Mechanical Engineering Chapter 16 Homework Determine the angular velocity of the propeller in

650

16–21.

The motor turns the disk with an angular velocity of

v

=

(

5t

2+

3t

)

rad

>

s, where t is in seconds. Determine the

magnitudes of the velocity and the n and t components of

acceleration of the point A on the disk when

t=3 s

.

SOLUTION

Angular Motion. At

t=3 s

,

=

t=3 s

=

Motion of Point A. The magnitude of the velocity is

u

150 mm

A

Ans:

651

16–22.

SOLUTION

Angular Motion: The angular velocity and acceleration of gear Bmust be

determined first. Here, and .Then,

Since gear Cis attached to gear Bdna neht,

.Realizing that and ,then

aCrC=aDrD

vCrC=vDrD

aC=aB=0.8 rad>s2

vC=vB=8 rad>s

aArA=aBrB

vArA=vBrB

If the motor turns gear Awith an angular acceleration of

when the angular velocity is ,

determine the angular acceleration and angular velocity of

gear D.

vA=20 rad>saA=2 rad>s2

A

B

C

D

50 mm

100 mm

40 mm

Ans:

652

16–23.

If the motor turns gear Awith an angular acceleration of

when the angular velocity is ,

determine the angular acceleration and angular velocity of

gear D.

vA=60 rad>saA=3 rad>s2

SOLUTION

Angular Motion: The angular velocity and acceleration of gear Bmust be

determined first. Here, and .Then,

aArA=aBrB

vArA=vBrB

A

B

C

D

50 mm

100 mm

40 mm

Ans:

653

*16–24.

Ans:

SOLUTION

Angular Motion:The angular velocity of gear Aat must be determined

first. Applying Eq. 16–2, we have

t=1.5 s

The gear Aon the drive shaft of the outboard motor has a

radius .and the meshed pinion gear Bon the

propeller shaft has a radius .Determine the

angular velocity of the propeller in ,if the drive shaft

rotates with an angular acceleration ,

where tis in seconds.The propeller is originally at rest and

the motor frame does not move.

a=(400t3) rad>s2

t=1.5 s

rB=1.2 in

rA=0.5 in

2.20 in.

P

B

A

654

16–25.

SOLUTION

Angular Motion:The angular velocity of gear Aat must be determined

first. Applying Eq. 16–2, we have

The angular acceleration of gear Aat is given by

However, and where and are the angular

velocity and acceleration of propeller.Then,

Motion of P:The magnitude of the velocity of point Pcan be determined using

Eq. 16–8.

The tangential and normal components of the acceleration of point Pcan be

determined using Eqs. 16–11 and 16–12, respectively.

aB

vB

aArA=aBrB

vArA=vBrB

t=0.75 s

dv=adt

t=0.75 s

2.20 in.

P

B

A

Ans:

655

16–26.

The pinion gear Aon the motor shaft is given a constant

angular acceleration If the gears Aand B

have the dimensions shown, determine the angular velocity

and angular displacement of the output shaft C, when

starting from rest. The shaft is fixed to Band turns

with it.

t=2s

a=3 rad>s2.

SOLUTION

v=v0+act

C

125 mm

35 mm

A

B

Ans:

656

16–27.

SOLUTION

(3002t)(0.7) =ap(1.4)

aArA=aBrB

The gear Aon the drive shaft of the outboard motor has a

radius and the meshed pinion gear Bon the

propeller shaft has a radius Determine the

angular velocity of the propeller in if the drive

shaft rotates with an angular acceleration

where tis in seconds.The propeller is

originally at rest and the motor frame does not move.

a=13002t2rad>s2,

t=1.3 s

rB=1.4 in.

rA=0.7 in.

2.2 in.

P

B

A

657

*16–28.

The gear A on the drive shaft of the outboard motor has a

radius rA

=

0.7 in. and the meshed pinion gear B on the

propeller shaft has a radius rB

=

1.4 in. Determine the

magnitudes of the velocity and acceleration of a point P

located on the tip of the propeller at the instant t

=

0.75 s.

the drive shaft rotates with an angular acceleration

a

=

(300

1t

) rad

>

s2, where t is in seconds. The propeller is

originally at rest and the motor frame does not move.

Ans:

SOLUTION

Angular Motion: The angular velocity of gear A at

t=0.75 s

must be determined

first. Applying Eq. 16–2, we have

The angular acceleration of gear A at

t=0.75 s

is given by

Motion of P: The magnitude of the velocity of point P can be determined using

Eq.16–8.

The tangential and normal components of the acceleration of point P can be

determained using Eqs. 16–11 and 16–12, respectively.

2.2 in.

P

B

A

658

16–29.

A stamp S, located on the revolving drum, is used to label

canisters. If the canisters are centered 200 mm apart on the

conveyor, determine the radius of the driving wheel A

and the radius of the conveyor belt drum so that for each

revolution of the stamp it marks the top of a canister. How

many canisters are marked per minute if the drum at Bis

rotating at ? Note that the driving belt is

twisted as it passes between the wheels.

vB=0.2 rad>s

rB

rA

200mm

A

r

A

r

B

S

SOLUTION

For the drum at B:

l=2p(rB)

l=2p(rA)

Ans:

rA=31.8

mm

rB=31.8

mm

659

16–30.

SOLUTION

(rB)max =(rA)max =5022 mm

At the instant shown, gear A is rotating with a constant

angular velocity of vA=6 rad>s. Determine the largest

angular velocity of gear B and the maximum speed of

point C.

100 mm

B

C

A

100 mm

B

ω

A= 6 rad/s

ω

Ans:

16–31.

Determine the distance the load Wis lifted in using

the hoist. The shaft of the motor Mturns with an angular

velocity , where tis in seconds.v=100(4 +t) rad>s

t

=5s

300 mm

30 mm

50 mm 225 mm

B

A

E

C

W

D

40 mm

M

SOLUTION

Angular Motion: The angular displacement of gear Aat must be determined

first. Applying Eq. 16–1, we have

Here,.Then, the angular displacement of gear Bis given by

rAuA=rBuB

t=5s

Ans:

661

Ans:

*16–32.

A

B

125 mm

200 mm

vA

vB

T

h

e

d

r

i

v

i

ng

b

e

l

t

i

s tw

i

ste

d

so t

h

at pu

ll

ey Brotates

i

n t

h

e

oppo

site direction to that of drive wheel A.If Ahas a

con

stant angular acceleration of ,determine

the

tangential and normal components of acceleration of a

point

located at the rim of Bwhen ,starting from rest.t=3 s

aA=30 rad>s2

SOLUTION

Motion

of Wheel A: Since the angular acceleration of wheel Ais constant, its

an

gular velocity can be determined from

M

otion of Wheel B: Since wheels Aand Bare connected by a nonslip belt, then

and

16–33.

A

B

125 mm

200 mm

vA

vB

The driving belt is twisted so that pulley Brotates in the

opposite direction to that of drive wheel A. If the angular

displacement of Ais rad, where tis in

seconds, determine the angular velocity and angular

acceleration of Bwhen t=3 s.

uA=(5t3+10t2)

SOLUTION

Motion of Wheel A: The angular velocity and angular acceleration of wheel Acan

be determined from

and

When ,

Motion of Wheel B: Since wheels Aand Bare connected by a nonslip belt, then

t=3 s

663

16–34.

Forashort time amotorofthe random-orbit sander drives

the

gear Awith an angular velocity of

where tis in seconds.This gear is

connected

to gear B,which is fixed connected to the shaft

CD

.The end of this shaft is connected to the eccentric

spindle

EF and pad P,which causes the pad to orbit around

shaft

CD at aradius of 15 mm. Determine the magnitudes

of

the velocity and the tangential and normal components

of

acceleration of the spindle EF when after

starting

from rest.

t=2s

vA=401t3+6t2rad>s,

SOLUTION

aArA=aBrB

vArA=vBrB

40 mm

10 mm

15 mm

A

B

C

D

E

F

V

Ans:

664

16–35.

If the shaft and plate rotates with a constant angular velocity

of ,determine the velocity and acceleration of

point Clocated on the corner of the plate at the instant

shown. Express the result in Cartesian vector form.

v=14 rad>s

SOLUTION

We will first express the angular velocity of the plate in Cartesian vector form.The

unit vector that defines the direction of is

Since is constant

fo noitarelecca dna yticolev ehT.nesohc si,ecneinevnoc roF

point Ccan be determined from

and

rC=[–0.3i+0.4j] m

v

xy

C

O

D

z

0.2 m

0.3 m

0.4 m

0.6 m

A

v

a

Ans:

665

Ans:

*16–36.

SOLUTION

We will first express the angular velocity of the plate in Cartesian vector form.The

unit vector that defines the direction of and is

Thus,

fo noitarelecca dna yticolev ehT.nesohc si,ecneinevnoc roF

point Dcan be determined from

vD=v*rD

rD=[–0.3i+0.4j] m

av

v

At the instant shown, the shaft and plate rotates with an

angular velocity of and angular acceleration

of . Determine the velocity and acceleration of

point Dlocated on the corner of the plate at this instant.

Express the result in Cartesian vector form.

a=7 rad>s2

v=14 rad>s

xy

C

O

D

z

0.2 m

0.3 m

0.4 m

0.6 m

A

v

a

666

16–37.

The rod assembly is supported by ball-and-socket joints at

Aand B. At the instant shown it is rotating about the yaxis

with an angular velocity and has an angular

acceleration Determine the magnitudes of

the velocity and acceleration of point Cat this instant.

Solve the problem using Cartesian vectors and

and 16–13.

a=8 rad>s2.

v=5 rad>s

SOLUTION

vC=v*r

0.3 m

z

x

y

A

C

B

0.4 m

AV

Eqs. 16–9

Ans:

667

16–38.

The sphere starts from rest at

u=0°

and rotates with an

angular acceleration of a

=

(4u

+

1) rad

>

s

2

, where

u

is in

radians. Determine the magnitudes of the velocity and

acceleration of point P on the sphere at the instant

u=6

rad.

Ans:

P

r  8 in.

30fi

SOLUTION

v dv=a du

u=6 rad

=

668

16–39.

SOLUTION

Position coordinate equation:

Time derivatives:

T

h

e en

d

Aof t

h

e

b

ar

i

s mov

i

ng

d

ownwar

d

a

l

ong t

h

e s

l

otte

d

guide with a constant velocity Determine the angular

velocity and angular acceleration of the bar as a

function of its position y.

AV

vA.

y

B

r

v

A

V,A

U

Ans:

669

Ans:

*16–40.

At the instant

u=60°

, the slotted guide rod is moving

to the left with an acceleration of 2 m

>

s2 and a velocity of

5 m

>

s. Determine the angular acceleration and angular

velocity of link AB at this instant.

SOLUTION

Position Coordinate Equation. The rectilinear motion of the guide rod can be

related to the angular motion of the crank by relating

x and u

using the geometry

shown in Fig. a, which is

Time Derivatives. Using the chain rule,

Here

x

#

=v

,

x

$

=a

,

u

#

=v

and

u

$

=a

when

u =60°

. Realizing that the velocity

Subsequently, Eq. (2) gives

The negative sign indicates that

A

is directed in the negative sense of

u

.

200 mm

v  5 m/s

a  2 m/s2

A

B

u