dynamics lecture7 [compatibility mode] 2
Post on 08-Apr-2018
218 Views
Preview:
TRANSCRIPT
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 1/38
Plane Motion of Rigid Bodies:
Forces and Accelerations
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 2/38
Introduction
• In this chapter and in Chapters 17 and 18, we will be
concerned with the kinetics of rigid bodies, i.e., relations
between the forces acting on a rigid body, the shape and mass
of the body, and the motion produced.
• Results of this chapter will be restricted to:
- plane motion of rigid bodies, and
- rigid bodies consisting of plane slabs or bodies which
• Our approach will be to consider rigid bodies as made of large
numbers of particles and to use the results discussed previously
for the motion of systems of particles. Specifically,
GG H M amF
== ∑∑ and
are symmetrical with respect to the reference plane.
• D’Alembert’s principle is applied to prove that the external
forces acting on a rigid body are equivalent a vector
attached to the mass center and a couple of moment
am
.α I
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 3/38
Equations of Motion for a Rigid Body
• Consider a rigid body acted upon
by several external forces.
• For the motion of the mass center
G of the body with respect to the
• Assume that the body is made of
a large number of particles.
,
amF =∑
• For the motion of the body with
respect to the centroidal frame
Gx’y’z’,GG H M
=∑
• System of external forces is
equipollent to the system
consisting of .and G H am
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 4/38
Angular Momentum of a Rigid Body in Plane Motion
• Angular momentum of the slab may be
computed by
( )
( )[ ]
( )ω
ω
mr
mr r
mvr H
ii
n
iiii
n
iiiiG
′=
′××′=
′×′=
∑
∑
∑
=
=
∆
∆
∆
2
1
1
• Consider a rigid slab in
plane motion.
I =
• After differentiation,
α ω
I I H G ==
• Results are also valid for plane motion of bodieswhich are symmetrical with respect to the
reference plane.
• Results are not valid for asymmetrical bodies or
three-dimensional motion.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 5/38
Plane Motion of a Rigid Body: D’Alembert’s Principle
α I M amF amF G y y x x
=== ∑∑∑
• Motion of a rigid body in plane motion is
completely defined by the resultant and moment
resultant about G of the external forces.
• The external forces and the collective effective
forces of the slab particles are equipollent (reduce
to the same resultant and moment resultant) and
equ va ent ave t e same e ect on t e o y .
• The most general motion of a rigid body that is
symmetrical with respect to the reference plane
can be replaced by the sum of a translation and a
centroidal rotation.
• d’Alembert’s Principle: The external forces
acting on a rigid body are equivalent to the
effective forces of the various particles forming
the body.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 6/38
Axioms of the Mechanics of Rigid Bodies
• The forces act at different points on
a rigid body but but have the same magnitude,
direction, and line of action.
F F ′and
• The forces produce the same moment about
any point and are therefore, equipollent
.
• This proves the principle of transmissibility
whereas it was previously stated as an axiom.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 7/38
Problems Involving the Motion of a Rigid Body
• The fundamental relation between the forces
acting on a rigid body in plane motion and
the acceleration of its mass center and the
angular acceleration of the body is illustratedin a free-body-diagram equation.
• The techniques for solving problems of
static e uilibrium may be a lied to solve
problems of plane motion by utilizing- d’Alembert’s principle, or
- principle of dynamic equilibrium
• These techniques may also be applied to
problems involving plane motion of
connected rigid bodies by drawing a free-
body-diagram equation for each body and
solving the corresponding equations of
motion simultaneously.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 8/38
Sample Problem 16.1
SOLUTION:
• Calculate the acceleration during the
skidding stop by assuming uniform
acceleration.
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces.
At a forward speed of 10 m/s, the truck
brakes were applied, causing the wheels
to stop rotating. It was observed that the
truck to skidded to a stop in 7 m.
Determine the magnitude of the normal
reaction and the friction force at each
wheel as the truck skidded to a stop.
• Apply the three corresponding scalar
equations to solve for the unknown
normal wheel forces at the front and rear
and the coefficient of friction between
the wheels and road surface.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 9/38
Sample Problem 16.1
m7s
ft300 == xv
SOLUTION:
• Calculate the acceleration during the skidding stop
by assuming uniform acceleration.
( )
( )m72s
m100
2
2
0
2
0
2
a
x xavv
+
=
−+=
s
m14.7−=a
• Draw a free-body-diagram equation expressing the
equivalence of the external and effective forces.
• Apply the corresponding scalar equations.
0=−+ W N N B A∑∑ =eff y y F F
( )
( )
728.0
81.9
14.7===
−=−
=+−
−=−−
g
a
agW W
N N
amF F
k
k
B Ak
B A
µ
µ
µ ( )∑∑ = eff x x F F
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 10/38
Sample Problem 16.1
• Apply the corresponding scalar equations.
( ) ( ) ( )
W N
g
aW a
g
W W N
am N W
B
B
B
659.0
2.15.112
2.15.112
1
m2.1m6.3m5.1
=
+=
+=
=+−
( )∑∑ =eff A A M M
W N W N B A
341.0=−=
( )W N N Arear 341.021
21 == W N rear 1705.0=
( )W N N V front 659.021
21 ==
W N front 32595.0=
( )( )W N F rear k rear 1705.0728.0== µ W F rear 124.0=
( )( )W N F front k front 3295.0728.0== µ
W F front 24.0=
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 11/38
Sample Problem 16.2
SOLUTION:
• Note that after the wire is cut, all
particles of the plate move along parallel
circular paths of radius 150 mm. Theplate is in curvilinear translation.
• Draw the free-body-diagram equation
expressing the equivalence of the
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.
external and effective forces.• Resolve into scalar component equations
parallel and perpendicular to the path of
the mass center.
• Solve the component equations and themoment equation for the unknown
acceleration and link forces.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 12/38
Sample Problem 16.2
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.
•
parallel and perpendicular to the path of the masscenter.
( )∑∑ =eff t t F F
=°
=°
30cos
30cos
mg
amW
( °= 30cosm/s81.9 2a
2sm50.8=a 60o
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 13/38
Sample Problem 16.2
• Solve the component equations and the moment
equation for the unknown acceleration and link
forces.
( )eff GG M M ∑∑ =
( )( ) ( )( )
( )( ) ( )( ) 0mm10030cosmm25030sin
mm10030cosmm25030sin
=°+°
°−°
DF DF
AE AE
F F
F F
2sm50.8=a 60o
AE DF
DF AE
F F 1815.0
..
−=
( )∑∑ =eff nn F F
( )( )2sm81.9kg8619.0
030sin1815.0
030sin
=
=°−−
=°−+
AE
AE AE
DF AE
F
W F F
W F F
T F AE N9.47=
( )N9.471815.0−= DF F C F DF N70.8=
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 14/38
Sample Problem 16.3
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.
A pulley weighing 6 kg and having a radius
of gyration of 200 mm is connected to two
blocks as shown.
Assuming no axle friction, determine the
angular acceleration of the pulley and the
acceleration of each block.
• Draw the free-body-diagram equationexpressing the equivalence of the
external and effective forces on the
complete pulley plus blocks system.
• Solve the corresponding momentequation for the pulley angular
acceleration.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 15/38
Sample Problem 16.3
2=
note:
SOLUTION:
• Determine the direction of rotation by evaluating the net
moment on the pulley due to the two blocks.
( ) ( )( ) 0mm502kg5.2mm150 =−=∑ gm M BG
rotation is counterclockwise.kg167.4= Bm
• Relate the acceleration of the blocks to the angularacceleration of the pulley.
( )α
α
m25.0=
= A A r a
( )α
α
m15.0=
= B B r a
( )( )2mkg24.0
kg2.0kg6
⋅=
=
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 16/38
Sample Problem 16.3
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
complete pulley and blocks system.
• Solve the corresponding moment equation for the pulley
angular acceleration.
( )∑∑ =eff GG M M
( )( )( ) ( )( )( )
m25.0.81m/s9kg5.2m15.0.81m/s9kg5 22−
( )
( ) 2
2
2
sm15.0
sm25.0
mkg24.0
α
α
=
=
⋅=
B
A
a
a
I
( )( ) ( )( )25.25.05.215.05.0524.01312.63575.7
..
−++=−
−=
α α α
A A B B
2srad41.2=α
( )( )2srad2.41.15m0=
= α B B r a2sm362.0= Ba
( )( )2srad2.41.25m0=
= α A Ar a
2sm603.0= Aa
Then,
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 17/38
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the external
and effective forces on the disk.
• Solve the three corresponding scalar
equilibrium equations for the horizontal,
A cord is wrapped around a
homogeneous disk of mass 15 kg.
The cord is pulled upwards with a
force T = 180 N.
Determine: (a) the acceleration of the
center of the disk, (b) the angular
acceleration of the disk, and (c) the
acceleration of the cord.
,
disk.
• Determine the acceleration of the cord by
evaluating the tangential acceleration of
the point A on the disk.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 18/38
Sample Problem 16.4
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of the external and effective forces on the
disk.
=
( )∑∑ =eff x x F F
xam=0 0= xa
• Solve the three scalar equilibrium equations.
eff y y
( )( )kg15
sm81.9kg15-N1802
=−
=
=−
m
W T a
amW T
y
y
2
sm19.2=
ya( )∑∑ =eff GG M M
( )
( )( )m5.0kg15
N18022
2
21
−=−=
==−
mr
T
mr I Tr
α
α α
2srad0.48=α
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 19/38
Sample Problem 16.4
• Determine the acceleration of the cord by evaluating the
tangential acceleration of the point A on the disk.
( )( )( )22 srad48m5.0sm19.2 +=
+== t G At Acord aaaa
2sm2.26=cord a
2sm19.2= ya
2srad0.48=α
0= xa
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 20/38
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation
expressing the equivalence of the
external and effective forces on thesphere.
• Solve the three corresponding scalar
e uilibrium e uations for the normal
A uniform sphere of mass m and radiusr is projected along a rough horizontal
surface with a linear velocity v0. The
coefficient of kinetic friction between
the sphere and the surface is µ k .Determine: (a) the time t 1 at which the
sphere will start rolling without sliding,
and (b) the linear and angular velocities
of the sphere at time t 1.
reaction from the surface and the linearand angular accelerations of the sphere.
• Apply the kinematic relations for
uniformly accelerated motion to
determine the time at which thetangential velocity of the sphere at the
surface is zero, i.e., when the sphere
stops sliding.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 21/38
Sample Problem 16.5
SOLUTION:
• Draw the free-body-diagram equation expressing the
equivalence of external and effective forces on the
sphere.
• Solve the three scalar equilibrium equations.
∑∑ =eff y y F F
0=−W N mgW N ==
eff x x
=−
=−
mg
amF
k µ ga k µ −=
( ) ( )α µ
α 2
32mr r mg
I Fr
k =
=
r
gk µ α
2
5=
( )∑∑ =eff GG M M
NOTE: As long as the sphere both rotates and slides,
its linear and angular motions are uniformly
accelerated.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 22/38
Sample Problem 16.5
• 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.
( )t gvt avv k µ −+=+= 00
t r
gt k
+=+=
µ α ω ω
2
500
ga k µ −=
r
gk µ α
2
5=
1102
5t
r
gr gt v k
k
=−
µ µ
g
vt
k µ
01
7
2=
=
=
g
v
r
gt
r
g
k
k k
µ
µ µ ω 0
117
2
2
5
2
5
r
v01
7
5=ω
==
r
vr r v 0
117
5ω 07
51 vv =
At the instant t 1
when the sphere stops sliding,
11 r v =
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 23/38
Constrained Plane Motion
• Most engineering applications involve rigid
bodies which are moving under given
constraints, e.g., cranks, connecting rods, and
non-slipping wheels.
• Constrained plane motion: motions with
definite relations between the components of
acceleration of the mass center and the angular
.
• Solution of a problem involving constrainedplane motion begins with a kinematic analysis.
• e.g., given θ, ω, and α , find P, N A, and N B.
- kinematic analysis yields
- application of d’Alembert’s principle yieldsP, N A, and N B.
.and y x aa
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 24/38
Constrained Motion: Noncentroidal Rotation
• Noncentroidal rotation: motion of a body is
constrained to rotate about a fixed axis that does
not pass through its mass center.
• Kinematic relation between the motion of the mass
center G and the motion of the body about G,
2ω α r ar a nt ==
• The kinematic relations are used to eliminate
from equations derived from
d’Alembert’s principle or from the method of
dynamic equilibrium.
nt aa and
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 25/38
Constrained Plane Motion: Rolling Motion
• For a balanced disk constrained to
roll without sliding,
α θ r ar x =→=
• Rolling, no sliding: N F s µ ≤ α r a =
Rolling, sliding impending:
N F s= α r a =
Rotating and sliding:
N F k = α r a , independent
• For the geometric center of an
unbalanced disk,
α r aO
=
The acceleration of the mass center,
( ) ( )nOGt OGO
OGOG
aaa
aaa
++=
+=
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 26/38
Sample Problem 16.6
kg3
mm85
kg4
=
=
=
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
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.
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.
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 27/38
Sample Problem 16.6
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
222
rad/s8=
2srad40=α
kg3mm85
kg4
=
=
=
OB
E
E
mk
m
mN156.1
sram.g
⋅=
== α α E E E m
( ) ( ) ( )( )
N0.24
srad40m200.0kg32
=
== α r mam OBt OBOB
( ) ( )( )( )
N4.38
srad8m200.0kg3 22
=
== ω r mam OBnOBOB
( )( )
mN600.1
srad40m.4000kg322
1212
121
⋅=
== α α Lm I OBOB
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 28/38
Sample Problem 16.6
• 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.
( )eff OO M M ∑∑ =
( ) ( ) ( )
( )( ) mN600.1m200.0N0.24mN156.1
m200.0m120.0
⋅++⋅=
++= α α OBt OBOB E I am I F
=
N4.29
N2.39
=
=
OB
E
W
W
mN156.1 ⋅=α E I
( ) N0.24=t OBOB am
( ) N4.38=nOBOB am
mN600.1 ⋅=α OB I
( )eff x x F F ∑∑ =
( ) N0.24==t OBOB x am R
N0.24= x R
eff y y F F ∑∑ =
( )
N4.38N4.29N2.39N0.63 =−−−
=−−−
y
OBOBOB E y
R
amW W F R
N0.24= y R
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 29/38
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the
sphere , expressing the equivalence of the
external and effective forces.• With the linear and angular accelerations
related, solve the three scalar equations
derived from the free-body-equation for
A sphere of weight W is released withno initial velocity and rolls without
slipping on the incline.
Determine: a) the minimum value of
the coefficient of friction, b) thevelocity of G after the sphere has
rolled 3 m and c) the velocity of G if
the sphere were to move 10 ft down a
frictionless incline.
the angular acceleration and the normaland tangential reactions at C .
• Calculate the velocity after 3 m of uniformly accelerated motion.
• Assuming no friction, calculate the linear
acceleration down the incline and the
corresponding velocity after 3 m.
• Calculate the friction coefficient required
for the indicated tangential reaction at C .
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 30/38
Sample Problem 16.8
SOLUTION:
• Draw the free-body-equation for the sphere , expressing
the equivalence of the external and effective forces.
• With the linear and angular accelerations related, solve
the three scalar equations derived from the free-body-equation for the angular acceleration and the normal
and tangential reactions at C.
= M M
α r a =
( ) ( )
( ) ( )
α α
α α
α θ
+
=
+=
+=
2
2
52
5
2
sin
r
g
W r r
g
W
mr r mr
I r amr W
r
g
7
sin5 θ α =
( )
7
30sinsm81.95
7
30sin5
2°
=
°==
gr a α
2sm504.3=a
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 31/38
Sample Problem 16.8
• Solve the three scalar equations derived from the free-
body-equation for the angular acceleration and the
normal and tangential reactions at C.
( )∑∑ =eff x x F F
W W F
ggW
amF W
143.030sin7
2
7sin5
sin
=°=
=
=−
θ
θ
r
g
7
sin5 θ α =
2
sm504.3== α r a
∑∑ = eff y y F F W W N
W N 866.030cos
0cos=°=
=− θ
• Calculate the friction coefficient required for the
indicated tangential reaction at C .
W
W
N
F
N F
s
s
866.0
143.0==
=
µ
µ
165.0=s µ
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 32/38
Sample Problem 16.8
• Calculate the velocity after 10 ft of uniformly
accelerated motion.
( )
( )( )m3sm504.320
2
2
0
2
0
2
+=
−+= x xavv
sm59.4=v
• Assuming no friction, calculate the linear acceleration
and the corres ondin velocit after 10 ft.
r
g
7
sin5 θ α =
2
sft50.11== α r a
( )∑∑ = eff GG M M 00 == α α I
( )∑∑ =eff x x F F
( ) 22sm905.430sinsm81.9
sin
=°=
==
a
ag
W amW θ
( )
( )( )m3sm905.420
2
2
0
2
0
2
+=
−+= x xavv
sft42.5=v
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 33/38
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the
wheel , expressing the equivalence of the
external and effective forces.
• Assuming rolling without slipping and
therefore, related linear and angular
accelerations, solve the scalar e uations
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 µ s = 0.20 and µ k = 0.15,
determine the acceleration of G and
the angular acceleration of the wheel.
for the acceleration and the normal andtangential reactions at the ground.
• Compare the required tangential reaction
to the maximum possible friction force.
• If slipping occurs, calculate the kineticfriction force and then solve the scalar
equations for the linear and angular
accelerations.
S l P bl 16 9
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 34/38
Sample Problem 16.9
SOLUTION:
• Draw the free-body-equation for the
wheel , expressing the equivalence of the
external and effective forces.
• Assuming rolling without slipping and
therefore, related linear and angular
accelerations, solve the scalar e uations
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 µ s = 0.20 and µ k = 0.15,
determine the acceleration of G and
the angular acceleration of the wheel.
for the acceleration and the normal andtangential reactions at the ground.
• Compare the required tangential reaction
to the maximum possible friction force.
• If slipping occurs, calculate the kineticfriction force and then solve the scalar
equations for the linear and angular
accelerations.
Sample Problem 16 9
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 35/38
Sample Problem 16.9
−= =
Without slipping,
• Compare the required tangential reaction to the
maximum possible friction force.
( ) N1.98N5.49020.0max === N F s µ
F > F max , rolling without slipping is impossible.
• Calculate the friction force with slipping and solve the
scalar equations for linear and angular accelerations.
. ... ==== k k µ
( )∑∑=
eff GG M M
( )( ) ( )( )
( )2
2
srad94.18
mkg245.0
m060.0.0N200m100.0N6.73
−=
⋅=
−
α
α
2srad94.18=α
( )∑∑ =eff x x F F
( )akg50N6.73N200 =−2sm53.2=a
Sample Problem 16 10
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 36/38
Sample Problem 16.10
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 the
The extremities of a 1.2-m rodweighing 25 kg 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.
externa an e ect ve orces.
• Solve the three corresponding scalar
equations for the angular acceleration and
the reactions at A and B.
Sample Problem 16 10
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 37/38
Sample Problem 16.10
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
A B A B aaa
+=
With the corresponding vector triangle and,4α = A Ba
The acceleration of G is now obtained from
AG AG aaaa
+== α 6.0 where = AGa
Resolving into x and y components,
α α
α α α
520.060sin6.0
339.160cos6.0639.1
−=°−=
=°−=
y
x
a
a
t e aw o s gns y e s
α α 47.1639.1 == B A aa
S l P bl 16 10
8/7/2019 Dynamics Lecture7 [Compatibility Mode] 2
http://slidepdf.com/reader/full/dynamics-lecture7-compatibility-mode-2 38/38
Sample Problem 16.10
• Draw the free-body-equation for the rod , expressing
the equivalence of the external and effective forces.
• Solve the three corresponding scalar equations for the
angular acceleration and the reactions at A and B.
( )( ) ( )( ) ( )( )2
3520.00.1334.15.3381.925
=
++= α α α
( )∑∑ =eff E E M M
2
srad33.2=α
( )∑∑ =eff x x F F
( )( )
N4.110
33.25.3345sin
=
=°
B
B
R
R
N4.110= B R
45o
∑∑ =eff y y F F
( ) ( )( )33.20.1324545cos4.110 −=−°+ A R
N6.136= A R
( )
( )
( ) α α
α α
α α
0.13520.025
5.33339.125
3
mkg3
m2.1kg12
25
2
22
121
−=−=
==
=
⋅=
==
y
x
am
am
I
ml I
top related