DF015 CHAPTER 8
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CHAPTER 8: Rotation of a rigid body
(8 Hours)
DF025 CHAPTER 8
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At the end of this chapter, students should be able to: a) Define and use:
angular displacement () average angular velocity (av)
instantaneous angular velocity () average angular acceleration (av)
instantaneous angular acceleration ().
Learning Outcome:
8.1 Rotational kinematics (2 hours)
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b) Relate parameters in rotational motion with their corresponding quantities in linear motion. Write and use;
c) Use equations for rotational motion with constant angular acceleration;
, , .
r
vrararvrθs ct
22 ; ; ;
αθωω 20
2 2αtωω 0 20 ttωθ
2
1
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8.1 Rotational kinematics a)(i) Angular displacement, is defined as an angle through which a point or line has
been rotated in a specified direction about a specified axis. The S.I. unit of the angular displacement is radian (rad). Figure 7.1 shows a point P on a rotating compact disc (CD)
moves through an arc length s on a circular path of radius r about a fixed axis through point O.
Figure 7.1
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From Figure 7.1, thus
Others unit for angular displacement is degree () and revolution (rev). Conversion factor :
Sign convention of angular displacement : Positive – if the rotational motion is anticlockwise. Negative – if the rotational motion is clockwise.
360rad 2πrev 1
r
sθ OR rθs
where radianin nt)displaceme(angular angle :θlength arc :s
circle theof radius :r
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Average angular velocity, av
is defined as the rate of change of angular displacement. Equation :
Instantaneous angular velocity, is defined as the instantaneous rate of change of angular
displacement. Equation :
a)(ii)(iii) Angular velocity
t
θ
tt
θθω
12
12
av
dt
dθ
t
θ0t
limit
where radianin nt displacemeangular final :2θ
interval time :tradianin nt displacemeangular initial :1θ
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It is a vector quantity. The unit of angular velocity is radian per second (rad s-1) Others unit is revolution per minute (rev min1 or rpm)
Conversion factor:
Note : Every part of a rotating rigid body has the same angular
velocity.
Direction of the angular velocity Its direction can be determine by using right hand grip rule
where
11 s rad 30
s rad 60
2 rpm 1
Thumb : direction of angular velocity
Curl fingers : direction of rotation
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Figures 7.2 and 7.3 show the right hand grip rule for determining the direction of the angular velocity.
Figure 7.2
Figure 7.3
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The angular displacement, of the wheel is given by
where in radians and t in seconds. The diameter of the wheel is 0.56 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 1 :
ttθ 2 5
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Solution :
a. At time, t1 =2.2 s :
At time, t2 =4.8 s :
m 0.282
0.56
2
dr
2.22.25 2 1θ
rad 221θ
1261rad π
180 rad 22
1θ
4.84.85 2 2θ
rad 1102θ
6303rad π
180 rad 110
2θ
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Solution :
b. By applying the equation of arc length,
Therefore
c. The average angular velocity in rad s1 is given by
m 0.282
0.56
2
dr
rθs
m 24.6s
12rrs 221100.28 s
12
12
ttt
θω
av
1av s rad 33.9 ω
2.24.8
22110av
ω
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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 :
1av min rev 324 ω
min 1
s 60
rad 2
rev 1
s 1
rad 33.9avω
OR rpm 324
ttdt
dω 25
dt
dθω
110 tω
13.010 ω1s rad 29 ω
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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 diver’s average angular (rotational) velocity during a dive.
(Given g = 9.81 m s2)
Solution :
Example 2 :
0yu00θ
m 10
waterrev 2.51θ
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Solution :
From the diagram,
Thus
Therefore the diver’s average angular velocity is s 1.43t
rad 5ππ22.5 1θm 10ys
2yy gttus
2
1
29.812
1010 t
t
θθω 01
av
1av s rad 11.0 ω
1.43
05πav
ω
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Average angular acceleration, av
is defined as the rate of change of angular velocity. Equation :
Instantaneous angular acceleration, is defined as the instantaneous rate of change of angular
velocity. Equation :
a)(iv)(v) Angular acceleration
t
ω
tt
ωω
12
12av
dt
dω
t
ωα
t
0
limit
where locityangular ve final :2ω
interval time :tlocityangular ve initial :1ω
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Figure 7.4
It is a vector quantity. The unit of angular acceleration is rad s2. Note:
If the angular acceleration, is positive, then the angular
velocity, is increasing. If the angular acceleration, is negative, then the angular
velocity, is decreasing.
Direction of the angular acceleration If the rotation is speeding up, and in the same direction as
shown in Figure 7.4.
α
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Figure 7.5
If the rotation is slowing down, and have the opposite direction as shown in Figure 7.5.
Example 3 :
The instantaneous angular velocity, of the flywheel is given
by
where in radian per second and t in 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.
α
23 tt8ω
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Solution :
a. At time, t1 =2.2 s :
At time, t2 =4.8 s :
Therefore the average angular acceleration is
2.22.28 23 1ω1s rad 80.3 1ω
2av s rad 301 α
4.84.88 23 2ω1s rad 862 2ω
12
12
tt
ωωα
av
2.24.8
80.3862av
α
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Solution :
b. The instantaneous angular acceleration as a function of time is
At time, t =3.0 s :
238 ttdt
dα
dt
dωα
ttα 224 2
3.023.024 2 α2s rad 210 α
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Exercise 8.1a :1. If a disc 30 cm in diameter rolls 65 m along a straight line
without slipping, calculate
a. 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; 138 rad
2. During a certain period of time, the angular displacement of a swinging door is described by
where is in radians and t is in seconds. Determine the angular displacement, angular speed and angular acceleration
a. at time, t =0,
b. at time, t =3.00 s.
ANS. : 5.00 rad, 10.0 rad s1, 4.00 rad s2; 53.0 rad, 22.0 rad s1, 4.00 rad s2
22.0010.05.00 ttθ
DF025 CHAPTER 8
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At the end of this chapter, students should be able to: Relate parameters in rotational motion with their
corresponding quantities in linear motion. Write and use;
Learning Outcome:8.1.b) Relationship between linear and rotational motion (½ hour)
r
vrararvrθs ct
22 ; ; ;
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8.1.b) Relationship between linear and rotational motion 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 7.6.
v
s
y
xr
P
O
Figure 8.6
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Point P moves in a circle of radius r with the tangential velocity
v where its magnitude is given by
The direction of the linear (tangential) velocity always tangent to the circular path.
Every particle on the rigid body has the same angular speed (magnitude of angular velocity) but the tangential speed is not the same because the radius of the circle, r is changing depend on the position of the particle.
dt
dsv
dt
drv
rθs
rv
and
Simulation 8.1
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ta
caa
x
y
P
O
If the rigid body is gaining the angular speed then the tangential velocity of a particle also increasing thus two component of acceleration are occurred as shown in Figure 7.7.
Relationship between tangential acceleration, at and
angular acceleration,
Figure 8.7
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The components are tangential acceleration, at and
centripetal acceleration, ac given by
but
The vector sum of centripetal and tangential acceleration of
a particle in a rotating body is resultant (linear) acceleration, a given by
and its magnitude,
dt
dvat
dt
drat
rat
rωv and
vrr
va 2
2
c
ct aaa
2c
2t aaa
Vector form
DF025 CHAPTER 8
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At the end of this chapter, students should be able to: Write and use equations for rotational motion with
constant angular acceleration;
Learning Outcome:
8.1.c) Rotational motion with uniform angular acceleration (1/2 hour)
αtωω 0
αθωω 20
2 2
20 ttωθ
2
1
DF025 CHAPTER 8
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8.1.c) Rotational motion with uniform angular acceleration Table 8.1 shows the symbols used in linear and rotational
kinematics.
Table 8.1
Linear motion Quantity
Rotational motion
s θDisplacement
u 0ωInitial velocity
v ωFinal velocity
a αAcceleration
t tTime
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Table 8.2 shows the comparison of linear and rotational motion with constant acceleration.
Linear motion Rotational motion
constanta
atuv
constantα
αtωω 0
2atuts2
1 2
0 αttωθ2
1
asuv 22 2 αθωω 20
2 2
tuvs 2
1 tωωθ 02
1
where in radian. Table 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 :
a. The initial angular velocity is
and the angular acceleration of the wheels is given by
Example 4 :
s 5.00 ,s m 2.00 ,m 0.48 ,s m 17.0 21 taru
0rωu 0ω0.4817.0 1s rad 35.4 0ω
α0.482.00 rαa
2s rad 4.17 α
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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
2
2
1αttωθ 0
rad 229θ
s 5.00 ,s m 2.00 ,m 0.48 ,s m 17.0 21 taru
25.004.172
15.0035.4 θ
rev 36.5rad 2π
rev 1 rad 229
θ
αtωω 0
1s rad 56.3 ω
5.004.1735.4 ω
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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. The wheels 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 :
Example 5 :
,m 0.35 2
0.70 ,rad 60π2π30 rθ
,s m 13.9s 3600
h 1
km 1
m10
h 1
km 50.0 13
u
13
s m 9.72s 3600
h 1
km 1
m10
h 1
km 35.0
v
DF025 CHAPTER 8
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Solution :
a. The initial angular speed of the wheels is
and the final angular speed of the wheels is
therefore
b. The car stops thus
Hence
0rωu 0ω0.3513.9 1s rad 39.7 0ω
rωv ω0.359.72 1s rad 27.8 ω
αθωω 0 222
2s rad 2.13 α 60π239.727.8 22 α
0ω 1s rad 27.8 0ωand
αtωω 0
s 13.1t
t2.1327.80
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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. The angular 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 :
a. Given t =4.00 s, thus
Example 6 :
,s rad 0.300π2π0.150 ,m 0.400 1 0ωr2s rad 1.50π2π0.750 α
αtωω 0 1s rad 19.8 ω
4.001.50π0.300π ω
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Solution :
b. The number of revolutions of the blade is
c. The tangential speed of a point is given by
2
2
1αttωθ 0
rad 41.5θ
24.001.502
14.000.300 θ
rev 6.61rad 2π
rev 1 rad 41.5
θ
rωv 19.80.400v
1s m 7.92 v
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Solution :
d. The magnitude of the resultant acceleration is
22tc aaa
2s m 157 a
2
22
rαr
va
2
22
1.50π0.4000.400
7.92
a
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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 :
a. The period of second-hand of the clock is T = 60 s, hence
Example 7 :
Tω
2π
1s rad 0.11 ω
60
2πω
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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 zero.
3600
2πω
13 s rad 101.74 ω
4104.32
2π
ω
14 s rad 101.45 ω
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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 speed of 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 :
The radius of the coin is
Example 8 :
m 102.40 2d
1s rad 18 0ω
s
2s rad 1.90 α
1s rad 0 ω
m 101.202
2d
r
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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
0rωu 18101.20 2u
1s m 0.216 u1s m 0 v
rαa 1.90101.20 2 a
22 s m 102.28 a
asuv 222 s22 102.2820.2160
m 1.02s
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Exercise 8.1b&c :1. A disk 8.00 cm in radius rotates at a constant rate of 1200 rev
min-1 about its central axis. Determine
a. 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. : 126 rad s1; 3.77 m s1; 1.26 103 m s2; 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. : 262 rad s1; 46 m s1, 1.2 104 m s2
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Exercise 8.1b&c :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 is 98.0 rad s-1. Calculate the constant angular acceleration of the wheel.
ANS. : 13.6 rad s2
4. A wheel rotates with a constant angular acceleration of 3.50 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. : 11.0 rad; 1.75 rev; 9.00 rad s1
DF025 CHAPTER 8
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Exercise 8.1b&c :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 angular acceleration 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 7.8.
a. What is the wheel’s 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. : 0.40 rad s1; 18
Figure 8.8
x
y
PO
DF025 CHAPTER 8
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At the end of this chapter, students should be able to:
a) Define and use torque,
b) State and use conditions for equilibrium of rigid body:
Examples of problems :
Fireman ladder leaning on a wall, see-saw, pivoted / suspended horizontal bar.
Sign convention for moment or torque :
+ve : anticlockwise
ve : clockwise
Learning Outcome:8.2 Equilibrium of a uniform rigid body (2 hours)
τ
0 , 0 , 0 τFF yx
DF025 CHAPTER 8
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8.2 Equilibrium of a rigid bodyNon-concurrent forces
is defined as the forces whose lines of action do not pass through a single common point.
The forces cause the rotational motion on the body. The combination of concurrent and non-concurrent forces cause
rolling motion on the body. (translational and rotational motion)
Figure 5.11 shows an example of non-concurrent forces.
2F
3F
1F
Figure 8.2
4F
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Torque (moment of a force), The magnitude of the torque is defined as the product of a
force and its perpendicular distance from the line of action of the force to the point (rotation axis).
OR
Because of
where r : distance between the pivot point (rotation axis) and the point of application of force.
Thus
Fdτ
force theof magnitude :Farm)(moment distancelar perpendicu : d
torque theof magnitude : τwhere
sinrd
sin FrrF
and between angle : where
OR Fr
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It is a vector quantity. The dimension of torque is
The unit of torque is N m (newton metre), a vector product unlike the joule (unit of work), also equal to a newton metre, which is scalar product.
Torque is occurred because of turning (twisting) effects of the forces on a body.
Sign convention of torque: Positive - turning tendency of the force is anticlockwise. Negative - turning tendency of the force is clockwise.
The value of torque depends on the rotation axis and the magnitude of applied force.
22 TMLdF
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Case 1 : Consider a force is applied to a metre rule which is pivoted at
one end as shown in Figures 5.12a and 5.12b.
Figure 8.12a
F
F
θ
Figure 8.12b
Pivot point (rotation axis)
Fdτ
θrd sin
θFrFdτ sin
(anticlockwise)
(anticlockwise)r
Point of action of a force
Line of action of a force
d
DF025 CHAPTER 8
48
O
Figure 8.13
2θ
Case 2 : Consider three forces are applied to the metre rule which is
pivoted at one end (point O) as shown in Figures 5.13.
Caution : If the line of action of a force is through the rotation axis
then
1F
1θ
111 θrd sin
321 ττττ O
Therefore the resultant (nett) torque is
3F
2F
1r
0sin 333333 θrFdFτ
222 θrd sin
111111 θrFdFτ sin
222222 θrFdFτ sin
2r
2211 dFdFτ O
θFrτ sin0τ
and 0θSimulation 8.2
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Determine a resultant torque of all the forces about rotation axis, O in the following problems.
a.
Example 4 :
m 5
N 102F
m 5 N 301F
m 3
m 3
N 203F
m 10
m 6O
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b.
Example 4 :
m 5
N 102F
m 5
N 301F
m 3
m 3
N 254F
N 203F
m 10
α
m 6O β
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51
m 5m 5
m 10
m 6O
Solution :
a.
Force Torque (N m), o=Fd=Frsin
1F
90330
2F
50510
N 102FN 301F
N 203F
m 31d
m 52d
3F
0The resultant torque:
m N 405090 Oτ(clockwise)
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52
m 5
m 10
m 3
m 6
m 5
Solution :
b.
Force Torque (N m), o=Fd=Frsin
1F
90330
2F
51.50.515520sin βrF33F
0 The resultant torque:51.590 Oτ
(clockwise)
N 102F
N 301F
0.51553
3sin
22
β
O
N 203F
N 254Fα
β
m 31dβ
m 5r
4F
0
3d
m N 38.5 Oτ
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8.2 Equilibrium of a rigid body Rigid body is defined as a body with definite shape that
doesn’t change, so that the particles that compose it stay in fixed position relative to one another even though a force is exerted on it.
If the rigid body is in equilibrium, means the body is translational and rotational equilibrium.
There are two conditions for the equilibrium of forces acting on a rigid body. The vector sum of all forces acting on a rigid body must
be zero.
0nettFF
OR
0 , 0 , 0 zyx FFF
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The vector sum of all external torques acting on a rigid body must be zero about any rotation axis.
This ensures rotational equilibrium. This is equivalent to the three independent scalar
equations along the direction of the coordinate axes,
Centre of gravity, CG is defined as the point at which the whole weight of a body
may be considered to act. A force that exerts on the centre of gravity of an object will
cause a translational motion.
0nettτ
0 , 0 , 0 zyx τττ
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Figures 5.14 and 5.15 show the centre of gravity for uniform (symmetric) object i.e. rod and sphere
rod – refer to the midway point between its end.
sphere – refer to geometric centre.
2
l
2
l
CG
CGl
Figure 5.14
Figure 5.15
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5.3.4 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 diagram of the system to help
conceptualize the problem. Sketch a separate free body diagram for each body. Choose a convenient coordinate axes for each body and
construct 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 body :
Solve the equations for the unknowns.
0xF 0yF; and 0τ
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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 5.16.
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.
(Given g =9.81 m s2)
Example 5 :
1W2W
A BO35 cm
75 cm
Figure 5.16
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58
Solution :
The free body diagram of the beam :
Let point O as the rotation axis.
N 23 ;kg 3 2Wm
0.75 mA B
OCG
1W
2W
N
gm
0.35 m
0.55 m 0.55 m
0.20 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 029.4
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Solution :
Since the beam remains at rest thus the system in equilibrium.
a. Hence
b.
N 2.891W
0 yFand
0Oτ
05.888.050.75 1W
N 55.3N
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 wall as shown in Figure 5.17. 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 m
up the ladder , Calculate the coefficient
of static friction between ladder and
floor.
(Given g =9.81 m s2)
Example 6 :
A
B
smooth wall
rough floor
Figure 5.17
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61
Solution :
a. The free body diagram of the ladder :
Let point B as the rotation axis.
kg 60 ;kg 5.0 fl mm
A
B
CG
gm f
1N
gml
2N
α
m 8.0m 10
m 3.0
m 5.0
Force x-comp. (N)
y-comp. (N)
Torque (N m), B=Fd=Frsin
gml
1N1N
0.810
8sin α
sf
gm f
49.1 0.6
10
6sin β
2N
sf
0
5890
2N
0
0
m 6.0
αβ βsin5.049.1
β
β
147
0
βsin3.05891060
αN1 sin101N8
0
0 sf
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62
Solution :
Since the ladder in equilibrium thus
0 Bτ
081060147 1N
N 1511N
0 xF
0 s1 fNN 151sfHorizontal force:
0 yF
058949.1 2NN 6382NVertical force:
DF025 CHAPTER 8
63
m 10
A
B
m 8.0
m 6.0
m 5.0 α
β
Solution :
b. The free body diagram of the ladder :
Let point B as the rotation axis.
0.6sin 0.8;sin βα
gm f
gml
sf
m 7.0
Force x-comp. (N)
y-comp. (N)
Torque (N m), B=Fd=Frsin
gml
1N1N
2s Nμ
gm f
49.1
2N
sf
0
5890
2N
0
0
α
βsin5.049.1
β
β147
0
βsin7.05892474
αN1 sin101N8
0
0
2N
1N
DF025 CHAPTER 8
64
Solution :
Consider the ladder stills in equilibrium thus
0 Bτ
082474147 1N
N 3281N
0 xF
0 2s1 NμN
0 yF
058949.1 2NN 6382N
0638328 sμ0.514sμ
DF025 CHAPTER 8
65
Figure 5.18
A floodlight of mass 20.0 kg in a park is supported at the end of a 10.0 kg uniform horizontal beam that is hinged to a pole as shown in Figure 5.18. 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.
(Given g =9.81 m s2)
Example 7 :
DF025 CHAPTER 8
66
Solution :
a. The free body diagram of the beam :
b. Let point O as the rotation axis.
kg 10.0 ;kg 20.0 bf mm
Force x-comp. (N) y-comp. (N) Torque (N m), o=Fd=Frsingm f
l1960 196
O CG
gm f
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
DF025 CHAPTER 8
67
Solution :
b. The floodlight and beam remain at rest thus
i.
ii.
0 Oτ
00.549.1196 TlllN 490T
0 xF
0cos xS30T
N 424xS
0 yF
030sin98.1196 yST
N 49.1yS
DF025 CHAPTER 8
68
Solution :
b. ii. Therefore the magnitude of the force is
and its direction is given byN 427S
2y
2x SSS
22S 49.1424
x
y
S
Sθ 1tan
6.61θ
424
49.1tan 1θ
from the +x-axis anticlockwise
DF025 CHAPTER 8
69
Exercise 8.2 :
Use gravitational acceleration, g = 9.81 m s2
1.
Figure 5.19 shows the forces, F1 =10 N, F2= 50 N and F3=
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 resultant torque about point D.
ANS. : -3.7 N m
D
AB
C γ
1F
3F
2F
Figure 8.19
a
b
DF025 CHAPTER 8
70
Figure 5.20
Exercise 8.2 :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 5.20. 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 m
DF025 CHAPTER 8
71
3.
A traffic light hangs from a structure as show in Figure 5.21. The uniform aluminum pole AB is 7.5 m long has a mass of 8.0 kg. 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 N
Figure 5.21
Exercise 8.2 :
DF025 CHAPTER 8
72
4.
A uniform 10.0 N picture frame is supported by two light string
as shown in Figure 5.22. The horizontal force, F is applied for holding 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 N
Exercise 8.2 :
Figure 5.22
F
50.0
cm 15.0
cm 30.0
DF025 CHAPTER 8
73
8.3 Rotational Dynamics (1 hour)
Centre of mass (CM) is defined as the point at which the whole mass of a body
may be considered to be concentrated.
Its coordinate (xCM, yCM) is given the expression below:
n
1ii
n
1iii
CM
m
xm
x
n
1ii
n
1iii
CM ;
m
ym
y
where particle i theof mass : thim
particle i theof coordinate : thi xx
particle i theof coordinate : thi yy
DF025 CHAPTER 8
74
Two masses, 2 kg and 4 kg are located on the x-axis at x =2 m
and x =5 m respectively. Determine the centre of mass of this system.
Solution :
Example 9 :
0 m 2x m 5
kg 4 kg; 2 21 mm1m 2m
21
22112
1ii
2
1iii
CM mm
xmxm
m
xm
x
42
5422CM
x
0 from m 4CM xx
CM
m 4
2 m from m1
OR
DF025 CHAPTER 8
75
A system consists of three particles have the following masses and coordinates :
(1) 2 kg, (1,1) ; (2) 4 kg, (2,0) and (3) 6 kg, (2,2).
Determine the coordinate of the centre of mass of the system.
Solution :
The x coordinate of the CM is
Example 10 :
kg 3 kg; 2 kg; 1 321 mmm
321
3322113
1ii
3
1iii
CM mmm
xmxmxm
m
xm
x
642
262412CM
x 1.83CM x
DF025 CHAPTER 8
76
Solution :
The y coordinate of the CM is
Therefore the coordinate of the CM is
321
3322113
1ii
3
1iii
CM mmm
ymymym
y
ym
y
642
260412CM
y 1.17CM y
1.17,1.83
DF025 CHAPTER 8
77
Figure 7.9 shows a rigid body about a fixed axis O with angular
velocity .
is defined as the sum of the products of the mass of each particle and the square of its respective distance from the rotation axis.
Moment of inertia, I
1m
2mnm
3m
1r2r
3rnr
O
Figure 7.9
DF025 CHAPTER 8
78
OR
It is a scalar quantity. Moment of inertia, I in the rotational kinematics is analogous
to the mass, m in linear kinematics. The dimension of the moment of inertia is M L2. The S.I. unit of moment of inertia is kg m2. The factors which affect the moment of inertia, I of a rigid body:
a. the mass of the body,b. the shape of the body,c. the position of the rotation axis.
n
1i
2ii
2nn
233
222
211 ... rmrmrmrmrmI
axisrotation about body rigid a of inertia ofmoment : Iparticle of mass : m
axisrotation the toparticle thefrom distance : r
where
DF025 CHAPTER 8
79
Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
2CM MRI
2CM 2
1MRI
Hoop or ring or thin cylindrical
shell
Solid cylinder or disk
CM
CM
DF025 CHAPTER 8
80
CM
Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
2CM 12
1MLI
Uniform rod or long thin rod with
rotation axis through the
centre of mass.
CM
2CM 5
2MRI Solid Sphere
DF025 CHAPTER 8
81
Moments of inertia of various bodies Table 7.3 shows the moments of inertia for a number of objects
about axes through the centre of mass.
Shape Diagram Equation
2CM 3
2MRI Hollow Sphere or
thin spherical shell
CM
Table 7.3
DF025 CHAPTER 8
82
Four spheres are arranged in a rectangular shape of sides 250 cm and 120 cm as shown in Figure 7.10.
The spheres are connected by light rods . Determine the moment of inertia of the system about an axis
a. through point O,
b. along the line AB.
Example 11 :
cm 250
cm 60
cm 60
kg 2 kg 3
kg 4kg 5
OA B
Figure 7.10
DF025 CHAPTER 8
83
Solution :
a. rotation axis about point O,
Since r1= r2= r3= r4= r thus
and the connecting rods are light therefore
kg 5 kg; 4 kg; 3 kg; 2 4321 mmmm
1r m 0.6
1m 2m
3m4m
O
2r
4r 3r
m 1.25
m 1.391.250.6 22 r
244
233
222
211O rmrmrmrmI 54321.39 2
43212
O mmmmrI2
O m kg 27.0I
DF025 CHAPTER 8
84
Solution :
b. rotation axis along the line AB,
r1= r2= r3= r4= r=0.6 m therefore
kg 5 kg; 4 kg; 3 kg; 2 4321 mmmm
244
233
222
211AB rmrmrmrmI
54320.6 2AB I
2AB m kg 5.04I
1m 2m
3m4m
A B1r 2r
4r 3r
43212
AB mmmmrI
DF025 CHAPTER 8
85
Parallel-Axis Theorem (Steiner’s Theorem) States that the moment of inertia, I about any axis parallel to
and a distance, d away from the axis through the centre of
mass, ICM is given by
2CM MdII
axisrotation new aabout inertia ofmoment :I
body rigid theof mass : M
CM through axisan about inertia ofmoment :CMI
axis original theand axis new ebetween th distance : d
where
DF025 CHAPTER 8
86
Determine the moment of inertia of a solid cylinder of radius R and
mass M about an axis tangent to its edge and parallel to its symmetry axis as shown in Figure 7.11.
Given the moment of inertia of the solid cylinder about axis through the centre of mass is
Example 12 :
Figure 7.11
CM
d
2CM 2
1MRI
DF025 CHAPTER 8
87
CM
Solution :
From the diagram, d = RBy using the parallel axis theorem,
CM
Initiald
Final
2CM MdII
22
2
1MRMRI
2
2
3MRI
2CM 2
1MRI
DF025 CHAPTER 8
88
Relationship between torque, and angular acceleration, Consider a force, F acts on a rigid body freely pivoted on an
axis through point O as shown in Figure 7.12.
The body rotates in the anticlockwise direction and a nett torque is produced.
Torque,
1m
2m
nm1r
2r
nr
O
1a
na
2a
F
Figure 7.12
DF025 CHAPTER 8
89
A particle of mass, m1 of distance r1 from the rotation axis O will
experience a nett force F1 . 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
111 amF
αrmF 111
αra 11 and
90sin111 Fr 2
111 rm
n
iiirm
1
2
2222
211 ... nnrmrmrm
I
Irmn
iii
1
2 and
DF025 CHAPTER 8
90
From the equation, the nett torque acting on the rigid body is proportional to the body’s angular acceleration.
Note :
I, eNett torqu
maFforce,Nett
is analogous to the
DF025 CHAPTER 8
91
Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure 7.13.
Calculate,
a. the nett torque on the disc.
b. the magnitude of angular acceleration influence by the disc.
( Use the moment of inertia, )
Example 13 :
Figure 7.13
2CM 2
1MRI
1F
O
cm 30.0
2F
DF025 CHAPTER 8
92
Solution :
a. The nett torque on the disc is
b. By applying the relationship between torque and angular
acceleration,
kg 5.00 ;m 0.30 MR
21 2121 FFRRFRF
3.1060.530.0 m N 1.41
2
2
1MR I
2s rad 6.27
230.000.5
2
141.1
DF025 CHAPTER 8
93
A wheel of radius 0.20 m is mounted on a frictionless horizontal
axis. The moment of inertia of the wheel about the axis is
0.050 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 magnitude P = 3.0 N is applied to the block as
shown in Figure 7.14. Assume the string does not slip on the
wheel.
a. Sketch a free body diagram of the wheel and the block.
b. Calculate the magnitude of the angular acceleration of the wheel.
Example 14 :
Figure 7.14
DF025 CHAPTER 8
94
Solution :
a. Free body diagram :
for wheel,
for block,
kg 2.0 N; 3.0 ;m kg 0.050 ;m 0.20 2 mPIR
W
T
S
N
T
bW
P
a
DF025 CHAPTER 8
95
Solution :
b. For wheel,
For block,
By substituting eq. (1) into eq. (2), thus
Iατ
IαRT R
IαT (1)
maF maTP (2)
kg 2.0 N; 3.0 ;m kg 0.050 ;m 0.20 2 mPIR
Rαa maR
IαP
and
mRαR
IαP
αα0.202.0
0.20
0.0503.0
2s rad 4.62 α
DF025 CHAPTER 8
96
An object of mass 1.50 kg is suspended from a rough pulley of radius 20.0 cm by light string as shown in Figure 7.15. 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. (Given g = 9.81 m s-2)
Example 15 :
Figure 7.15
R
kg 1.50
DF025 CHAPTER 8
97
Solution :
a. Free body diagram :
for pulley,
for block,
W
a
T
S Iατ
IαRT R
aα and
R
aIRT
2R
IaT (1)
T
gm
maF
maTmg (2)
DF025 CHAPTER 8
98
Solution :
a. By substituting eq. (1) into eq. (2), thus
b. By using the relationship between a and , hence
2s m 7.36 a
maR
Iamg
2
kg; 1.50 ;m kg 0.020 ;m 0.20 2 mIRs 0.3 0; tu
Rαa
aa
1.500.20
0.0209.811.50
2
α0.207.36 2s rad 36.8 α
DF025 CHAPTER 8
99
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
N 3.68T
kg; 1.50 ;m kg 0.020 ;m 0.20 2 mIRs 0.3 0; tu
atuv 0.37.360 v
m 0.331s
2R
IaT
20.20
7.360.020T
1s m 2.21 v(downwards)
2at2
1uts
20.37.3602
1s
DF025 CHAPTER 8
100
Exercise 8.3 :Use gravitational acceleration, g = 9.81 m s2
1. Three odd-shaped blocks of chocolate have following masses and 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. : (0.044 m, 0.056 m)
2. Figure 7.16 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 perpendicular
to the plane
a. through point A, and
b. through point B.
ANS. : 0.141 kg m2; 0.211 kg m2
cm 80
cm 80
g 150 g 150
g 07
g 70
cm 40
A
B
Figure 7.16
DF025 CHAPTER 8
101
2s m 2.00
2T
1T
Exercise 8.3 :3. A 5.00 kg object placed on a
frictionless horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object as in Figure 7.17. The pulley has a radius of 0.250 m and moment of inertia I. The block on the table is moving with a constant acceleration 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. Determine I.
ANS. : 10.0 N, 70.3 N; 1.88 kg m2
Figure 7.17
DF025 CHAPTER 8
102
At the end of this chapter, students should be able to: a) Solve problems related to:
Rotational kinetic energy,
work,
power,
Learning Outcome:
8.4 Work and Energy of Rotational Motion(2 hours)
2
2
1IωK r
τωP
τθW
DF025 CHAPTER 8
103
Rotational kinetic energy and powerRotational kinetic energy, Kr
Consider a rigid body rotating about the axis OZ as shown in Figure 7.18.
Every particle in the body is in the circular motion about point O.
1m
2m
nm
3m
1r2r
3r
nr
O
1v
2v
3vnv
Z
Figure 7.18
DF025 CHAPTER 8
104
The rigid body has a rotational kinetic energy which is the total of kinetic energy of all the particles in the body is given by
2nn
233
222
211r vmvmvmvmK
2
1...
2
1
2
1
2
1
22nn
2233
2222
2211r ωrmωrmωrmωrmK
2
1...
2
1
2
1
2
1
2nn
233
222
211
2r rmrmrmrmωK ...
2
1
2r IωK
2
1
n
1i
2ii
2r rmωK
2
1Irm
n
1i
2ii
and
DF025 CHAPTER 8
105
From the formula for translational kinetic energy, Ktr
After comparing both equations thus
For rolling body without slipping, the total kinetic energy of
the body, K is given by
2tr mv
2
1K
is analogous to vI is analogous to m
rtr KKK
energy kinetic onal translati: trK
energy kinetic rotational : rK
where
DF025 CHAPTER 8
106
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 sphere rolls 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 16 :
2mRI5
2CM
DF025 CHAPTER 8
107
Solution :
a. From the principle of conservation of energy,
kg 10.0 ;m 0.15 mR
fi EE
Kmgh 25sinmgsK
J 31.1K
25sin0.759.8110.0K
m 0.75s
25sinsh
v CM 25
R
DF025 CHAPTER 8
108
Solution :
b. The linear speed of the sphere is given by
c. By using the relationship between v and , thus
kg 10.0 ;m 0.15 mR
rtr KKK 22 IωmvK2
1
2
1
R
vω
1s m 2.11 v
and
222
R
vmRmvK
5
2
2
1
2
1
2mvK10
7
1s rad 14.1 ω
2v10.010
731.1
Rωv ω0.152.11
DF025 CHAPTER 8
109
The pulley in the Figure 7.19 has a radius of 0.120 m and a moment of inertia 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.
(Given g = 9.81 m s2)
Example 17 :
kg 2.00
kg 5.00
m 7.00
Figure 7.19
DF025 CHAPTER 8
110
Solution :
The moment of inertia of the pulley,m 7.00 m; 0.120 kg; 2.00 ;kg 5.00 21 hRmm
292
2432 m kg 105.5
cm 1
m10
g 1
kg 10cm 1g 0.055
I
2m
1m
m 7.00
Initial
2m
1mm 7.00
v
v
Final
1i UE 2r2tr1trf UKKKE
DF025 CHAPTER 8
111
Solution :
By using the principle of conservation of energy, thus
fi EE
2r2tr1tr1 UKKKU
ghmIωvmvmghm 222
22
11 2
1
2
1
2
1
2
212
21 2
1
2
1
R
vImmvghmm
2
92
0.120105.5
2
12.005.00
2
17.009.812.005.00
v
v
1s m 7.67 v
m; 0.120 kg; 2.00 ;kg 5.00 Rmm 2129 m kg 105.5 m; 7.00 Ih
DF025 CHAPTER 8
112
Consider a tangential force, F acts on the solid disc of radius R freely pivoted on an axis through O as shown in Figure 7.20.
The work done by the tangential force is given by
Work, W
Figure 7.20
F
ds
O
dR
R
FdsdW FRdθdW
2
1
θ
θτdθ dW
2
1
θ
θτdθ W
Rdθds and
DF025 CHAPTER 8
113
If the torque is constant thus
Work-rotational kinetic energy theorem states
12W 2
1
dW
W
torque: τntdisplacemeangular in change : Δθ
where
done work : W
irfrr KKKW
20
2
2
1
2
1IωIωW
is analogous to the FsW
DF025 CHAPTER 8
114
From the definition of instantaneous power,
Caution : The unit of kinetic energy, work and power in the
rotational kinematics is same as their unit in translational kinematics.
Power, P
dt
dWP τdθdW and
dt
τdθP
τω P
ωdt
dθand
is analogous to the FvP
DF025 CHAPTER 8
115
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 its centre. 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.
(Given g = 9.81 m s2)
Solution :
Example 18 :
F
m 2.40R
DF025 CHAPTER 8
116
Solution :
a. By applying the relationship between nett torque and angular
acceleration, thus
Use the equation of rotational motion with uniform angular
acceleration,
Therefore the rotational kinetic energy for 15.0 s is
Iατ IαRF α210018.02.40
22 s rad 102.06 α
αtωω 0 15.0102.060 2ω1s rad 0.309 ω
2
2
1IωK r
J 100rK
20.30921002
1rK
N; 18.0 ;m kg 2100 ;m 2.40 2 FIR0 s; 15.0 0 ωt
DF025 CHAPTER 8
117
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
τθW
20 2
1αttωθ
t
WPav
rad 2.32θ
2.3218.02.40W
N; 18.0 ;m kg 2100 ;m 2.40 2 FIR0 s; 15.0 0 ωt
22 15.0102.062
10 θ
RFθW
J 100W
15.0
100avP
W6.67avP
DF025 CHAPTER 8
118
At the end of this chapter, students should be able to: Define and use the formulae of angular momentum,
State and use the principle of conservation of angular momentum
Learning Outcome:
8.5 Conservation of angular momentum (1 hour)
IωL
DF025 CHAPTER 8
119
Conservation of angular momentumAngular momentum, is defined as the product of the angular velocity of a body
and its moment of inertia about the rotation axis.
OR
It is a vector quantity. Its dimension is M L2 T1
The S.I. unit of the angular momentum is kg m2 s1.
L
where
IL momentumangular : L
body a of inertia ofmoment : Ibody a oflocity angular ve : ω
is analogous to the mvp
DF025 CHAPTER 8
120
The relationship between angular momentum, L with linear momentum, p is given by vector notation :
magnitude form :
Newton’s second law of motion in term of linear momentum is
hence we can write the Newton’s second law in angular form as
and states that the vector sum of all the torques acting on a rigid body is proportional to the rate of change of angular momentum.
vmrprL
θmvrθrpL sinsin where
vrθ
with between angle the:axisrotation the toparticle thefrom distance : r
dt
pdFF nett
dt
Ldττ nett
DF025 CHAPTER 8
121
states that the total angular momentum of a system about an rotation axis is constant if no external torque acts on the system.
OR
Therefore
Principle of conservation of angular momentum
constantI
0 dt
Ldτ
0Ld
If the 0 τ
if L-LdL
fi LL
and
DF025 CHAPTER 8
122
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 7.21. 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 19 :
0ω
Before
R
After
R
Figure 7.21
2
2
1MR
DF025 CHAPTER 8
123
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,
kg 50 kg; 200 ;s rad 4.0 ;m 3.00 10
bw mmωR
22 3.002002
1
2
1 RmI ww
1s rad 2.67 ω
ωIIωI bww 0
22 3.0050 RmI bb
fi LL
2m kg 900wI
2m kg 450bI
ω4509004.0900
DF025 CHAPTER 8
124
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 the
friction between the surface of the disc with the bag of
sand.
kg 50 kg; 200 ;s rad 4.0 ;m 3.00 10
bw mmωR
220 4.0900
2
1
2
1 ωIK wir
22 2.674509002
1
2
1 ωIIK bwfr
J 7200irK
frir KK J 4812frK
DF025 CHAPTER 8
125
A raw egg and a hard-boiled egg are rotating about the same axis of rotation with the same initial angular velocity. Explain which egg will rotate longer.
Solution :
The answer is hard-boiled egg.
Example 20 :
DF025 CHAPTER 8
126
Solution :
Reason
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 the I is 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.
2mrI
constantI
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127
Before After
0
A student sits on a freely rotating stool holding two weights, each of mass 3.00 kg as shown in Figure 7.22.
When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad s1. The moment of inertia of the student and stool is 3.00 kg m2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis. Determine
a. the new angular speed of the student.
b. the kinetic energy of the rotating
system before and after he pulls
the weights inward.
Example 21 :
Figure 7.22
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128
Before
0
After
ar ar
Solution : 210 m kg 3.00 ;s rad 0.750 ;kg 3.00
ssw Iωm
brbr
wmwm
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129
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
wssi III 22bwbwssi rmrmII
22 bwssi rmII 22 m kg 9.001.003.0023.00 iI
22 awssf rmII 22 m kg 3.540.3003.0023.00 fI
;m kg 3.00 ;s rad 0.750 ;kg 3.00 210
ssw Iωmm 0.300 ;m 1.00 ab rr
1s rad 1.91 ω
ωIωI fi 0
fi LL
ω3.540.7509.00
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130
Solution :
b. The initial rotational kinetic energy is given by
and the final rotational kinetic energy is
202
1ωIK iir
20.7509.002
1irK
J 2.53irK
2
2
1ωIK ffr
21.913.542
1frK
J 6.46frK
;m kg 3.00 ;s rad 0.750 ;kg 3.00 210
ssw Iωmm 0.300 ;m 1.00 ab rr
DF025 CHAPTER 8
131
Exercise 8.5 :Use gravitational acceleration, g = 9.81 m s2
1. A woman of mass 60 kg stands at the rim of a horizontal turntable 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. : 0.360 rad s1 ,U think; 99.9 J
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132
Exercise 8.5 :2. Determine the angular momentum of the Earth
a. about its rotation axis (assume the Earth is a uniform solid
sphere), and
b. about its orbit around the Sun (treat the Earth as a particle
orbiting the Sun).
Given the Earth’s mass = 6.0 x 1024 kg, radius = 6.4 x 106 m and is 1.5 x 108 km from the Sun.
ANS. : 7.1 x 1033 kg m2 s1; 2.7 x 1040 kg m2 s1
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 with angular velocity about one end. (Given the moment of inertia of thin rod about the axis through the CM is )
ANS. : 4.71 x 106 kg m2 s1
2
12
1ML
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133
Linear Motion Relationship Rotational MotionSummary:
m
rv dt
d
dt
dva
dt
d
maF I
FsW W
dt
dsv
ra
sinrF
FvP P
mvp IL sinrpL
I2i
n
1iirmI
DF015 CHAPTER 8
134
THE END…Next Chapter…
CHAPTER 9 :Simple Harmonic Motion