Mechanical Engineering Chapter 21 Homework Has Weight Lbft And Supported Pin

1149

*21–40.

SOLUTION

In general

M=d

dt (Hxi+Hyj+Hzk)

Derive the scalar form of the rotational equation of motion

about the xaxis if and the moments and products of

inertia of the body are not constant with respect to time.

æZV

Ans:

1150

21–41.

Der

i

ve t

h

e sca

l

ar form of t

h

e rotat

i

ona

l

equat

i

on of

motion about the xaxis if and the moments and

products of inertia of the body are constant with respect

to time.

æZV

SOLUTION

In general

Substitute and expanding the cross product yields

Substitute Hx,Hyand Hzusing Eq. 21–10. For the icomponent

For constant inertia, expanding the time derivative of the above equation yields

Æ=Æ

xi+Æ

yj+Æ

zk

M=d

dt (Hxi+Hyj+Hzk)

Ans:

1151

21–42.

SOLUTION

In general

Substitute and expanding the cross product yields

Æ=Æ

xi+Æ

yj+Æ

zk

Derive the Euler equations of motion for , i.e.,

Eqs

. 21–26.

æZ

V

1152

21–43.

The 4-lb bar rests along the smooth corners of

an open box. At the instant shown, the box has a

velocity and an acceleration

Determine the x, y, zcomponents of force which the corners

exert on the bar.

a=5–6j6ft>s2.v=53j6ft>s

SOLUTION

[1]

Solving Eqs. [1] to [4] yields:

©Fx=m(aG)x;Ax+Bx=0

z

2ft

B

A

Ans:

Bz=4 lb

1153

*21–44.

The uniform rectangular plate has a mass of and

is given a rotation of about its bearings at A

and B. If and , determine the vertical

reactions at the instant shown. Use the x,y,zaxes shown

and note that Izx =-

amac

12 bac2–a2

c2+a2b.

c=0.3 ma=0.2 m

v=4 rad>s

m

=2kg

SOLUTION

Substitute the data,

vx=0,

vy=0, vz=-4

x

A

B

V

ca

y

z

Ans:

1154

21–45.

z

A

B

y

x

1.5 ft

1 ft

v fi 30 rad/s

30

0.5 ft

If the shaft AB is rotating with a constant angular velocity

of , determine the X,Y,Zcomponents of

reaction at the thrust bearing Aand journal bearing Bat

the instant shown. The disk has a weight of 15 lb.Neglect

the weight of the shaft AB.

v=30 rad>s

SOLUTION

The rotating xyz frame is set with its origin at the plate’s mass center,Fig. a.This

frame will be fixed to the disk so that its angular velocity is and the x,y, and

zaxes will always be the principle axes of inertia of the disk. Referring to Fig. b,

Since is always directed towards the axis and has a constant magnitude,

The mass moments of intertia of the disk about the x,y,zaxes are

Applying the equations of motion,

+Yv

Æ=v

Solving Eqs. (1) through (4),

1155

1 m

2 m

1 m

A

x

y

B

z

v

21–46.

The assembly is supported by journal bearings at A and B,

which develop only y and z force reactions on the shaft. If

the shaft is rotating in the direction shown at

V

=52i6 rad>s,

determine the reactions at the bearings

when the assembly is in the position shown. Also, what is

the shaft’s angular acceleration? The mass per unit length of

each rod is

5 kg>m.

SOLUTION

Equations of Motion. The inertia properties of the assembly are

with v

#

x=2 rad>s

, vy

=

vz

=0

and v

#

y=v

#

z=0 by referring to the FBD of the

assembly, Fig. a,

Ans:

#

1156

21–47.

The assembly is supported by journal bearings at A and B,

which develop only y and z force reactions on the shaft. If

the shaft A is subjected to a couple moment

M=540i6 N #m,

and at the instant shown the shaft has an

angular velocity of V

=52i6 rad>s,

determine the

reactions at the bearings of the assembly at this instant.

Also, what is the shaft’s angular acceleration? The mass per

unit length of each rod is

5 kg>m.

1 m

2 m

1 m

A

x

y

B

z

v

SOLUTION

Equations of Motions. The inertia properties of the assembly are

with v

x=2 rad>s

, vy

=

vz

=0

and v

#

y

=

v

#

z

=0

by referring to the FBD of the

assembly, Fig. a,

Ans:

v

#

=

9.285 rad

>

s

2

1157

*21–48.

SOLUTION

©Fz=m(aG)z;–5+Az=5

Æ

#=(v

#

xyz)+Æ*v=-2i+0

Ix=Iz=1

12

A

5

32.2

B

(3)2=0.1165 ft4

The man sits on a swivel chair which is rotating with a

constant angular velocity of . He holds the uniform

5-lb rod AB horizontal. He suddenly gives it an angular

acceleration of measured relative to him, as

shown. Determine the required force and moment

components at the grip, A, necessary to do this. Establish

axes at the rod’s center of mass G, with upward, and

directed along the axis of the rod towards A.

+y+z

2 rad>s2,

3 rad>s

3 ft 2 ft

A

B

3 rad/s

2 rad/s2

1158

21–49.

The rod assembly is supported by a ball-and-socket joint at

Cand a journal bearing at D, which develops only xand y

force reactions.The rods have a mass of 0.75 kg/m.

Determine the angular acceleration of the rods and the

components of reaction at the supports at the instant

as shown.v=8 rad>s

SOLUTION

Eqs. 21–24 become

Thus,

–Dy(4) –7.3575(0.5) =0.75(8)2

Æ=v=8k

=8rad/s

ω

2m

1m

2m

z

B

A

D

50 N ⋅m

1159

21–50.

The bent uniform rod ACD has a weight of 5 lb

>

ft and is

supported at A by a pin and at B by a cord. If the vertical

shaft rotates with a constant angular velocity v=20 rad>s,

determine the x, y, z components of force and moment

developed at A and the tension in the cord.

SOLUTION

wx=wy=0

The center of mass is located at

Eqs. 21–24 reduce to

1 ft

0.5 ft

1 ft

y

C

B

A

z

Ans:

TB=47.1 lb

1160

21–51.

The uniform hatch door,having a mass of 15 kg and a mass

center at G,issupported in the horizontal plane by bearings at

Aand B.If a vertical force is applied to the door

as shown, determine the components of reaction at the

bearings and the angular acceleration of the door.The bearing

at Awill resist a component of force in the ydirection,

whereas the bearing at Bwill not. For the calculation, assume

the door to be a thin plate and neglect the size of each

bearing.The door is originally at rest.

F=300 N

SOLUTION

Eqs. 21–25 reduce to

(1)

Bz–Az=-440

©Mx=0; 300(0.25 –0.03) +Bz(0.15) –Az(0.15) =0

y

30 mm

z

x

G

B

A

30 mm

100 mm

150 mm

100 mm

200 mm

F

Ans:

1161

*21–52.

SOLUTION

vx=0, vy=-10 rad/s,

vz=0

The 5-kg circular disk is mounted off center on a shaft

which is supported by bearings at Aand B. If the shaft is

rotating at a constant rate of , determine the

vertical reactions at the bearings when the disk is in the

position shown.

v=10 rad>s

100 mm

20 mm

BA

G

100mm

100 mm

ω

Ans:

1162

21–53.

SOLUTION

Iy=

1

12 a10

32.2b(4)2=0.4141 slug #ft2 Iz=0

Applying Eq. 21–25 with

Also,

Two uniform rods, each having a weight of 10 lb, are pin

connected to the edge of a rotating disk. If the disk has

a constant angular velocity vD=4

rad>s, determine

the angle u made by each rod during the motion, and the

components of the force and moment developed at the

pin A. Suggestion: Use the x, y, z axes oriented as shown.

θ

ω

D = 4 rad/s

y

θ

2 ft

x

z

A

G

B

1.75 ft

Ans:

Mz=0

1163

21–54.

The 10-kg disk turns around the shaft AB, while the shaft

rotates about BC at a constant rate of vx

=5 rad>s.

If the

disk does not slip, determine the normal and frictional force

it exerts on the ground. Neglect the mass of shaft AB.

SOLUTION

Kinematics. The instantaneous axis of zero velocity (IA) is indicated

in Fig. a. Here the resultant angular velocity is always directed along

IA. The fixed reference frame is set to coincide with the rotating xyz

frame using the similar triangle,

Thus,

Vx

The direction of

Vb

does not change with reference to the xyz rotating frame if this

frame rotates with

Ω=

vx

=5–5i6 rad>s

. Then

x

0.4 m

y

z

2 m A

C

B

vx fi 5 rad/s

21–54. Continued

Equations of Motion. The mass moments of inertia of the disk about the x, y and

zaxes are

By referring to the FBD of the disk, Fig. b,

ΣM

y

=I

y

v

#

y

–(I

z

–I

x

)

vzvx;

ΣM

1165

21–55.

The 20-kg disk is spinning on its axle at vs

=30 rad>s,

while the forked rod is turning at v

1=6 rad>s.

Determine

the x and z moment components the axle exerts on the disk

during the motion.

SOLUTION

Solution I

Kinematics. The fixed reference XYZ frame is set coincident with the rotating xyz

Then

Equation of Motion. Although the disk spins about the y axis, but the mass moment

of inertia of the disk remain constant with respect to the xyz frame.

x

O

y

200 mm

z

A

vs fi 30 rad/s

v1 fi 6 rad/s

1166

21–55. Continued

With

𝛀 ≠ V

with

Ωx=0

,

Ω

y

=0

and

Ωz=–6 rad>s

Solution II

Here, the xyz frame is set to rotate with

𝛀=

V

=530j–6k6 rad>s

. Setting

another

x′y′z′

frame coincide with xyz and XYZ frame to have an angular velocity

Ans:

1167

*21–56.

SOLUTION

Applying the third of Eq. 21–25 with

Also,

The 4-kg slender rod AB is pinned at Aand held at Bby a

cord.The axle CD is supported at its ends by ball-and-socket

joints and is rotating with a constant angular velocity of

. Determine the tension developed in the cord and

the magnitude of force developed at the pin A.

2 rad>s

ω

D

2m

C

B

y

z40°

Ans:

1168

21–57.

The blades of a wind turbine spin about the shaft Swith a

constant angular speed of , while the frame precesses

about the vertical axis with a constant angular speed of .

Determine the x,y, and zcomponents of moment that the

shaft exerts on the blades as a function of . Consider each

blade as a slender rod of mass mand length l.

u

vp

vs

SOLUTION

The rotating xyz frame shown in Fig. awill be attached to the blade so that it rotates

The angular acceleration of the blade with respect to the XYZ frame can be

obtained by setting another frame having an angular velocity of

Since ,.Thus,

Also, the x,y, and zaxes will remain as principle axes of inertia for the blade.Thus,

Applying the moment equations of motion and referring to the free-body diagram

shown in Fig. a,

v

#

x¿y¿z¿=v

#

Æ=v

x¿y¿z¿

v

#

z

x

y

S

u

s

p

©Mx=Ixv

#

x–

A

Iy–Iz

B

vyvz;Mx=2

3ml2(–vsvpcos u)–a2

3ml2–0b(vs)(vpcos u)

v

Ans: