HW6 Solutions Planar Kinetics of a Rigid Body: Force and

Introduction to Dynamics (N. Zabaras)

HW6 Solutions Planar Kinetics of a Rigid

Body: Force and AccelerationProf. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

February 14, 2016

1

mailto:[email protected]

http://www.zabaras.com/


21

2GI m r

2

38000 [ (0.25 ) (0.01 )] 15.71d d d

kgm V m m kg

m

2 2

2 2

2

1( )

2

1(15.71 )(0.25 ) (15.71 )(0.25 )

2

1.473 kg.m

d O d d dI m r m d

kg m kg m

2

38000 [ (0.125 ) (0.01 )] 3.93h h h

kgm V m m kg

m

2

( ) ( )

1.473 0.276 1.2 .

O d O h OI I I

kg m

Problem 1

2

2 2

2 2

2

1( )

2

1(3.93 )(0.125 ) (3.93 )(0.25 )

2

0.276 kg.m

h O h h hI m r m d

kg m kg m

Determine IOthrough an axis

perpendicular

to the page and

passing through O

Hole

38000d

kg

m


mm=125 kg

mR= 75 kg

a= ?

NB=?

FB=?

ms=? Minimum to

lift front wheel

Problem 2

3

The motorcycle shown has a mass of 125 kg and a center

of mass at G1 while the rider has a mass of 75 kg and a

center of mass at G2.

Determine the minimum coefficient of static friction

between the wheels and the pavement in order for the

rider to do a "wheely,'‘ i.e., lift the front wheel off the

ground as shown in the photo. What acceleration is

necessary to do this? Neglect the mass of the wheels and

assume that the front wheel is free to roll.


( ) ; (75 125 )x G x B GF m a F kg kg a

( ) ; 735.75 1226.25 0y G y BF m a N N N

28.94 /

1962

1790

G

B

B

a m s

N N

F N

1790( ) min 0.912

1962

Bs

B

F

Nm

mm=125 kg

mR= 75 kg

a= ?

NB=?

FB=?

ms=? Minimum to

Lift front wheel

Problem 2

4

( ) ; (735.75 )(0.4 ) (1226.25 )(0.8 )

(75 ) (0.9 ) (125 )(0.6 )

B B

G G

M N m N m

kg a m kg a m

Μ k


m = 80 kg

mB= 0.8

NA= ?

NB= ?

When rear wheel locks for break

a = ? Deceleration

NA NB

( ) ; 0.8 80x G x B GF m a N a

mBNB

( ) ; 80(9.81) 0y G y B AF m a N N

( ) ; (0.95) 80(9.81)(0.55) 80 (1.2)A A B GM N a Μ k

22.26 /

226

559

G

B

A

a m s

N N

N N

aG

Problem 3

5


m= 80 kg

mB= ?

NA=?

NB=?

a =?

When traveling at constant velocity

and no break was applied

( ) ; 80(9.81) 0y G y B AF m a N N

0; (0.95) 80(9.81)(0.55) 0A BM N

NA NB

mBNB

a = 0

20 /

330

454

G

B

A

a m s

N N

N N

0; 0x k BF Nm

6

Problem 4


m= 80 kg

NA=?

NB=?

a =?

mk=? minimum

When rider applies the front break

and back wheel start to lift off the ground

NA NB=0

mkNA

( ) ; 80x G x k A GF m a N am

( ) ; 80(9.81) 0y G y AF m a N

( ) ; 80(9.81)(0.55) 80 (1.2)A A GM a Μ k

24.5 /

785

0.458

G

A

k

a m s

N N

m

0BN

Problem 5

7

aG


2( ) 4.9 / 1.32G t B Da m s T T kN

mBD=100kg

mAB=mCD= Neglect

Consider instant: q = 30o, w = 6 rad/s

Compute: TA= ? TD= ? aG=?

2 2 2( ) (6) (0.5) 18 /G na ω r m s ( ) 981cos30 100(18)n G n B DF m a T T

( ) 981sin 30 100( )t G t G tF m a a

0 cos30 (0.4) cos30 (0.4) 0G B DM T T

Problem 6

8

BD undergoes

curvilinear

translation.

Each

point on it

having a

circular

path of radius

0.5 m but with

different

center.


Problem 7

The thin plate of mass 8 kg is held

in place as shown.

Neglecting the mass of the links,

determine immediately after the

wire has been cut (a) the

acceleration of the plate, and (b) the

force in each link.

SOLUTION:

• Note that after the wire is cut, all

particles of the plate move along

parallel circular paths of radius 150

mm. The plate is in curvilinear

translation.

• Draw the free-body-diagram

equation expressing the equivalence

of the external and effective forces.

• Resolve into scalar component

equations parallel and perpendicular

to the path of the mass center.

• Solve the component equations and

the moment equation for the

unknown acceleration and link

forces.

9


Problem 7SOLUTION:

• Note that after the wire is cut, all particles of

the plate move along parallel circular paths

of radius 150 mm. The plate is in curvilinear

translation.

• Draw the free-body-diagram equation

expressing the equivalence of the external

and effective forces.

• Resolve the diagram equation into

components parallel and perpendicular to

the path of the mass center.

efftt FF

30cos

30cos

mg

amW

30cosm/s81.9 2a

2sm50.8a 60o

10

At the instant

BH is cut, the

velocity of the

plate is zero.

Thus we only

have

tangential

acceleration

as shown.


Problem 7

2sm50.8a 60o

• Solve the component equations and the

moment equation for the unknown acceleration

and link forces.

effGG MM

0mm10030cosmm25030sin

mm10030cosmm25030sin

DFDF

AEAE

FF

FF

AEDF

DFAE

FF

FF

1815.0

06.2114.38

effnn FF

2sm81.9kg8619.0

030sin1815.0

030sin

AE

AEAE

DFAE

F

WFF

WFF

TFAE N9.47

N9.471815.0DFF CFDF N70.8

11


Problem 8

A pulley weighing 12 lb and having a

radius of gyration of 8 in. is connected

to two blocks as shown.

Assuming no axle friction, determine

the angular acceleration of the pulley

and the acceleration of each block.

SOLUTION:

• Determine the direction of rotation

by evaluating the net moment on

the pulley due to the two blocks.

• Relate the acceleration of the

blocks to the angular acceleration

of the pulley.


equation expressing the

equivalence of the external and

effective forces on the complete

pulley plus blocks system.

• Solve the corresponding moment

equation for the pulley angular

acceleration.

12


Problem 8

• Relate the acceleration of the blocks to the

angular acceleration of the pulley.

ft1210

AA ra

ft126

BB ra

2

2

2

22

sftlb1656.0

ft12

8

sft32.2

lb12

kg

WkmInote:

SOLUTION:

• Determine the direction of rotation by evaluating the

net moment on the pulley due to the two blocks.

lbin10in10lb5in6lb10 GM

rotation is counterclockwise.

13


Problem 8

• Draw the free-body-diagram equation expressing

the equivalence of the external and effective forces

on the complete pulley and blocks system.

2

126

2

1210

2

sft

sft

sftlb1656.0

B

A

a

a

I

effGG MM

1210

1210

2.325

126

126

2.3210

1210

126

1210

126

1210

126

1656.0510

ftftftlb5ftlb10

AABB amamI

• Solve the corresponding moment equation for the

pulley angular acceleration.

2srad374.2

2

126 srad2.374ft

BB ra2sft187.1Ba

2

1210 srad2.374ft

AA ra

2sft978.1Aa

Then,

14


Problem 9

A uniform sphere of mass m and radius

r is projected along a rough horizontal

surface with a linear velocity v0. The

coefficient of kinetic friction between the

sphere and the surface is mk.

Determine:

(a) the time t1 at which the sphere will

start rolling without sliding, and

(b) the linear and angular velocities of

the sphere at time t1.

SOLUTION:


equation expressing the equivalence

of the external and effective forces

on the sphere.

• Solve the three corresponding scalar

equilibrium equations for the normal

reaction from the surface and the

linear and angular accelerations of

the sphere.

• Apply the kinematic relations for

uniformly accelerated motion to

determine the time at which the

tangential velocity of the sphere at

the surface is zero, i.e., when the

sphere stops sliding.

15


Problem 9SOLUTION:

• Draw the free-body-diagram equation expressing

the equivalence of external and effective forces

on the sphere.

• Solve the three scalar equilibrium equations.

effyy FF

0WN mgWN

effxx FF

k

F ma

mg mam

ga km

225k

Fr I

mg r mr

m

r

gkm

2

5

effGG MM

NOTE: As long as the sphere both rotates and

slides, its linear and angular motions are

uniformly accelerated.

16


Problem 9

ga km

r

gkm

2

5

• Apply the kinematic relations for uniformly accelerated

motion to determine the time at which the tangential

velocity of the sphere at the surface is zero, i.e., when

the sphere stops sliding.

tgvtavv km 00

tr

gt k

mww

2

500

1102

5t

r

grgtv k

k

mm

g

vt

km0

17

2

g

v

r

gt

r

g

k

kk

m

mmw 0

117

2

2

5

2

5

r

v01

7

5w

r

vrrv 0

117

5w 07

51 vv

At t1 when the sphere stops sliding, vC=0, and C

becomes the instantaneous center of rotation, i.e.

11 wrv

17


2 2 2(60 )(0.25 ) 3.75 .o oI mk kg m kg m

( ) ; 20(9.81) 20y G yF m a T a

; (0.4)a r a 2 2106 4.52 / 11.3 /T N a m s rad s

m=60 kg

Radius of gyration

kO=0.25

mb=20 kg

=? Drum

a=?

( ) ; 0x G x xF m a O

2; (0.4 ) [3.75 . ]o oM I T m kg m

Problem 10

18

(for the weight only)

(for the pulley only)


2 250; ( )(8 / ) (0.5f )

32.2n G nF m r O rad s tw

50; 50 ( )( )(0.5f )

32.2t G tF m r O t

; (0.5) 80 Ib.ft (0.559)G G tM I O 249.7 36.1 111 /n tO N O Ib rad s

W=50 Ib

kG=0.6 ft

w= 8 rad/s

Pin reaction =?

Problem 11

19

2 2 250( )(0.6) 0.559 slug.ft32.2

G GI mk


Problem 12

The portion AOB of the mechanism

is actuated by gear D and at the

instant shown has a clockwise

angular velocity of 8 rad/s and a

counterclockwise angular

acceleration of 40 rad/s2.

Determine: a) tangential force

exerted by gear D, and b)

components of the reaction at shaft

O.

kg 3

mm 85

kg 4

OB

E

E

m

k

m

SOLUTION:

• Draw the free-body-equation for

AOB, expressing the equivalence of

the external and effective forces.

• Evaluate the external forces due to

the weights of gear E and arm OB

and the effective forces associated

with the angular velocity and

acceleration.

• Solve the three scalar equations

derived from the free-body-

equation for the tangential force at

A and the horizontal and vertical

components of reaction at shaft O.

20


Problem 12

rad/s 8w

2srad40

kg 3

mm 85

kg 4

OB

E

E

m

k

m

SOLUTION:

• Draw the free-body-equation for AOB.

• Evaluate the external forces due to the weights

of gear E and arm OB and the effective forces.

N4.29sm81.9kg3

N2.39sm81.9kg4

2

2

OB

E

W

W

mN156.1

srad40m085.0kg4 222

EEE kmI

N0.24

srad40m200.0kg3 2

rmam OBtOBOB

N4.38

srad8m200.0kg322

wrmam OBnOBOB

mN600.1

srad40m.4000kg3 22

1212

121

LmI OBOB

21


Problem 12

N4.29

N2.39

OB

E

W

W

mN156.1 EI

N0.24tOBOB am

N4.38nOBOB am

mN600.1 OBI

• Solve the three scalar equations derived from the

free-body-equation for the tangential force at A

and the horizontal and vertical components of

reaction at O.

effOO MM

mN600.1m200.0N0.24mN156.1

m200.0m120.0

OBtOBOBE IamIF

N0.63F

effxx FF

N0.24tOBOBx amR

N0.24xR

effyy FF

N4.38N4.29N2.39N0.63

y

OBOBOBEy

R

amWWFR

N0.24yR

22


Problem 13

A cord is wrapped around the

inner hub of a wheel and pulled

horizontally with a force of 200 N.

The wheel has a mass of 50 kg

and a radius of gyration of 70 mm.

Knowing ms = 0.20 and mk = 0.15,

determine the acceleration of G

and the angular acceleration of

the wheel.

SOLUTION:

• Draw the free-body-equation for the

wheel, expressing the equivalence of


• Assuming rolling without slipping and

therefore, related linear and angular

accelerations, solve the scalar

equations for the acceleration and the

normal and tangential reactions at the

ground.

• Compare the required tangential

reaction to the maximum possible

friction force.

• If slipping occurs, calculate the kinetic

friction force and then solve the scalar

equations for the linear and angular

accelerations.

23


Problem 13SOLUTION:

• Draw the free-body-equation for the wheel,.

Assume rolling without slipping,

m100.0

ra

2

22

mkg245.0

m70.0kg50

kmI

• Assuming rolling without slipping, solve the scalar

equations for the acceleration and ground

reactions.

22

2

22

sm074.1srad74.10m100.0

srad74.10

mkg245.0m100.0kg50mN0.8

m100.0m040.0N200

a

Iam

effCC MM

effxx FF

N5.490sm074.1kg50

0

2

mgN

WN

effxx FF

N3.146

sm074.1kg50N200 2

F

amF

24


Problem 13

N3.146F N5.490N

Without slipping,

• Compare the required tangential reaction to the

maximum possible friction force.

N1.98N5.49020.0max NF sm

F > Fmax , rolling without slipping is

impossible.

• Calculate the friction force with slipping and solve

the scalar equations for linear and angular

accelerations.

N6.73N5.49015.0 NFF kk m

effGG MM

2

2

srad94.18

mkg245.0

m060.0.0N200m100.0N6.73

2srad94.18

effxx FF

akg50N6.73N200 2sm53.2a

25


Problem 14

The extremities of a 4-ft rod

weighing 50 lb can move freely

and with no friction along two

straight tracks. The rod is

released with no velocity from

the position shown.

Determine: a) the angular

acceleration of the rod, and b)

the reactions at A and B.

SOLUTION:

• Based on the kinematics of the

constrained motion, express the

accelerations of A, B, and G in terms

of the angular acceleration.

• Draw the free-body-equation for the

rod, expressing the equivalence of


• Solve the three corresponding scalar

equations for the angular acceleration

and the reactions at A and B.

26


Problem 14

SOLUTION:

• Based on the kinematics of the constrained

motion, express the accelerations of A, B, and

G in terms of the angular acceleration.

Express the acceleration of B as

ABAB aaa

With the corresponding vector triangle

and the law of signs yields

,4ABa

90.446.5 BA aa

The acceleration of G is now obtained from

AGAG aaaa

2 where AGa

Resolving into x and y components,

732.160sin2

46.460cos246.5

y

x

a

a

27

30

30

90

60

B Aa


Problem 14• Draw the free-body-equation for the rod,

expressing the equivalence of the external and

effective forces.

69.2732.12.32

50

93.646.42.32

50

07.2

sftlb07.2

ft4sft32.2

lb50

12

1

2

2

2

2

121

y

x

am

am

I

mlI

• Solve the three corresponding scalar equations

for the angular acceleration and the reactions at

A and B.

2srad30.2

07.2732.169.246.493.6732.150

effEE MM

2srad30.2

effxx FF

lb5.22

30.293.645sin

B

B

R

R

lb5.22BR

45o

effyy FF

30.269.25045cos5.22 AR

lb9.27AR

28

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HW6 Solutions Planar Kinetics of a Rigid Body: Force and