad

>

s

1171

*21–60.

S

h

ow t

h

at t

h

e angu

l

ar ve

l

oc

i

ty of a

b

o

d

y,

i

n terms of

Euler angles ,,and ,can be expressed as

, where i,j, and kare directed along the x,y,

zaxes as shown in Fig. 21–15d.

(f

#

cos u+c

#

)k

v=(f

#

sin usin c+u

#

cos c)i+(f

#

sin ucos c–u

#

sin c)j+

cuf

SOLUTION

#

1172

21–61.

SOLUTION

Ans.

Ans.g=cos–1(0.7071) =45°

b=cos–1(–0.6124) =128°

a=cos–10.3536 =69.3°

u=0.3536i–0.6124j+0.7071k

u=(1 sin 45°) sin 30° i–(1 sin 45°) cos 30°j+1cos 45° k

At

hi

n ro

d

i

s

i

n

i

t

i

a

ll

y co

i

nc

id

ent w

i

t

h

t

h

e Zax

i

s w

h

en

i

t

i

s

given three rotations defined by the Euler angles ,

,and .If these rotations are given in the

order stated, determine the coordinate direction angles ,,

of the axis of the rod with respect to the X,Y, and Zaxes.

Are these directions the same for any order of the

rotations? Why?

g

ba

c=60°u=45°

f=30°

21–62.

SOLUTION

Ans.vP=27.9 rad>s

(0.450)(9.81)(0.125) +(0.180)(9.81)(0.080) =1

2(0.450)(0.035)2vP(90)

The gyroscope consists of a uniform 450-g disk Dwhich is

attached to the axle AB of negligible mass. The supporting

frame has a mass of 180 g and a center of mass at G. If the

disk is rotating about the axle at , determine

the constant angular velocity at which the frame

precesses about the pivot point O. The frame moves in the

horizontal plane.

vp

vD=90 rad>s

25 mm

35 mm

25 mm 20 mm

80 mm

ω

p

ABG

O

21–63.

The toy gyroscope consists of a rotor R which is attached

to the frame of negligible mass. If it is observed that the

frame is precessing about the pivot point O at vp=2 rad>s,

determine the angular velocity vR

of the rotor. The stem

OA moves in the horizontal plane. The rotor has a mass of

200 g and a radius of gyration kOA =20 mm about OA.

SOLUTION

p

ω

30 mm

R

ω

O

A

*21–64.

The top consists of a thin disk that has a weight of 8lb and

a radius of 0.3 ft. The rod has a negligible mass and a length

of 0.5 ft. If the top is spinning with an angular velocity

vs=300 rad>s, determine the steady-state precessional

angular velocity vp of the rod when u=40°.

SOLUTION

#2 sin

#

0.3 ft

0.5 ft

ω

p

ω

s

θ

21–65.

SOLUTION

0.3 ft

0.5 ft

ω

p

ω

s

θ

Solve Prob. 21–64 when

u

= 90°.

21–66.

The propeller on a single-engine airplane has a mass of 15 kg

and a centroidal radius of gyration of 0.3 m computed about

the axis of spin. When viewed from the front of the airplane,

the propeller is turning clockwise at about the spin

axis.If the airplane enters a vertical curve having a radius

of80mand is traveling at ,determine the

gyroscopic bending moment which the propeller exerts on

the bearings of the engine when the airplane is in its

lowest position.

200 km>h

350 rad>s

SOLUTION

Ans.M

x=

328 N

#

m

Mx=[15(0.3)2](0.694)(350)

©Mx=IzÆyvz

Æy=55.56

80 =0.694 rad>s

v=200 km/h =200(103)

3600 =55.56 m>s

vs=350 rad>s=vz

r80 m

21–67.

A wheel of mass mand radius rrolls with constant spin

about a circular path having a radius a. If the angle of

inclination is determine the rate of precession. Treat the

wheel as a thin ring. No slipping occurs.

u,

V

SOLUTION

or

(1)

Also,

(2)

(3)

Thus,

Applying

Using Eqs. (1), (2) and (3), and eliminating , we have

Ans.f

#

=(2gcot u

a+rcos u)1/2

2gcos u=af

#2sin u+rf

#2sin ucos u

maf

#2sin ur–mgrcos u=mr

2

2(

–f

#2a

r) sin u–mr

2

2(f

#2sin ucos u)

maf

#2rsin u–mgrcos u=mr

2

2(–f

#

) sin u(a+rcos u

r)f

#

+mr2

2(

–f

#

a

r)f

#

sin u

c

#

Frsin u–Nrcos u=mr

2

2(–f

#

c

#

sin u)+(mr

2–mr2

2)(–c

#

+f

#

cos u)(f

#

sin u)

©Mx=Ixv

#

x+(Iz–Iy)vzvy

v

#

x=-f

#

c

#

sin u,v

#

y=v

#

z=0

v

#=f

#

*c

#

=-

f

#

c

#

sin u

vx=0, vy=f

#

sin u,vz¿=-c

#

+f

#

cos u

v=f

#

sin uj+(–c

#

+f

#

cos u)k

Ix=Iy=mr2

2,Iz=mr2

#2)

v=f

#

+c

#

c

#

=(a+rcos u

r)f

#

(r)c

=(a+rcos u)f

a

r

.

u

V

f

1179

*21–68.

The conical top has a mass of 0.8 kg, and the moments

of inertia are Ix

=

Iy

=

3.5

(

10

–3

)

kg

#

m

2

and

Iz

=

0.8

(

10

–3

)

kg

#

m

2

. If it spins freely in the ball-and

socket joint at A with an angular velocity vs

=750 rad>s,

compute the precession of the top about the axis of the

shaft AB.

SOLUTION

s=750 rad>s

y

B

x

100 mm

A