Transcript Chapter 7:Rotation of a Rigid Body

PHYSICS
CHAPTER 8
CHAPTER 8:
Rotational of rigid body
(8 Hours)
1
PHYSICS
CHAPTER 8
WHAT IS ROTATIONAL OF RIGID BODY ?
The study of the motions of a rigid body under the
influence of forces and torques.
WHAT IS RIGID BODY ?
an object or system of particles in which
the distances between particles are
fixed and remain constant.
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Learning Outcome:
8.1 Rotational Kinematics (2 hour)
At the end of this chapter, students should be able to:
 Define and use:

angular displacement ()

average angular velocity (av)

instantaneous angular velocity ()

average angular acceleration (av)


instantaneous angular acceleration ().
Relate parameters in rotational motion with their
corresponding quantities in linear motion. Write and use,
2
v
s  r , v  r , at  r , ac  r 2 
r
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CHAPTER 8
Learning Outcome:

Use equations for rotational motion with constant angular
acceleration,
  o   t
1 2
  ot   t and
2
 2  o 2  2
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CHAPTER 8
8.1.1 Parameters in rotational motion
1) 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 8.1 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 8.1
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
From Figure 8.1, thus
s
θ
r
where

OR
s  rθ
θ : angle (angular displacement) in radian
s : arc length
r : radius of the circle
Others unit for angular displacement is degree () and
revolution (rev).
 Conversion factor :
1 rev  2π rad  360

Sign convention of angular displacement :
 Positive – if the rotational motion is anticlockwise.
 Negative – if the rotational motion is clockwise.
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2) Angular velocity
Average angular velocity, av
 is defined as the rate of change of angular displacement.
 Equation :
θ 2  θ1 θ
ωav 

t 2  t1
t
where
θ2 : final angular displacement in radian
θ1 : initial angular displacement in radian
t : time interval
Instantaneous angular velocity, 
 is defined as the instantaneous rate of change of angular
displacement.
 Equation :
θ dθ
  limit

t 0 t
dt
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PHYSICS



CHAPTER 8
It is a vector quantity.
The unit of angular velocity is radian per second (rad s-1)
Others unit is revolution per minute (rev min1 or rpm)
 Conversion factor:
2

1
1 rpm 
rad s 
rad s 1
60
30
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

Thumb
Curl fingers
: direction of angular velocity
: direction of rotation
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
CHAPTER 8
Figures 8.2 and 8.3 show the right hand grip rule for determining
the direction of the angular velocity.


Figure 8.2


Figure 8.3
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Example 8.1 :
The angular displacement, of the wheel is given by
θ  5t 2  t
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 s1 and in rev min1 (rpm),
between 2.2 s and 4.8 s,
d. the instantaneous angular velocity at time 3.0 s.
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Solution :
CHAPTER 8
d 0.56
r 
 0.28 m
2
2
a. At time, t1 =2.2 s :
At time, t2 =4.8 s :
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PHYSICS
Solution :
CHAPTER 8
d 0.56
r 
 0.28 m
2
2
b. By applying the equation of arc length,
c. The average angular velocity in rad s1 is given by
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Solution :
c. and the average angular velocity in rev min1 is
d. The instantaneous angular velocity as a function of time is
At time, t =3.0 s :
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3) Angular acceleration
Average angular acceleration, av
 is defined as the rate of change of angular velocity.
 Equation :
ω2  ω1 ω
 av 

t2  t1
t
where ω2 : finalangular velocity
ω1 : initialangular velocity
t : time interval
Instantaneous angular acceleration, 
 is defined as the instantaneous rate of change of angular
velocity.
 Equation :
ω dω
α  limit

t 0 t
dt
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


CHAPTER 8
It is a vector quantity.
The unit of angular acceleration is rad s2.
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 8.4.



α
Figure 8.4
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
CHAPTER 8
If the rotation is slowing down,  and  have the opposite
direction as shown in Figure 8.5.



α
Figure 8.5
Example 8.3 :
The instantaneous angular velocity,  of the flywheel is given
by
3
2
ω  8t  t
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.
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Solution :
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Solution :
b. The instantaneous angular acceleration as a function of time is
At time, t =3.0 s :
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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
θ  5.00  10.0t  2.00t 2
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 s1, 4.00 rad s2; 53.0 rad, 22.0 rad s1,
4.00 rad s2
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8.1.2 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 8.6.
y

v
P
r

O
s
x
Figure 8.6
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
CHAPTER 8
Point P moves in a circle of radius r with the tangential velocity
v where its magnitude is given by
ds
and s  rθ
v
dt
d
vr
dt
v  r


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.
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Relationship between tangential acceleration, at and
angular acceleration, 

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 8.7.
y
O

at
 P
a 
ac
x
Figure 8.7
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
CHAPTER 8
The components are tangential acceleration, at and
centripetal acceleration, ac given by
dv
at 
dt
but

and
v  rω
d
at  r
at  r
dt
v2
2
ac   r  v
r
The vector sum of centripetal and tangential acceleration of
a particle in a rotating body is resultant (linear) acceleration, a
 
given by 
a  at  ac Vector form
and its magnitude,

2
2
a  at  ac
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CHAPTER 8
8.1.3 Rotational motion with uniform
angular acceleration

Linear
motion
s
u
v
a
t
Table 8.1 shows the symbols used in linear and rotational
kinematics.
Quantity
Displacement
Initial velocity
Rotational
motion
θ
ω0
Acceleration
ω
α
Time
t
Final velocity
Quantity
Angular displacement
Angular velocity
(initial)
Angular velocity (final)
Angular acceleration
time
Table 8.1
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
CHAPTER 8
Table 8.2 shows the comparison of linear and rotational motion
with constant acceleration.
Linear motion
Rotational motion
a  constant
α  constant
v  u  at
ω  ω0  αt
1 2
s  ut  at
2
1 2
θ  ω0t  αt
2
v 2  u 2  2as
ω2  ω02  2αθ
1
s  v  u t
2
where  in radian.
1
θ  ω  ω0 t
2
Table 8.2
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Example 8.3 :
A car is travelling with a velocity of 17.0 m s1 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 s2 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 : u  17.0 m s 1 , r  0.48 m, a  2.00 m s 2 , t
a. The initial angular velocity is
 5.00 s
and the angular acceleration of the wheels is given by
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Solution : u  17.0 m s 1 , r  0.48 m, a  2.00 m s 2 , t  5.00 s
a. By applying the equation of rotational motion with constant
angular acceleration, thus
b. The angular speed of the wheels after 5.00 s is
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Example 8.5 :
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 :
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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
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Example 8.5 :
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 : r  0.400 m, ω0
 0.150  2π  0.300π rad s 1 ,
α  0.750  2π  1.50π rad s 2
a. Given t =4.00 s, thus
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Solution :
b. The number of revolutions of the blade is
c. The tangential speed of a point is given by
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Solution :
d. The magnitude of the resultant acceleration is
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Example 8.6 :
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 s1 and rolls in a straight line without slipping. If the
rotation slows down with an angular acceleration of magnitude
1.90 rad s2, calculate the distance travelled by the coin before
coming to rest.
Solution :
1
ω  18 rad s 1
ω  0 rad s
0
2
d  2.40  10 m
α  1.90 rad s 2
s
The radius of the coin is
d
r   1.20  10 2 m
2
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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
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Exercise 8.1B :
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 s1; 3.77 m s1; 1.26  103 m s2; 20.1 m
2. A 0.35 m diameter grinding wheel rotates at 2500 rpm.
Calculate
a. its angular velocity in rad s1,
b. the linear speed and the radial acceleration of a point on the
edge of the grinding wheel.
ANS. : 262 rad s1; 46 m s1, 1.2  104 m s2
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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 s2
4. A wheel rotates with a constant angular acceleration of
3.50 rad s2.
a. If the angular speed of the wheel is 2.00 rad s1 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 s1
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Learning Outcome:
8.2 Equilibrium of a uniform rigid body (2 hours)
At the end of this chapter, students should be able to:
 Define and use torque,
 State and use conditions for equilibrium of rigid body,
τ
F  0
and
  0
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8.2.1

CHAPTER 8


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
τ  Fd
where
Because of
τ : magnitude of the torque
F : magnitude of the force
d : perpendicular distance (moment arm)
d  r sin 
where r : distance between the pivot point (rotation
axis) and the point of application of force.
Thus
  rF sin 
OR
 
  r F



where  : angle between F and r
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PHYSICS


CHAPTER 8
It is a vector quantity.
The dimension of torque is
   F d   ML T
2




2
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.
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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 8.12a and 8.12b.

F τ  Fd
(anticlockwise)
d
Figure 8.12a
Line of action of a force
Pivot point
(rotation axis)
d  r sin θ
r
θ
Figure 8.12b

F
Point of action of a force
τ  Fd  Fr sin θ
(anticlockwise)
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Case 2 :
 Consider three forces are applied to the metre rule which is
pivoted at one end (point O) as shown in Figures 8.13.

F3 d1  r1 sin θ1
r2
O
d 2  r2 sin θ2

θ2
θ1
 r1
F2 Figure
8.13

F1
τ1  F1d1  F1r1 sin θ1
τ 2   F2 d 2   F2 r2 sin θ2
τ 3  F3 d 3  F3 r3 sin θ3  0
Therefore the resultant (nett)
torque is
τ
τ
O
 τ1  τ 2  τ 3
O
 F1d1  F2 d 2
Caution :
 If the line of action of a force is through the rotation axis
then
τ  Fr sin θ andθ  0
τ 0
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8.2.2 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.

F  F
nett
0
OR
F
x
 0,
F
y
 0,
F
z
0
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
CHAPTER 8
The vector sum of all external torques acting on a rigid
body must be zero about any rotation axis.




  τ nett  0
This ensures rotational equilibrium.
This is equivalent to the three independent scalar
equations along the direction of the coordinate axes,
τ
any axis
0
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.
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PHYSICS

CHAPTER 8
Figures 8.14 and 8.15 show the centre of gravity for uniform
(symmetric) object i.e. rod and sphere
 rod – refer to the midway point between its end.
l
CG
l
2
l
2
Figure 8.14

sphere – refer to geometric centre.
CG
Figure 8.15
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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 :
F
x

 0;
F
y
0
and
τ
any axis
0
Solve the equations for the unknowns.
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CHAPTER 8
Example 8.7 :
A
35
cm
O
75
cm
B
W1
W2
Figure 8.9
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 8.9.
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 s2)
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CHAPTER 8

N
Solution :
m  3 kg; W2  23 N
The free body diagram of the beam :
A
Let point O as the rotation axis.
Force

W1

W2

mg

N

W2
y-comp. (N)
0.35 m
0.20 m
O
0.55 m
CG

mg
0.55 m

W1
Torque (N m), o=Fd=Frsin
 W1
 W1 0.75  0.75W1
 23
 230.35  8.05
 39.81
 29.4
N
B
0.75 m
 29.40.20  5.88
0
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Solution :
Since the beam remains at rest thus the system in equilibrium.
a. Hence
b.
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CHAPTER 8
Example 8.9 :
A uniform ladder AB of length 10 m and
mass 5.0 kg leans against a smooth wall
as shown in Figure 8.10. The height of the A
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 s2)
smooth
wall
B
rough floor
Figure 8.10
PHYSICS
CHAPTER 8
Solution :
ml  5.0 kg; m f  60 kg
a. The free body diagram of the ladder :
Let point B as the rotation axis.
Force

ml g

mf g

N1

N2

fs
x-comp.
(N)
y-comp.
(N)
Torque (N m),
B=Fd=Frsin
A
α

N1
β
8.0 m CG
8
sin α   0.8
10
6
sin β   0.6
10
10 m

ml g β

mf g β
5.0 m
α

fs
3.0
 m
N2
B
6.0 m
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Solution :
Since the ladder in equilibrium thus
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sin α  0.8; sin β  0.6
Solution :
b. The free body diagram of the ladder :
Let point B as the rotation axis.
Torque (N m),
x-comp. y-comp.
Force
(N)
(N)
B=Fd=Frsin

ml g

mf g

N1

N2

fs
A
α

N1
β

mf g
β 10 m
7.0 m
8.0 m

ml g β

N2
5.0 m
α
 B
fs
6.0 m
52
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Solution :
Consider the ladder stills in equilibrium thus
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Example 8.9 :
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 8.11. 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 s2)
Figure 8.11
54
PHYSICS
CHAPTER 8
Solution :
m f  20.0 kg; mb  10.0 kg
a. The free body diagram of the
 beam :

T
30 
S
O
0.5l
CG 
l
mb g

mf g
b. Let point O as the rotation axis.
Force x-comp. (N) y-comp. (N)

mf g

mb g

T

S
Torque (N m), o=Fd=Frsin
55
PHYSICS
CHAPTER 8
Solution :
b. The floodlight and beam remain at rest thus
i.
56
PHYSICS
CHAPTER 8
Solution :
b. ii. Therefore the magnitude of the force is
and its direction is given by
57
PHYSICS
CHAPTER 8
Exercise 8.2 :
Use gravitational acceleration,
 g = 9.81 m s2
1.
F1
B
a
A

F2
b
D
C
γ

F3
Figure 8.12
Figure 8.12 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
58
PHYSICS
CHAPTER 8
2.
Figure 8.13
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 8.13. 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
59
PHYSICS
CHAPTER 8
3.
Figure 8.14
A traffic light hangs from a structure as show in Figure 8.14. 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
60
PHYSICS
CHAPTER 8
Exercise 5.2 :
4.
30.0 cm
15.0 cm
50.0

F
Figure 8.15
A uniform 10.0 N picture frame is supported by two light string
as shown in Figure 8.15. 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
61
PHYSICS
CHAPTER 8
Learning Outcome:
8.3 Rotational dynamics (1 hours)
At the end of this chapter, students should be able to:

Define the moment of inertia of a rigid body about an
axis,
I   mi ri 2

State and use torque,
  I
62
PHYSICS
CHAPTER 8
8.3.1 Moment of inertia, I

Figure 8.16 shows a rigid body about a fixed axis O with angular
velocity .
m1
mn
rn
r1
Or
3

r2 m2
m3
Figure 8.16

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.
63
PHYSICS
OR
CHAPTER 8
I
2
m1r1

2
m2 r2

2
m3 r3
2
 ...mn rn

n
m r
2
i i
i 1
where I : moment of
inertia of a rigid body about rotation axis
m : mass of particle
r : distance from the particle to the rotation axis

It is a scalar quantity.

Moment of inertia, I in the rotational kinematics is analogous
to the mass, m in linear kinematics.
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.
64
PHYSICS
CHAPTER 8
Moments of inertia of various bodies
 Table below shows the moments of inertia for a number of
objects about axes through the centre of mass.
Shape
Diagram
Equation
Hoop or ring or
thin cylindrical
shell
CM
I CM  MR
Solid cylinder or
disk
CM
I CM
2
1
 MR 2
2
65
PHYSICS
Shape
Uniform rod or
long thin rod with
rotation axis
through the
centre of mass.
CHAPTER 8
Diagram
CM
Solid Sphere
CM
Equation
I CM
1
 ML2
12
I CM
2
 MR 2
5
66
PHYSICS
Shape
Hollow Sphere or
thin spherical
shell
CHAPTER 8
Diagram
CM
Equation
I CM
2
 MR 2
3
67
PHYSICS
CHAPTER 8
Example 8.10 :
Four spheres are arranged in a rectangular shape of sides 250 cm
and 120 cm as shown in Figure.
2 kg
3 kg
60 cm
A
O
5 kg
250 cm
B
60 cm
4 kg
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.
68
PHYSICS
CHAPTER 8
Solution : m1  2 kg; m2  3 kg; m3
a. rotation axis about point O,
 4 kg; m4  5 kg
m2
m1
r2
r1
O
r4
m4
0.6 m
1.25 m
r3
m3
Since r1= r2= r3= r4= r thus
and the connecting rods are light therefore
69
PHYSICS
CHAPTER 8
Solution : m1  2 kg; m2  3 kg; m3
b. rotation axis along the line AB,
m1
 4 kg; m4  5 kg
m2
r2
r1
B
A
r4
m4
r1= r2= r3= r4= r=0.6 m therefore
r3
m3
70
PHYSICS
CHAPTER 8
8.3.2 Torque,
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.
a1 m1
mn
r1
an rn
O

r2

F
a2
m2
The body rotates in the anticlockwise direction and a nett torque
is produced.
71
PHYSICS

CHAPTER 8
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
F1  m1a1 and a1  r1α
F1  m1r1α

The torque on the mass m1 is

The total (nett) torque on the rigid body is given by
 1  r1F1 sin 90
2
 1  m1r1 
2
2
2


m
r


m
r


...

m
r
 11
2 2
n n
n
 n

2
2
and
m
r


m
r




i i  I
 
i i
i 1
i 1

  I
72
PHYSICS
CHAPTER 8

From the equation, the nett torque acting on the rigid body is
proportional to the body’s angular acceleration.

Note :
Nett torqu e ,   I
is analogous to the
Nett force,  F  ma
73
PHYSICS
CHAPTER 8
Example 8.11 :
Forces, F1 = 5.60 N and F2 = 10.3 N are applied tangentially to a

F2
disc with radius 30.0 cm and the mass 5.00 kg as shown in Figure.
O

F1
30.0 cm
Calculate,
a. the nett torque on the disc.
b. the magnitude of angular acceleration
by the disc.
1 influence
2
( Use the moment of inertia, I CM  MR )
2
74
PHYSICS
CHAPTER 8
Solution : R  0.30 m; M  5.00 kg
a. The nett torque on the disc is
b. By applying the relationship between torque and angular
acceleration,
75
PHYSICS
CHAPTER 8
Example 8.12 :
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. 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.
76
PHYSICS
CHAPTER 8
Solution : R  0.20 m; I
a. Free body diagram :
for wheel,
 0.050 kg m 2 ; P  3.0 N; m  2.0 kg
for block,
77
PHYSICS
Solution : R  0.20 m; I
b. For wheel,
CHAPTER 8
 0.050 kg m 2 ; P  3.0 N; m  2.0 kg
For block,
By substituting eq. (1) into eq. (2), thus
78
PHYSICS
CHAPTER 8
Example 8.13 :
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. 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)
R
1.50 kg
79
PHYSICS
CHAPTER 8
Solution :
a. Free body diagram :
for pulley,
for block,
80
PHYSICS
Solution :
CHAPTER 8
R  0.20 m; I  0.020 kg m 2 ; m  1.50 kg;
u  0; t  0.3 s
a. By substituting eq. (1) into eq. (2), thus
b. By using the relationship between a and , hence
81
PHYSICS
Solution :
CHAPTER 8
R  0.20 m; I  0.020 kg m 2 ; m  1.50 kg;
u  0; t  0.3 s
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
82
PHYSICS
CHAPTER 8
Exercise 8.3 :
Use gravitational acceleration, g = 9.81 m s2
1. Figure 7.16 shows four masses that are held at
the corners of a square by a very light
70 g
40 cm
frame. Calculate the moment of inertia
80 cm
of the system about an axis perpendicular
B
to the plane
A
150 g
a. through point A, and
150 g
b. through point B.
80 cm
ANS. : 0.141 kg m2; 0.211 kg m2
70 g
83
PHYSICS
CHAPTER 8
2. 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 8.17. 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 s2.
a. Sketch free body diagrams of
both objects and pulley.
b. Calculate T1 and T2 the tensions
in the string.
2.00 m s 2
T1
T2
Figure 8.17
c. Determine I.
ANS. : 10.0 N, 70.3 N; 1.88 kg m2
84
PHYSICS
CHAPTER 8
Learning Outcome:
8.4 Work and energy of rotational motion(2 hours)
At the end of this chapter, students should be able to:

Solve problems related to:
1 2

Rotational kinetic energy, K r  I 


Work, W  
Power, P  
2
85
PHYSICS
CHAPTER 8
8.4 Work and energy of rotational motion.
8.4.1 Rotational kinetic energy, Kr

Consider a rigid body rotating about the axis OZ as shown in
Figure.
Z
v1 m1
mn
r1
vn rn
O

r3
r2
m3
v2
m2
v3
Every particle in the body is in the circular motion about point O.
86
PHYSICS

CHAPTER 8
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
1
1
1
1
2
2
2
K r  m1v1  m2 v2  m3v3  ...  mn vn2
2
2
2
2
1
1
1
1
2 2
2 2
2 2
K r  m1r1 ω  m2 r2 ω  m3 r3 ω  ...  mn rn2 ω2
2
2
2
2
1 2
K r  ω m1r12  m2 r22  m3 r32  ...  mn rn2
2
n


1 2 n
2
2
K r  ω  mi ri  and  mi ri   I
2  i 1
 i 1






1 2
K r  Iω
2
87
PHYSICS

CHAPTER 8
From the formula for translational kinetic energy, Ktr
1 2
K tr  mv
2

After comparing both equations thus
 is analogous to v
I is analogous to m

For rolling body without slipping, the total kinetic energy of
the body, K is given by
K  K tr  K r
where
K tr : translati onal kinetic energy
K r : rotational kinetic energy
88
PHYSICS
CHAPTER 8
Example 8.14 :
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.
2
(Given the moment of inertia of solid sphere is I CM  mR 2and
5
the gravitational acceleration, g = 9.81 m s2)
89
PHYSICS
CHAPTER 8
Solution :
R  0.15 m; m  10.0 kg
s  0.75 m
R
h  s sin 25
v
CM
25 
a. From the principle of conservation of energy,
90
PHYSICS
CHAPTER 8
Solution : R  0.15 m; m  10.0 kg
b. The linear speed of the sphere is given by
c. By using the relationship between v and , thus
91
PHYSICS
CHAPTER 8
8.4.2 Work, W

Consider a tangential force, F acts on the solid disc of radius R
freely pivoted on an axis through O as shown in Figure 8.19.
ds
d
R
R
O

F
Figure 8.19

The work done by the tangential force is given by
dW  Fds and ds  Rdθ
dW  FRdθ
 dW  
θ2
θ1
τdθ
W

θ2
θ1
τdθ
92
PHYSICS

CHAPTER 8
If the torque is constant thus
2
W    d
1
W    2  1 
W  
is analogous to the W
 Fs
where τ : torque
Δθ : change in angular displacement
W : work done

Work-rotational kinetic energy theorem states
W  K r  K r  f  K r i
1 2 1 2
W  Iω  Iω0
2
2
93
PHYSICS
CHAPTER 8
8.4.3 Power, P

From the definition of instantaneous power,
dW
and dW  τdθ
P
dt
τdθ and dθ
ω
P
dt
dt
P  τω is analogous to the P  Fv

Caution :
 The unit of kinetic energy, work and power in the
rotational kinematics is same as their unit in translational
kinematics.
94
PHYSICS
CHAPTER 8
Example 8.15 :
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 s2)
Solution :
R  2.40 m

F
95
PHYSICS
CHAPTER 8
Solution : R  2.40 m; I
 2100 kg m 2 ; F  18.0 N;
t  15.0 s; ω0  0
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
96
PHYSICS
CHAPTER 8
Solution : R  2.40 m; I
 2100 kg m 2 ; F  18.0 N;
t  15.0 s; ω0  0
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
97
PHYSICS
CHAPTER 8
Learning Outcome:
8.5 Conservation of angular momentum (1 hour)
At the end of this chapter, students should be able to:

Define and use the formulae of angular momentum,
L  Iω

State and use the principle of conservation of angular
momentum
98
PHYSICS
CHAPTER 8
8.5 Conservation of angular
momentum

8.5.1 Angular momentum,
L

is defined as the product of the angular velocity of a body
and its moment of inertia about the rotation axis.
OR
is analogous to the p  mv
L  I
where
L : angular momentum
I : moment of inertia of a body
ω : angular velocity of a body



It is a vector quantity.
Its dimension is M L2 T1
The S.I. unit of the angular momentum is kg m2 s1.
99
PHYSICS

CHAPTER 8
The relationship between angular momentum, L with linear
momentum, p is given by
   

vector notation : L  r  p  r  mv
magnitude form :
L  rp sin θ  mvr sin θ
where

r : distance from the particle

 to the rotation axis
θ : the angle between r with v
Newton’s second law of motion in term
of linear momentum is

 
dp
F  Fnett 
dt

hence we can write the Newton’s second
 law in angular form as

 
dL
τ  τ nett 
dt
and states that a vector sum of all the torques acting on a
rigid body is proportional to the rate of change of angular
momentum.
100
PHYSICS
CHAPTER 8
8.5.2 Principle of conservation of angular momentum

states that a total angular momentum of a system about an
rotation axis is constant if no external torque acts on the
system.
OR
I  constant
Therefore
If the


τ 0

 dL
τ
0
 dt
dL  0 and dL   L f - Li


 Li   L f

101
PHYSICS
CHAPTER 8
Example 8.16 :
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. A
50 kg bag of sand falls onto the disc at the edge of the wooden
disc.
R

ω0
R
Before
After
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?
1
(Use the moment of inertia of disc is
MR2 )
2
102
PHYSICS
CHAPTER 8
Solution : R  3.00 m; ω0  4.0 rad s 1 ; mw
a. The moment of inertia of the disc,
 200 kg; mb  50 kg
The moment of inertia of the bag of sand,
By applying the principle of conservation of angular momentum,
103
PHYSICS
CHAPTER 8
Solution : R  3.00 m; ω0  4.0 rad s 1 ; mw
b. The initial rotational kinetic energy,
 200 kg; mb  50 kg
The final rotational kinetic energy,
thus
It is because
104
PHYSICS
CHAPTER 8
Example 8.17 :
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 that
is 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 s1.
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 :
105
PHYSICS
CHAPTER 8
Solution :
m  1.25 kg; ω0  18.5 rad s 1; I ss  5.43 kg m 2 ;
rb  0.759 m; ω  22.2 rad s 1;

0
rb
rb
m
m
ra ra
Before
After
106
PHYSICS
Solution :
CHAPTER 8
m  1.25 kg; ω0  18.5 rad s 1; I ss  5.43 kg m 2 ;
rb  0.759 m; ω  22.2 rad s 1;
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
107
PHYSICS
CHAPTER 8
 18.5 rad s 1; I ss  5.43 kg m 2 ;
rb  0.759 m; ω  22.2 rad s 1;
Solution : m  1.25 kg; ω0
b. The initial rotational kinetic energy is given by
and the final rotational kinetic energy is
108
PHYSICS
CHAPTER 8
Example 8.18 :
The pulley in the Figure 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 s2)
5.00 kg
7.00 m
2.00 kg
109
PHYSICS
CHAPTER 8
Solution : m1  5.00 kg; m2  2.00 kg; R  0.120 m; h  7.00 m
The moment of inertia of the pulley,
3
4
2




10
kg
10
m
2
9
2


I  0.055 g  1 cm 

5.5

10
kg
m
 1 cm2 
1
g





m2 v
m1
7.00 m
m2
Initial
v m1
7.00 m
Final
110
PHYSICS
CHAPTER 8
 5.00 kg; m2  2.00 kg; R  0.120 m;
h  7.00 m; I  5.5 10 9 kg m 2
Solution : m1
By using the principle of conservation of energy, thus
111
PHYSICS
CHAPTER 8
Exercise 8.5:
Use gravitational acceleration, g = 9.81 m s2
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 s1
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 s1 ,U think; 99.9 J
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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 s1; 2.7 x 1040 kg m2 s1
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 1 ML2 )
12
ANS. : 4.71 x 106 kg m2 s1
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CHAPTER 8
Summary:
Linear Motion
ds
v
dt
dv
a
dt
Relationship
Rotational Motion
a  r
n
m
 F  ma
p  mv
I   mi ri 2
i 1
  rF sin 
L  rp sin 
W  Fs
P  Fv
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