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PHYSICS CHAPTER 8
CHAPTER 8:CHAPTER 8:Rotational of rigid bodyRotational of rigid body
(8 Hours)(8 Hours)
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PHYSICS CHAPTER 8
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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
a) Define and describea) Define and describe::
angular displacement (angular displacement ()) average angular velocity (average angular velocity (avav)) instantaneous angular velocity (instantaneous angular velocity ()) average angular acceleration (average angular acceleration (
avav))
instantaneous angular acceleration (instantaneous angular acceleration ().).b) Relateb) Relate parameters in rotational motion with their correspondingparameters in rotational motion with their corresponding
quantities in linear motion.quantities in linear motion. Write and useWrite and use ::
c)c) UseUse equations for rotational motion with constant angularequations for rotational motion with constant angular
acceleration.acceleration.
Learning Outcome:
8.1 Rotational Kinematics (2 hour)
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s=r;v=r;at=r;a
c=r
2=
v2
r
=0t =0 t1
2 t2
2
=022
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8.1 Parameters in rotational motion
8.1.1 Angular displacement, is defined as an angle through which a point or line hasan angle through which a point or line has
been rotated in a specified direction about a specified axis.been rotated in a specified direction about a specified axis.
The S.I. unit of the angular displacement is radian (rad)radian (rad).
Figure 8.1 shows a point Pon a rotating compact disc (CD)moves through an arc lengths on a circular path of radius rabout a fixed axis through point O.
Figure 8.1Figure 8.1
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Average angular velocity,Average angular velocity, avav is defined as the rate of change of angular displacementthe rate of change of angular displacement.
Equation :
Instantaneous angular velocity,Instantaneous angular velocity, is defined as the instantaneous rate of change of angularthe instantaneous rate of change of angular
displacementdisplacement.
Equation :
8.1.2 Angular velocity
av=
2
1
t2t1=
t
= limitt 0
t=d
dt
where 2: final angular displacement in radian
t: time interval
1: initial angular displacement in radian
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It is a vector quantityvector quantity.
The unit of angular velocity is radian per second (rad sradian per second (rad s-1-1))
Others unit is revolution per minute (rev minrevolution per minute (rev min11 or rpm)or rpm) Conversion factor:
Note :
Every partEvery part of a rotating rigid body has the same angularsame angular
velocityvelocity.
Direction of the angular velocityDirection of the angular velocity
Its direction can be determine by using right hand grip ruleright hand grip rule
where
1 rpm =2
60
rad s1=
30
rad s1
ThumbThumb : direction ofangular velocityangular velocity
Curl fingersCurl fingers : direction ofrotationrotation
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Figures 8.2 and 8.3 show the right hand grip rule for determining
the direction of the angular velocity.
Figure 8.2Figure 8.2
Figure 8.3Figure 8.3
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The angular displacement, of the wheel is given by
where in radians and tin seconds. The diameter of the wheel is0.56 m. Determine
a. the angle, in degree, at time 2.2 s and 4.8 s,b. the distance that a particle on the rim moves during that time
interval,
c. the average angular velocity, in rad s1 and in rev min1 (rpm),
between 2.2 s and 4.8 s,d. the instantaneous angular velocity at time 3.0 s.
Example 8.1 :
=5t2t
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Solution :Solution :
a. At time, t1=2.2 s :
At time, t2=4.8 s :
r=d
2
=0.56
2
=0.28 m
1=5 2. 2
2 2. 2
1=22 rad
2=5 4 . 8
24 . 8
2=110 rad
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Solution :Solution :
b. By applying the equation of arc length,
Therefore
c. The average angular velocity in rad s1 is given by
r=d
2
=0.56
2
=0.28 m
s=rs=r=r21
s=0.28 11022
av=
t
=21
t2t1
av=
110224 .82.2
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Solution :Solution :
c. and the average angular velocity in rev min1 is
d. The instantaneous angular velocity as a function of time is
At time, t=3.0 s :
av=
33 . 9 rad
1 s 1 rev
2 rad 60 s
1 min
=d
dt
5t2t
=d
dt
=10 t1
=10 3 .0 1
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A diver makes 2.5 revolutions on the way down from a 10 m high
platform to the water. Assuming zero initial vertical velocity,
calculate the divers average angular (rotational) velocity during a
dive.
(Giveng= 9.81 m s2)
Solution :Solution :
Example 8.2 :
uy=0
0=0
10 m
water
1=2. 5 rev
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Solution :Solution :
From the diagram,
Thus
Therefore the divers average angular velocity is
1=2. 52=5 rad
sy
=10 m
sy=uy t1
2gt2
10=01
29.81 t2
av=10
t
av=501.43
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Average angular acceleration,Average angular acceleration, avav is defined as the rate of change of angular velocitythe rate of change of angular velocity.
Equation :
Instantaneous angular acceleration,Instantaneous angular acceleration, is defined as the instantaneous rate of change of angularthe instantaneous rate of change of angular
velocityvelocity.
Equation :
8.1.3 Angular acceleration
av=
2
1
t2t1=
t
= limitt0
t=d
dt
where 2 : final angular velocity
t: time interval
1: initial angular velocity
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PHYSICS CHAPTER 8
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It is a vector quantityvector quantity.
The unit of angular acceleration is rad srad s22. Note:
If the angular acceleration, is positivepositive, then the angularvelocity, is increasingincreasing.
If the angular acceleration, is negativenegative, then the angularvelocity, is decreasingdecreasing.
Direction of the angular accelerationDirection of the angular acceleration
If the rotation is speeding upspeeding up, and in the same directionsame directionas shown in Figure 8.4.
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Figure 8.5Figure 8.5
If the rotation is slowing downslowing down, and have the oppositeoppositedirectiondirection as shown in Figure 8.5.
Example 8.3 :
The instantaneous angular velocity, of the flywheel is given
by
where in radian per second and tin seconds.
Determine
a. the average angular acceleration between 2.2 s and 4.8 s,
b. the instantaneous angular acceleration at time, 3.0 s.
=8t3
t2
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Solution :Solution :
a. At time, t1=2.2 s :
At time, t2=4.8 s :
Therefore the average angular acceleration is
1=8 2 . 2 32 . 2 2
1=80 .3 rad s1
2=8 4 . 8
34 . 8
2
av=
2
1
t2t1
av=86280.3
4.82.2
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Solution :Solution :
b. The instantaneous angular acceleration as a function of time is
At time, t=3.0 s :
=d
dt8t3t2
= ddt
=24 3.0 22 3.0
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Exercise 8.1 :
1. If a disc 30 cm in diameter rolls 65 m along a straight line
without slipping, calculatea. the number of revolutions would it makes in the process,
b. the angular displacement would be through by a speck of
gum on its rim.
ANS. : 69 rev; 138ANS. : 69 rev; 138
radrad2. During a certain period of time, the angular displacement of a
swinging door is described by
where is in radians and tis in seconds. Determine the
angular displacement, angular speed and angular acceleration
a. at time, t=0,
b. at time, t=3.00 s.
ANS. :ANS. : 5.00 rad, 10.0 rad s5.00 rad, 10.0 rad s11, 4.00 rad s, 4.00 rad s22; 53.0 rad, 22.0 rad s; 53.0 rad, 22.0 rad s11,,
4.00 rad s4.00 rad s22
=5.0010.0t2.00 t2
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8.1.2 Relationship between linear and
rotational motion8.1.2 Relationship between linear velocity, v and
angular velocity, When a rigid body is rotates about rotation axis O , every
particle in the body moves in a circle as shown in the Figure 8.6.
v
s
y
x
r
P
O
Figure 8.6Figure 8.6
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Point P moves in a circle of radius rwith the tangential velocity
v where its magnitude is given by
The directiondirection of the linear (tangential) velocitylinear (tangential) velocity always
tangent to the circular pathtangent to the circular path. Every particle on the rigid body has the same angular speedsame angular speed
(magnitude of angular velocity) but the tangential speedtangential speed is notnot
the samesame because the radiusradius of the circle, rris changingchangingdependdepend on the position of the particleposition of the particle.
v=ds
dt
v=rd
dt
s=r
v=r
and
Simulation 7.1
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at
ac
a
x
y
P
O
If the rigid bodyrigid body is gaining the angular speedgaining the angular speed then the
tangential velocitytangential velocity of a particle also increasingincreasing thus twotwo
component of accelerationacceleration are occurredoccurred as shown in
Figure 8.7.
8.1.2 Relationship between tangential acceleration,
atand angular acceleration,
Figure 8.7Figure 8.7
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The components are tangential acceleration,tangential acceleration, aattand
centripetal acceleration,centripetal acceleration, aaccgiven by
but
The vector sum of centripetal and tangential accelerationvector sum of centripetal and tangential acceleration of
a particle in a rotating body is resultant (linear) acceleration,resultant (linear) acceleration, aagiven by
and its magnitude,
at=dv
dt
a t=rd
dt
at=r
v=rand
ac=v
2
r=r2=v
a=ata
c
a=at2ac2Vector formVector form
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8.1.3 Rotational motion with uniform
angular acceleration Table 8.1 shows the symbols used in linear and rotationalkinematics.
Table 8.1Table 8.1
Linearmotion
QuantityRotational
motion
s DisplacementDisplacement
u 0Initial velocityInitial velocity
v Final velocityFinal velocity
a AccelerationAcceleration
t tTimeTime
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Table 8.2 shows the comparison of linear and rotational motion
with constant acceleration.
Linear motion Rotational motion
a=constant
v=uat
=constant
=0
t
s=ut1
2at2
=0 t1
2t2
v
2
=u2
2as 2
=02
2
s=1
2vu t =
1
20 t
where in radian. Table 8.2Table 8.2
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A car is travelling with a velocity of 17.0 m s1 on a straight
horizontal highway. The wheels of the car has a radius of 48.0 cm.
If the car then speeds up with an acceleration of 2.00 m s2 for
5.00 s, calculate
a. the number of revolutions of the wheels during this period,
b. the angular speed of the wheels after 5.00 s.Solution :Solution :
a. The initial angular velocity is
and the angular acceleration of the wheels is given by
Example 8.4 :
u=17 . 0 m s1
, r=0 . 48 m, a=2 . 00 m s2
, t=5 . 00 s
u=r0
17.0=0.480
2.00=0.48a=r
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Solution :Solution :
a. By applying the equation of rotational motion with constant
angular acceleration, thus
therefore
b. The angular speed of the wheels after 5.00 s is
=0 t1
2t2
=
229 rad
u=17 . 0 m s1
, r=0 . 48 m, a=2 . 00 m s2
, t=5 . 00 s
=35.4 5.001
24.17 5.002
=0t=35 .44.17 5.00
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The wheels of a bicycle make 30 revolutions as the bicycle
reduces its speed uniformly from 50.0 km h-1 to 35.0 km h-1. Thewheels have a diameter of 70 cm.
a. Calculate the angular acceleration.
b. If the bicycle continues to decelerate at this rate, determine the
time taken for the bicycle to stop.Solution :Solution :
Example 8.5 :
=302=60 rad, r=0.702
=0 .35 m,
u=50.0 km
1 h 103m
1 km 1 h3600 s =13. 9 m s1,v=
35.0 km
1 h 103m
1 km 1 h3600 s =9.72 m s1
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Solution :Solution :
a. The initial angular speed of the wheels is
and the final angular speed of the wheels is
therefore
b. The car stops thusHence
u=r013.9=0.350
v=r
9.72=0.352=0
22
27.8 2=39.722 60
=0 0=27.8 rad s1
and=
0t
0=27.8 2.13 t
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A blade of a ceiling fan has a radius of 0.400 m is rotating about a
fixed axis with an initial angular velocity of 0.150 rev s-1. Theangular acceleration of the blade is 0.750 rev s -2. Determine
a. the angular velocity after 4.00 s,
b. the number of revolutions for the blade turns in this time interval,
c. the tangential speed of a point on the tip of the blade at time,
t=4.00 s,
d. the magnitude of the resultant acceleration of a point on the tip
of the blade at t=4.00 s.
Solution :Solution :
a. Given t=4.00 s, thus
Example 8.6 :
r=0 .400 m,0=0.1502=0.300 rad s
1,
=0.7502=1.50 rad s2
=0t
=19. 8 rad s1
=0.3001.50 4.00
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Solution :Solution :
b. The number of revolutions of the blade is
c. The tangential speed of a point is given by
=0 t12t2
=41.5 rad
=0.300 4.001
21.50 4.00
2
v=r
v=0.400 19.8
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Solution :Solution :
d. The magnitude of the resultant acceleration is
a=ac2at2
a=
v 2
r
2
r 2
a=7.922
0.400 2
0.4001.50 2
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Calculate the angular velocity of
a. the second-hand,
b. the minute-hand and
c. the hour-hand,
of a clock. State in rad s-1.
d. What is the angular acceleration in each case?Solution :Solution :
a. The period of second-hand of the clock is T= 60 s, hence
Example 8.7 :
=
2
T =
2
60
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Solution :Solution :
b. The period of minute-hand of the clock is T= 60 min = 3600 s,
hence
c. The period of hour-hand of the clock is T= 12 h = 4.32 104 s,
hence
d. The angular acceleration in each cases is
=2
3600
=2
4.3210
4
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A coin with a diameter of 2.40 cm is dropped on edge on a
horizontal surface. The coin starts out with an initial angular speedof 18 rad s1 and rolls in a straight line without slipping. If the
rotation slows down with an angular acceleration of magnitude
1.90 rad s2, calculate the distance travelled by the coin before
coming to rest.
Solution :Solution :
The radius of the coin is
Example 8.8 :
d=2.40102
m
0=18 rad s1
s
=1 . 90 rad s2
=0 rad s1
r=d
2=1.2010
2m
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Solution :Solution :
The initial speed of the point at the edge the coin is
and the final speed is
The linear acceleration of the point at the edge the coin is given by
Therefore the distance travelled by the coin is
u=r0u=1.20102 18
v=0 m s
1
a=r
a=1.20102 1.90
v2=u22as0=0.216
22 2.28102 s
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Exercise 8.2 :
1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev
min-1 about its central axis. Determinea. its angular speed,
b. the tangential speed at a point 3.00 cm from its centre,
c. the radial acceleration of a point on the rim,
d. the total distance a point on the rim moves in 2.00 s.
ANS. :ANS. : 126 rad s126 rad s11; 3.77 m s; 3.77 m s11; 1.26; 1.26 101033 m sm s22; 20.1 m; 20.1 m
2. A 0.35 m diameter grinding wheel rotates at 2500 rpm.
Calculate
a. its angular velocity in rad s1,b. the linear speed and the radial acceleration of a point on the
edge of the grinding wheel.
ANS. :ANS. : 262 rad s262 rad s11; 46 m s; 46 m s11, 1.2, 1.2 101044 m sm s22
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Exercise 8.2 :
3. A rotating wheel required 3.00 s to rotate through 37.0
revolution. Its angular speed at the end of the 3.00 s interval is98.0 rad s-1. Calculate the constant angular acceleration of the
wheel.
ANS. :ANS. : 13.6 rad s13.6 rad s22
4. A wheel rotates with a constant angular acceleration of3.50 rad s2.
a. If the angular speed of the wheel is 2.00 rad s1 at t=0,
through what angular displacement does the wheel rotate in
2.00 s.b. Through how many revolutions has the wheel turned during
this time interval?
c. What is the angular speed of the wheel at t = 2.00 s?
ANS. :ANS. : 11.0 rad; 1.75 rev; 9.00 rad s11.0 rad; 1.75 rev; 9.00 rad s11
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Exercise 8.2 :
5. A bicycle wheel is being tested at a repair shop. The angular
velocity of the wheel is 4.00 rad s-1 at time t= 0 , and its angularacceleration is constant and equal 1.20 rad s-2. A spoke OP on
the wheel coincides with the +x-axis at time t = 0 as shown in
Figure 8.8.
a. What is the wheels angular velocity at t = 3.00 s?
b. What angle in degree does the spoke OP make with the
positive x-axis at this time?
ANS. :ANS. : 0.40 rad s0.40 rad s11; 18; 18
Figure 8.8Figure 8.8
x
y
P
O
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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
Define and useDefine and use torque.torque.
State and useState and use conditions for equilibrium of rigid body:conditions for equilibrium of rigid body:
Learning Outcome:
8.2 Equilibrium of a uniform rigid body (2 hour)
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=== 0,0,0 FF yx
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8.2.1 Torque (moment of a force),
The magnitude of the torquemagnitude of the torque is defined as the product of athe product of a
force and its perpendicular distance from the line of actionforce and its perpendicular distance from the line of action
of the force to the point (rotation axis)of the force to the point (rotation axis).
OR
Because of
where r: distance between the pivot point (rotationaxis) and the point of application of force.
Thus
Fd=
forcetheofmagnitude:Farm)(momentdistancelarperpendicu:d
torquetheofmagnitude:where
sinrd =
sin Fr=
rF
andbetweenangle:
where
OR Fr
=
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It is a vector quantityvector quantity.
The dimension of torque is
The unit of torqueunit of torque is N mN m (newton metre), a vector productvector product
unlike thejoule (unit of work)joule (unit of work), also equal to a newton metre,
which is scalar productscalar product.
Torque is occurred because ofturning (twisting) effects ofturning (twisting) effects of
the forcesthe forces on a body.
Sign convention of torque:
PositivePositive - turning tendency of the force is anticlockwiseanticlockwise.
NegativeNegative - turning tendency of the force is clockwiseclockwise. The value oftorque dependstorque depends on the rotation axisrotation axis and the
magnitude of applied forcemagnitude of applied force.
[ ] [ ][ ] 22TMLdF ==
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Case 1 :Case 1 :
Consider a force is applied to a metre rule which is pivoted at
one end as shown in Figures 8.9a and 8.9b.
Figure 8.9aFigure 8.9a
F
F
Figure 8.9bFigure 8.9b
Pivot point
(rotation axis)
Fd=
rd sin=
FrFd sin==
(anticlockwise)
(anticlockwise)r
Point of action of a force
Line of action of a force
d
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O
Figure 8.10Figure 8.10
2
Case 2 :Case 2 :
Consider three forces are applied to the metre rule which is
pivoted at one end (point O) as shown in Figures 8.10.
Caution :
If the line of action of a force is through the rotation axisline of action of a force is through the rotation axis
then
1F
1
111 rd sin=
321 ++= OTherefore the resultant (nett)
torque is
3F
2F 1
r
0sin=== 333333
rFdF
222 rd sin=
111111 rFdF sin==222222 rFdF sin==
2r
2211 dFdF =O
Fr sin=
0=
and
0=
Simulation 5.1
PHYSICS CHAPTER 8
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Determine a resultant torque of all the forces about rotation axis, O
in the following problems.a.
Example 8.9 :
m5
N10=2F
m5 N30=1F
m3
m3
N20=3F
m10
m6O
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b.
Example 8.9 :
m5
N10=2F
m5
N30=
1F
m3
m3
N25=4F
N20=3F
m10
m6O
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m5m5
m10
m6O
Solution :Solution :
a.
Force Torque (N m), o=Fd=Frsin
1F
( ) ( ) 90330 =
2F ( ) ( ) 50510 +=+
N10=2F
N30=1F
N20=3
F
m3=1d
m5=2d
3F
0The resultant torque:
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m5
m10
m3
m6
m5
Solution :Solution :
b.
Force Torque (N m), o=Fd=Frsin
1F
( ) ( ) 90330 =
2F
( )( ) ( ) 51.50.515520sin ==rF33F
0 The resultant torque:
N10=2F
N30=1F
0.515
53
3sin
22=
+
=
O
N20=3F
N25=4F
m3=1d
m5=r
4F
0
3d
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8.2 Equilibrium of a rigid body
8.2.1.1 Non-concurrent forces is defined as the forces whose lines of action do not passthe forces whose lines of action do not pass
through a single common point.through a single common point.
The forces cause the rotational motionrotational motion on the body.
The combination of concurrent and non-concurrent forces cause
rolling motionrolling motion on the body. (translational and rotationaltranslational and rotationalmotion)
Figure 8.11 shows an example of non-concurrent forces.
2F
3F
1F
Figure 8.11Figure 8.11
4F
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8.2.1.2 Equilibrium of a rigid body
Rigid bodyRigid body is defined as a body with definite shape thata body with definite shape that
doesnt change, so that the particles that compose it stay indoesnt change, so that the particles that compose it stay in
fixed position relative to one another even though a force isfixed position relative to one another even though a force is
exerted on itexerted on it.
If the rigid body is in equilibriumrigid body is in equilibrium, means the body is
translational and rotational equilibriumtranslational and rotational equilibrium.
There are two conditionstwo conditions for the equilibrium of forces acting on
a rigid body.
The vector sum of all forces acting on a rigid body mustThe vector sum of all forces acting on a rigid body must
be zero.be zero.
==0nettFF
OR
=== 0,0,0 zyx FFF
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The vector sum of all external torques acting on a rigidThe vector sum of all external torques acting on a rigid
body must be zero about any rotation axisbody must be zero about any rotation axis.
This ensures rotational equilibriumrotational equilibrium.
This is equivalent to the three independent scalar
equations along the direction of the coordinate axes,
Centre of gravity, CGCentre of gravity, CG is defined as the point at which the whole weight of a bodythe point at which the whole weight of a body
may be considered to actmay be considered to act.
A force that exerts on the centre of gravityexerts on the centre of gravity of an object will
cause a translational motiontranslational motion.
== 0nett
=== 0,0,0 zyx
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Figures 8.14 and 8.15 show the centre of gravity for uniformcentre of gravity for uniform
(symmetric) objectobject i.e. rod and sphere
rodrod refer to the midway point between its endmidway point between its end.
spheresphere refer to geometric centregeometric centre.
2
l
2
l
CG
CGl
Figure 8.12Figure 8.12
Figure 8.13Figure 8.13
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8.2.4 Problem solving strategies for equilibrium of
a rigid body
The following procedure is recommended when dealing with
problems involving the equilibrium of a rigid body:
Sketch a simple diagramSketch a simple diagram of the system to help
conceptualize the problem.
Sketch a separate free body diagramSketch a separate free body diagram for each body. Choose a convenient coordinate axesChoose a convenient coordinate axes for each body and
construct a tableconstruct a table to resolve the forces into their
components and to determine the torque by each force.
Apply the condition for equilibrium of a rigid bodyApply the condition for equilibrium of a rigid body :
SolveSolve the equationsequations for the unknowns.
= 0xF = 0yF; and = 0
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A hanging flower basket having weight, W2 =23 N is hung out over
the edge of a balcony railing on a uniform horizontal beam AB of
length 110 cm that rests on the balcony railing. The basket is
counterbalanced by a body of weight, W1 as shown in Figure 8.14.
If the mass of the beam is 3.0 kg, calculate
a. the weight, W1 needed,
b. the force exerted on the beam at point O.
(Giveng=9.81 m s2)
Example 8.10 :
1W2W
A BO35 cm 75 cm
Figure 8.14Figure 8.14
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Solution :Solution :
The free body diagram of the beam :
Let point O as the rotation axis.
N23;kg3 == 2Wm
0.75 mA B
OCG
1W
2W
N
gm
0.35 m
0.55 m 0.55 m
0.20 m0.20 m
Force y-comp. (N) Torque (N m), o=Fd=Frsin
1W
1W
gm ( ) ( )9.813 ( ) ( ) 5.880.2029.4 =
( ) 11 WW 0.750.75 =
2W
23 ( ) ( ) 8.050.3523 =+
N
N 0
29.4=
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Solution :Solution :
Since the beam remains at rest thus the system in equilibrium.
a. Hence
b. 0= yFand
= 0O05.888.050.75 =+ 1W
029.423 =+ NW1( ) 029.4232.89 =+ N
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A uniform ladder AB of length 10 m and
mass 5.0 kg leans against a smooth wallas shown in Figure 8.15. The height of the
end A of the ladder is 8.0 m from the
rough floor.
a. Determine the horizontal and vertical
forces the floor exerts on the end B of
the ladder when a firefighter of mass
60 kg is 3.0 m from B.
b. If the ladder is just on the verge of
slipping when the firefighter is 7.0 mup the ladder , Calculate the coefficient
of static friction between ladder and
floor.
(Giveng=9.81 m s2)
Example 8.11 :
A
B
smooth
wall
rough floor
Figure 8.15Figure 8.15
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Solution :Solution :
a. The free body diagram of the ladder :
Let point B as the rotation axis.
kg60;kg5.0 == fl mm
A
B
CG
gmf
1N
gml
2N
m8.0m10
m3.0
m5.0
Forcex-comp.
(N)y-comp.
(N)
Torque (N m),
B=Fd=Frsin
gml
1N1N
0.810
8sin ==
sf
gmf
49.1 0.6
10
6sin ==
2N
sf
0
5890
2N
0
0
m6.0
( ) ( ) sin5.049.1
147=
0
( ) ( ) sin3.05891060=
( ) N1 sin101N8=
0
0 sf
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Solution :Solution :
Since the ladder in equilibrium thus
0= B081060147 =+ 1N
N151=1N
0= xF 0= s1 fNHorizontal force:Horizontal force:
0=yF
058949.1 =+ 2NVertical force:Vertical force:
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m10
A
B
m8.0
m6.0
m5.0
Solution :Solution :
b. The free body diagram of the ladder :
Let point B as the rotation axis.
0.6sin0.8;sin ==
gmf
gml
sf
m7.0
Forcex-comp.
(N)y-comp.
(N)
Torque (N m),
B=Fd=Frsin
gml
1N1N
2sN
gmf
49.1
2N
sf
0
5890
2N
0
0
( ) ( ) sin5.049.1
147=
0
( ) ( ) sin7.05892474=
( ) N1 sin101N8=
0
0
2N
1N
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Solution :Solution :
Consider the ladder stills in equilibrium thus
0= B082474147 =+ 1N
N328=1N
0=
x
F
0= 2s1 NN
0= yF 058949.1 =+ 2NN638=2N
( ) ( ) 0638328 = s
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Figure 8.16Figure 8.16
A floodlight of mass 20.0 kg in a park is
supported at the end of a 10.0 kg uniformhorizontal beam that is hinged to a pole as
shown in Figure 8.16. A cable at an angle
30 with the beam helps to support the light.
a. Sketch a free body diagram of the beam.
b. Determine
i. the tension in the cable,
ii. the force exerted on the beam by the
pole.
(Giveng=9.81 m s2
)
Example 8.12 :
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Solution :Solution :
a. The free body diagram of the beam :
b. Let point O as the rotation axis.
kg10.0;kg20.0 == bf mm
Force x-comp. (N) y-comp. (N) Torque (N m), o=Fd=Frsin
gmf
( )l1960 196
O CG
gmf
gmb
T
S
30
l
l0.5
gmb ( ) ( ) ll 49.10.598.1 =0 98.1
T
TlTl 0.530sin =30cosT 30sinT
S
xS yS 0
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Solution :Solution :
b. The floodlight and beam remain at rest thus
i.
ii.
0= O 00.549.1196 =+ Tlll
0=xF
0cos =+ xS30T
N424=xS0=
y
F
030sin98.1196 =++ yST
N49.1=yS
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Solution :Solution :
b. ii. Therefore the magnitude of the force is
and its direction is given by
2y
2x SSS +=
( ) ( ) 22S 49.1424 +=
=
x
y
S
S 1tan
=
42449.1tan 1
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Exercise 8.3 :
Use gravitational acceleration,g= 9.81 m s2
1.
Figure 8.17 shows the forces,F1 =10 N,F2= 50 N andF3=
60 N are applied to a rectangle with side lengths, a = 4.0 cm
and b = 5.0 cm. The angle is 30. Calculate the resultanttorque about point D.
ANS. : -3.7 N mANS. : -3.7 N m
D
AB
C
1F
3F
2F
Figure 8.17Figure 8.17
a
b
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Figure 8.18Figure 8.18
Exercise 8.3 :
2.
A see-saw consists of a uniform board of mass 10 kg and
length 3.50 m supports a father and daughter with masses 60
kg and 45 kg, respectively as shown in Figure 8.18. The fulcrum
is under the centre of gravity of the board. Determine
a. the magnitude of the force exerted by the fulcrum on the
board,
b. where the father should sit from the fulcrum to balance the
system.
ANS. : 1128 N; 1.31 mANS. : 1128 N; 1.31 m
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3.
A traffic light hangs from a structure as show in Figure 8.19.
The uniform aluminum pole AB is 7.5 m long has a mass of 8.0kg. The mass of the traffic light is 12.0 kg. Determine
a. the tension in the horizontal massless cable CD,
b. the vertical and horizontal components of the force exerted
by the pivot A on the aluminum pole.
ANS. : 248 N; 197 N, 248 NANS. : 248 N; 197 N, 248 N
Figure 8.19Figure 8.19
Exercise 8.3 :
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4.
A uniform 10.0 N picture frame is supported by two light string
as shown in Figure 8.20. The horizontal force,Fis applied forholding the frame in the position shown.
a. Sketch the free body diagram of the picture frame.b. Calculate
i. the tension in the ropes,
ii. the magnitude of the horizontal force,F.
ANS. : 1.42 N, 11.2 N; 7.20 NANS. : 1.42 N, 11.2 N; 7.20 N
Exercise 8.3 :
Figure 8.20Figure 8.20
F
50.0
cm15.0
cm30.0
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Learning Outcome:
8.3 Rotational dynamics (1 hour)
http//:kms.m
atrik.e
du.m
y/
http//:kms.m
atrik.e
du.m
y/
At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
DefineDefine the moment of inertia of a rigid body about an axis,the moment of inertia of a rigid body about an axis,
State and useState and use torque,torque,
=
=n
1i
iirmI2
I =
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8.3.1 Centre of mass, moment of inertia
and torque8.3.1.1 Centre of mass (CM) is defined as the point at which the whole mass of a bodythe point at which the whole mass of a body
may be considered to be concentratedmay be considered to be concentrated.
Its coordinate ((xxCMCM
,,yyCMCM
)) is given the expression below:
xCM=i=1n
mixi
i=1
n
mi ; y CM=i=1
n
m i y i
i=1
n
mi
where mi : mass of the ith
particlexi :x coordinate of the i
thparticle
y i : y coordinate of the ith
particle
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Two masses, 3 kg and 5 kg are located on the y-axis at y=1 m and
y=5 m respectively. Determine the centre of mass of this system.Solution :Solution :
Example 8.13 :
0
1 m
=
5 mm
1=3 kg; m
2=5 kg
m1
m2
yCM=
i=1
2
miyi
i=1
2
mi
=m
1y
1m
2y
2
m1m 2
yCM=
3 1 5 5
35
CM3. 5 m
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A system consists of three spheres have the following masses and
coordinates :(1) 1 kg, (3,2) ; (2) 2 kg, (4,5) and (3) 3 kg, (3,0).
Determine the coordinate of the centre of mass of the system.
Solution :Solution :
Thex coordinate of the CM is
Example 8.14 :
m1=1 kg; m
2=2 kg; m
3=3 kg
xCM=
i=1
3
mi x i
i=1
3
mi
=m
1x
1m
2x
2m
3x
3
m1m
2m
3
xCM=
1 3 2 4 3 3
123
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Solution :Solution :
They coordinate of the CM is
Therefore the coordinate of the CM is
yCM=
i=1
3
miyi
i=1
3
yi
=m
1y
1m
2y
2m
3y
3
m1m
2m
3
yCM=
1 2 2 5 3 0
123
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Figure 8.21 shows a rigid body about a fixed axis O with angular
velocity .
is defined as the sum of the products of the mass of eachthe sum of the products of the mass of each
particle and the square of its respective distance from theparticle and the square of its respective distance from the
rotation axisrotation axis.
8.3.1.2 Moment of inertia,I
m1
m2
mn
m3
r1
r2r3
rn O
Figure 8.21Figure 8.21
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OR
It is a scalar quantityscalar quantity. Moment of inertia,Moment of inertia,II in the rotational kinematics is analogousanalogous
to the mass,mass, mm in linear kinematics. The dimensiondimension of the moment of inertia is M LM L22. The S.I. unitS.I. unit of moment of inertia is kg mkg m22.
The factorsfactors which affect the moment of inertia,Iof a rigid body:a. the massmass of the body,
b. the shapeshape of the body,
c. the positionposition of the rotation axisrotation axis.
I=m1r
12m
2r
22m
3r
32. ..m
nrn2=
i=1
n
miri2
I : moment of inertia of a rigid body about rotation axism : mass of particler : distance from the particle to the rotation axis
where
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Moments of inertia of various bodiesMoments of inertia of various bodies
Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
ICM=MR 2
ICM=1
2MR
2
Hoop or ring or
thin cylindrical
shell
Solid cylinder or
diskCM
CM
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CM
Moments of inertia of various bodiesMoments of inertia of various bodies
Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
ICM=1
12 ML2
Uniform rod or
long thin rod with
rotation axis
through the
centre of mass.
CM
ICM=2
5MR
2Solid Sphere
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Moments of inertia of various bodiesMoments of inertia of various bodies
Table 8.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
ICM=23MR2Hollow Sphere orthin spherical
shell
CM
Table 8.3Table 8.3
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Four spheres are arranged in a rectangular shape of sides 250 cm
and 120 cm as shown in Figure 8.22.
The spheres are connected by light rods . Determine the momentof inertia of the system about an axis
a. through point O,
b. along the line AB.
Example 8.15 :
250 cm
60 cm
60 cm
2 kg 3 kg
4 kg5 kg
OA B
Figure 8.22Figure 8.22
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Solution :Solution :
a. rotation axis about point O,
Since r1= r
2= r
3= r
4= r thus
and the connecting rods are light therefore
m1=2 kg; m
2=3 kg; m
3=4 kg; m
4=5 kg
r1
0. 6 m
m1
m2
m3m4
O
r2
r4
r3
1.25 m
r=0 . 6 21.25 2=1 .39 mIO=m1r12m2r22m3r32m4r42
IO=r
2 m1m2m3m4 =1.39 2 2345
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Solution :Solution :
b. rotation axis along the line AB,
r1= r
2= r
3= r
4= r=0.6 m therefore
m1=2 kg; m
2=3 kg; m
3=4 kg; m
4=5 kg
IAB=m1r12m2r2
2m3r3
2m4 r4
2
IAB=0 . 6
22345
m1
m2
m3m4
A B
r1
r2
r4 r3
IAB=r2
m1m2m3m4
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Relationship between torque,Relationship between torque, and angular acceleration,and angular acceleration, Consider a force,Facts on a rigid body freely pivoted on an
axis through point O as shown in Figure 8.23.
The body rotates in the anticlockwise direction and a nett torque
is produced.
8.3.2 Torque,
m1
m2
mn r1
r2
rn
O
a1
an
a2
F
Figure 8.23Figure 8.23
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A particle of mass, m1
of distance r1
from the rotation axis O will
experience a nett forceF1
. The nett force on this particle is
The torque on the mass m1 is
The total (nett) torque on the rigid body is given by
F1=m1a1F
1=m
1r
1
a1=r1 and
1=r
1F
1sin90
1=m1r12
=i=1n
mir
i
2
=m1r1
2m2 r2
2.. .mnrn
2
=I
i=1
n
m i ri2=Iand
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From the equation, the nett torquenett torque acting on the rigid body is
proportionalproportional to the bodys angular accelerationangular acceleration.
Note :
Nett torque , =I
Nett force,F=mais analogous to the
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Forces,F1 = 5.60 N andF2= 10.3 N are applied tangentially to a
disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure8.24.
Calculate,
a. the nett torque on the disc.
b. the magnitude of angular acceleration influence by the disc.
( Use the moment of inertia, )
Example 8.16 :
Figure 8.24Figure 8.24
ICM=1
2MR
2
F1
O
30.0 cm
F2
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Solution :Solution :
a. The nett torque on the disc is
b. By applying the relationship between torque and angular
acceleration,
R=0 . 30 m ; M=5 . 00 kg
=12 =RF1RF2=R F1F2 = 0.30 5.6010.3
=
1
2
MR2
=I
1.41=12 5.00 0.30 2
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A wheel of radius 0.20 m is mounted on a frictionless horizontal
axis. The moment of inertia of the wheel about the axis is0.050 kg m2. A light string wrapped around the wheel is attached
to a 2.0 kg block that slides on a horizontal frictionless surface. A
horizontal force of magnitudeP= 3.0 N is applied to the block as
shown in Figure 8.25. Assume the string does not slip on thewheel.
a. Sketch a free body diagram of the wheel and the block.
b. Calculate the magnitude of the angular acceleration of the
wheel.
Example 8.17 :
Figure 8.25Figure 8.25
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Solution :Solution :
a. Free body diagram :
for wheel,
for block,
R=0 .20 m ; I=0.050 kg m2; P=3.0 N; m=2.0 kg
W
TS
NT
Wb
P
a
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Solution :Solution :
b. For wheel,
For block,
By substituting eq. (1) into eq. (2), thus
=I
RT=I T=IR
(1)
F=ma PT=ma (2)
R=0 .20 m ; I=0.050 kg m2; P=3.0 N; m=2.0 kg
a=RPIR =ma andP
I
R =mR
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An object of mass 1.50 kg is suspended
from a rough pulley of radius 20.0 cm by lightstring as shown in Figure 8.26. The pulley
has a moment of inertia 0.020 kg m2 about
the axis of the pulley. The object is released
from rest and the pulley rotates without
encountering frictional force. Assume that
the string does not slip on the pulley. After
0.3 s, determine
a. the linear acceleration of the object,
b. the angular acceleration of the pulley,
c. the tension in the string,d. the liner velocity of the object,
e. the distance travelled by the object.
(Giveng= 9.81 m s-2)
Example 8.18 :
Figure 8.26Figure 8.26
R
1. 50 kg
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Solution :Solution :
a. Free body diagram :
for pulley,
for block,
W
a
T
S =IRT=I = aR
and
RT=I
a
R
T=IaR2 (1)T
m g
F=
mamgT=ma (2)
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Solution :Solution :
a. By substituting eq. (1) into eq. (2), thus
b. By using the relationship between a and , hence
mgIaR2 =ma
R=0 .20 m ; I=0. 020 kg m2; m=1 . 50 kg;
u=0; t=0. 3 s
a=R
1.50 9.81 0.020 a
0.202 =1.50 a
7.36=0.20
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Solution :Solution :
c. From eq. (1), thus
d. By applying the equation of liner motion, thus
e. The distance travelled by the object in 0.3 s is
R=0 .20 m ; I=0. 020 kg m2; m=1 . 50 kg;
u=0; t=0. 3 s
v=uat
v=0 7.36 0 . 3
T=IaR2
T= 0.020 7.36 0.20
2
s=ut12at2
s=01
27.36 0 .3 2
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Exercise 8.4 :
Use gravitational acceleration,g= 9.81 m s2
1. Three odd-shaped blocks of chocolate have following massesand centre of mass coordinates: 0.300 kg, (0.200 m,0.300 m);
0.400 kg, (0.100 m. -0.400 m); 0.200 kg, (-0.300 m, 0.600 m).
Determine the coordinates of the centre of mass of the system
of three chocolate blocks.
ANS. :ANS. : (0.044 m, 0.056 m)(0.044 m, 0.056 m)
2. Figure 8.27 shows four masses that are held at
the corners of a square by a very light
frame. Calculate the moment of inertia
of the system about an axis perpendicularto the plane
a. through point A, and
b. through point B.
ANS. :ANS. : 0.141 kg m0.141 kg m22; 0.211 kg m; 0.211 kg m22
80 cm
80 cm
150 g 150 g
70 g
70 g
40 cm
A
B
Figure 8.27Figure 8.27
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2 . 00 m s2
T2
T1
Exercise 8.4 :
3. A 5.00 kg object placed on a
frictionless horizontal table isconnected to a string that passesover a pulley and then is fastenedto a hanging 9.00 kg object as inFigure 8.28. The pulley has aradius of 0.250 m and moment of
inertiaI. The block on the table ismoving with a constantacceleration of 2.00 m s2.
a. Sketch free body diagrams of
both objects and pulley.
b. Calculate T1
and T2
the tensions
in the string.
c. DetermineI.
ANS. : 10.0 N, 70.3 N; 1.88 kg mANS. : 10.0 N, 70.3 N; 1.88 kg m22
Figure 8.28Figure 8.28
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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
Solve problem related to :Solve problem related to :
kinetic energy,kinetic energy,
work,work,
power,power,
Learning Outcome:8.4 Work and Energy of Rotational Motion (2
hours)
Kr=
1
2 I
2
P=
W=
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8.4 Rotational kinetic energy and power
8.4.1 Rotational kinetic energy,Kr Consider a rigid body rotating about the axis OZ as shown in
Figure 8.29.
Every particle in the body is in the circular motion about point O.
m1
m2
mn
m3
r1r
2
r3
rn
O
v1
v2
v3
vn
Z
Figure 8.29Figure 8.29
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The rigid body has a rotational kinetic energy which is the totaltotal
of kinetic energy of all the particles in the bodyof kinetic energy of all the particles in the body is given by
Kr=12m1 v1212m2 v2
212 m3v 32.. .12 mn vn
2
Kr=1
2m1r1
2
21
2m2r2
2
21
2m3 r3
2
2. ..1
2mn rn
2
2
Kr=122 m1r12m2r22m3r32. . .mnrn2
Kr=1
2I2
Kr=
1
22
i=1
n
miri2 i=1
n
mi ri2=Iand
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From the formula for translational kinetic energy,Ktr
After comparing both equations thus
Forrolling body without slippingrolling body without slipping, the total kinetic energy oftotal kinetic energy of
the body,the body,KKis given by
Ktr=1
2 mv2
is analogous to vvIIis analogous to mm
K=KtrKr
Ktr
: translational kinetic energy
Kr
: rotational kinetic energy
where
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A solid sphere of radius 15.0 cm and mass 10.0 kg rolls down an
inclined plane make an angle 25
to the horizontal. If the sphererolls without slipping from rest to the distance of 75.0 cm and the
inclined surface is smooth, calculate
a. the total kinetic energy of the sphere,
b. the linear speed of the sphere,
c. the angular speed about the centre of mass.(Given the moment of inertia of solid sphere is and
the gravitational acceleration,g= 9.81 m s2)
Example 8.19 :
ICM=2
5mR
2
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Solution :Solution :
a. From the principle of conservation of energy,
R=0 .15 m ; m=10 .0 kg
Ei=Efmgh=KK=mgs sin25
K=10.0 9.81 0.75 sin25
s=0 .75 m
h=ssin25
v CM 25
R
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k
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Solution :Solution :
b. The linear speed of the sphere is given by
c. By using the relationship between v and , thus
R=0 .15 m ; m=10 .0 kg
K=KtrKr K=1
2mv2
1
2 I2 =
v
Rand
K=1
2mv 2
1
2 25 mR2 vR 2
K= 710mv2
31.1=7
1010.0 v2
v=R 2.11=0.15
PHYSICS CHAPTER 8
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The pulley in the Figure 8.30 has a
radius of 0.120 m and a moment ofinertia 0.055 g cm2. The rope does not
slip on the pulley rim.
Calculate the speed of the 5.00 kg
block just before it strikes the floor.
(Giveng= 9.81 m s2)
Example 8.20 :
2.00 kg
5.00 kg
7.00 m
Figure 8.30Figure 8.30
PHYSICS CHAPTER 8
5 00 k 2 00 k R 0 120 h 7 00
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Solution :Solution :
The moment of inertia of the pulley,
m1=5 . 00 kg ;m
2=2. 00 kg; R=0 .120 m; h=7 . 00 m
I= 0 .055 g 1 cm210
3
kg1 g
104
m2
1 cm2 =5 .5109 kg m2
m2
m1
7.00 m
Initial
m2
m1
7.00 m
v
v
Final
Ei=U1 Ef=Ktr1Ktr2KrU2
PHYSICS CHAPTER 8
S 5 00 k 2 00 k R 0 120
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Solution :Solution :
By using the principle of conservation of energy, thus
Ei=EfU
1=K
tr1K
tr2K
rU
2
m1gh=1
2 m1 v2
1
2m2 v2
1
2 I2
m2gh
m1m2 gh=1
2v2 m1m2
1
2I vR
2
5.002.00 9.81 7.00=12v 2 5.002.00 12
5.5109 v
0.120 2
m1=5.00 kg ;m
2=2. 00 kg; R=0 .120 m;
h=7.00 m; I=5.5109
kg m2
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Consider a tangential force,Facts on the solid disc of radius R
freely pivoted on an axis through O as shown in Figure 8.31.
The work done by the tangential force is given by
8.4.2 Work, W
Figure 8.31Figure 8.31
F
ds
O
dR
R
dW=Fds
dW=FRd
dW=1
2d W=
1
2d
ds=Rdand
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If the torque is constant thus
Work-rotational kinetic energy theorem states
W=21
W=1
2d
W=
: torque : change in angular displacement
where
W : work done
W=Kr= KrfKri
W=1
2I2
1
2I0
2
is analogous to the W=Fs
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From the definition of instantaneous power,
Caution :
The unitunit ofkinetic energy, work and powerkinetic energy, work and powerin the
rotationalrotational kinematics is samesame as theirunitunit in translationaltranslationalkinematics.
8.4.3 Power,P
P=dWdt
dW=dand
P=d
dtP=
d
dt
=and
is analogous to the P=Fv
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A horizontal merry-go-round has a radius of 2.40 m and a
moment of inertia 2100 kg m2
about a vertical axle through itscentre. A tangential force of magnitude 18.0 N is applied to the
edge of the merry-go- round for 15.0 s. If the merry-go-round is
initially at rest and ignore the frictional torque, determine
a. the rotational kinetic energy of the merry-go-round,
b. the work done by the force on the merry-go-round,c. the average power supplied by the force.
(Giveng= 9.81 m s2)
Solution :Solution :
Example 8.21 :
FR=2 .40 m
PHYSICS CHAPTER 8
S l tiS l ti R 2 40 I 2100 k2
F 18 0 N
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Solution :Solution :
a. By applying the relationship between nett torque and angularacceleration, thus
Use the equation of rotational motion with uniform angular
acceleration,
Therefore the rotational kinetic energy for 15.0 s is
=IRF=I 2.40 18.0 =2100
=0t
=0 2.06102 15.0 =0.309 rad s
1
Kr=1
2I2
Kr=1
22100 0.309
2
R=2 .40 m ; I=2100 kg m2; F=18 .0 N;
t=15.0 s; 0=0
PHYSICS CHAPTER 8
S l tiSolution : R 2 40 I 2100 k2
F 18 0 N
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Solution :Solution :
b. The angular displacement, for 15.0 s is given by
By applying the formulae of work done in rotational motion, thus
c. The average power supplied by the force is
W=
=0 t1
2t2
Pav=Wt
W=2.40 18.0 2.32
R=2 .40 m ; I=2100 kg m ; F=18 .0 N;t=15.0 s;
0=0
=01
22.06102 15.0 2
W=RF
Pav=10015.0
PHYSICS CHAPTER 8
Learning Outcome:
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At the end of this chapter, students should be able to:At the end of this chapter, students should be able to:
Define and useDefine and use angular momentum,angular momentum,
State and useState and use the principle of conservation of angularthe principle of conservation of angular
momentummomentum
Learning Outcome:
8.5 Conservation of angular momentum (1 hour)
L=I
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8.5 Conservation of angular momentum
8.5.1 Angular momentum, is defined as the product of the angular velocity of a bodythe product of the angular velocity of a body
and its moment of inertia about the rotation axisand its moment of inertia about the rotation axis.
OR
It is a vectorIt is a vectorquantity.
Its dimension is M LM L22 TT11
The S.I. unit of the angular momentum is kg mkg m22 ss11.
L
where
L=I
L : angular momentumI : moment of inertia of a body : angular velocity of a body
is analogous to the p=mv
PHYSICS CHAPTER 8 The relationship between angular momentum L with linear
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The relationship between angular momentum,L with linearmomentum,p is given by
vector notation :
magnitude form :
Newtons second law of motion in term of linear momentum is
hence we can write the Newtons second law in angular form as
and states that a vector sum of all the torques acting on aa vector sum of all the torques acting on arigid body is proportional to the rate of change of angularrigid body is proportional to the rate of change of angularmomentummomentum.
L=
r
p=
r
mv
L=rpsin=mvrsinwhere
:the angle between {rwith {v r : distance from the particle to the rotation axis
F=Fnett=dpdt
=nett=dLdt
PHYSICS CHAPTER 88.5.2 Principle of conservation of angular
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states that a total angular momentum of a system about ana total angular momentum of a system about an
rotation axis is constant if no external torque acts on therotation axis is constant if no external torque acts on thesystemsystem.
OR
Therefore
8.5.2 Principle of conservation of angular
momentum
I=constant
=dL
dt=0
dL=0
If the=0
dL=LfLiLi=Lf
and
PHYSICS CHAPTER 8
Example 8 22 :
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A 200 kg wooden disc of radius 3.00 m is rotating with angular
speed 4.0 rad s-1
about the rotation axis as shown in Figure 8.32. A 50 kg bag of sand falls onto the disc at the edge of the
wooden disc.
Calculate,
a. the angular speed of the system after the bag of sand falling
onto the disc. (treat the bag of sand as a particle)
b. the initial and final rotational kinetic energy of the system.
Why the rotational kinetic energy is not the same?
(Use the moment of inertia of disc is )
Example 8.22 :
0
Before
R
After
R
Figure 8.32Figure 8.32
1
2
MR2
PHYSICS CHAPTER 8
Solution :Solution : R=3 00 m ; =4 0 rad s1 ; m =200 kg; m =50 kg
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Solution :Solution :
a. The moment of inertia of the disc,
The moment of inertia of the bag of sand,
By applying the principle of conservation of angular momentum,
R=3 .00 m ;0=4 .0 rad s ; mw=200 kg; mb=50 kg
Iw=1
2 mwR2
=1
2 200 3.00 2
Iw
0=
I
wI
b
Ib
=mb
R2= 50 3.002
Li=Lf
Iw=900 kg m2
Ib=450 kg m2
900 4.0 =900450
PHYSICS CHAPTER 8
Solution :Solution : R=3 00 m ; =4 0 rad s1 ; m =200 kg; m =50 kg
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Solution :Solution :
b. The initial rotational kinetic energy,
The final rotational kinetic energy,
thus
It is because the energy is lost in the form of heat from thethe energy is lost in the form of heat from the
friction between the surface of the disc with the bag offriction between the surface of the disc with the bag ofsand.sand.
R=3 .00 m ;0=4 .0 rad s ; mw=200 kg; mb=50 kg
Kri=1
2 Iw02
=1
2 900 4 .0 2
Krf=1
2 IwIb 2
=
1
2 900450 2.67
2
Kri=7200 J
KriKrf
PHYSICS CHAPTER 8
Example 8 23 :
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A raw egg and a hard-boiled egg are rotating about the same
axis of rotation with the same initial angular velocity. Explainwhich egg will rotate longer.
Solution :Solution :
The answer is hard-boiled egghard-boiled egg.
Example 8.23 :
PHYSICS CHAPTER 8
Solution :Solution :
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Solution :Solution :
ReasonReason
Raw egg :When the egg spins, its yolk being denser moves away from the
axis of rotation and then the moment of inertia of the egg increases
because of
From the principle of conservation of angular momentum,
If theIis increases hence its angular velocity, will decreases.
Hard-boiled egg :
The position of the yolk of a hard-boiled egg is fixed. When the egg
is rotated, its moment of inertia does not increase and then its
angular velocity is constant. Therefore the egg continues to spin.
I=mr2
I=constant
PHYSICS CHAPTER 8
Example 8 24 :
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A student on a stool rotates freely with an angular speed of 2.95 rev
s
1
. The student holds a 1.25 kg mass in each outstretched arm thatis 0.759 m from the rotation axis. The moment of inertia for the
system of student-stool without the masses is 5.43 kg m2. When the
student pulls his arms inward, the angular speed increases to 3.54
rev s1.
a. Determine the new distance of each mass from the rotation axis.
b. Calculate the initial and the final rotational kinetic energy of the
system.
Solution :Solution :
Example 8.24 :
0=
2.95 rev
1 s 2 rad
1 rev =18.5 rad s1
=3 .54 rev
1 s 2 rad
1 rev =22. 2 rad s1
PHYSICS CHAPTER 8
Solution :Solution : m=1 25 kg ; 0=18 5 rad s1
; I =5 43 kg m2;
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123Before
0
After
ra
ra
Solution :Solution :
rb
rb
mm
m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ;=22.2 rad s
1;
PHYSICS CHAPTER 8
Solution :Solution : m=1. 25 kg ; 0=18.5 rad s1
; I =5. 43 kg m2;
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Solution :Solution :
a. The moment of inertia of the system initially is
The moment of inertia of the system finally is
By using the principle of conservation of angular momentum,
thus
Ii=I
ssI
mIi=Issmrb2mrb2 =I
ss2mr
b2
Ii=5.43 2 1.25 0.759
2=6 . 87 kg m2
If=Iss2mra2
= 5.43 2 1.25 ra
2
Ii
0=I
f
Li=Lf
6.87 18.5 =5.432.5ra222.2
m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ; =22.2 rad s
1;
If=5.432.5r
a2
PHYSICS CHAPTER 8
Solution :Solution : m=1. 25 kg ; 0=18.5 rad s1
; I ss=5. 43 kg m2;
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Solution :Solution :
b. The initial rotational kinetic energy is given by
and the final rotational kinetic energy is
Kri=1
2Ii
02
=1
26.87 18.5
2
Kri=1.18103 J
K
rf=
1
2If
2
=1
25.432.5 0.344 222.22
m 1. 25 kg ; 0 18.5 rad s ; Iss 5. 43 kg m ;rb=0.759 m ; =22.2 rad s
1;
PHYSICS CHAPTER 8
Exercise 8.5 :
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Use gravitational acceleration,g= 9.81 m s2
1. A woman of mass 60 kg stands at the rim of a horizontalturntable having a moment of inertia of 500 kg m2 and a radius
of 2.00 m. The turntable is initially at rest and is free to rotate
about the frictionless vertical axle through its centre. The
woman then starts walking around the rim clockwise (as viewed
from above the system) at a constant speed of 1.50 m s1
relative to the Earth.
a. In the what direction and with what value of angular speed
does the turntable rotate?
b. How much work does the woman do to set herself and the
turntable into motion?
ANS. :ANS. : 0.360 rad s0.360 rad s11 ,U think; 99.9 J,U think; 99.9 J
PHYSICS CHAPTER 8
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2. Determine the angular momentum of the Earth
a. about its rotation axis (assume the Earth is a uniform solidsphere), and
b. about its orbit around the Sun (treat the Earth as a particle
orbiting the Sun).
Given the Earths mass = 6.0 x 1024 kg, radius = 6.4 x 106 m
and is 1.5 x 108 km from the Sun.
ANS. :ANS. : 7.1 x 107.1 x 103333 kg mkg m22 ss11; 2.7 x 10; 2.7 x 104040 kg mkg m22 ss11
3. Calculate the magnitude of the angular momentum of the
second hand on a clock about an axis through the centre of the
clock face. The clock hand has a length of 15.0 cm and a mass
of 6.00 g. Take the second hand to be a thin rod rotating withangular velocity about one end. (Given the moment of inertia of
thin rod about the axis through the CM is )
ANS. :ANS. : 4.71 x 104.71 x 1066 kg mkg m22 ss11
1
12ML
2
PHYSICS CHAPTER 8Summary:
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Linear Motion Relationship Rotational Motion
m
v=r =ddt
a=dv
dt=
d
dt
F=ma =I
W=Fs W=
v=dsdt
a=r
=rFsin
P=Fv P=
p=mv L=IL=rpsin
II=i=1
n
miri2
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THE END
Next ChapterCHAPTER 9 :
Simple Harmonic Motion (SHM)
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