Download kinematics of rotation of rigid bodies

KINEMATICS OF ROTATION OF RIGID BODIES
Definitions of a Rigid Body:



7.1.
A rigid body is an object or system of particles in which the distances between
particles are fixed and remain constant.
A body that has a definite and unchanging shape and size.
A body that is none deformable, one in which the separation between all pairs of
particles remain constant.
Kinematics rotation of rigid bodies
7.1.1 Definition of angular displacement, ()

When a rigid body rotates about a fixed axis, the angular displacement is the
angle  swept out by a line passing through at any point on the body and
intersecting the axis of rotation perpendicularly.

The angle
axis
through
which
a
rigid
object
y
r
1
S.I Unit :
:
:
radian (rad)
degree ()
revolution (rev)
Equation :
 ( in radians) =
2
Arch length
s
=
Radius
r
x
rotates
about
a
fixed
 (in degree) =
=
circumference of a circle
Radius of a circle
2r
r
= 2  rad
Thus, 2 rad corresponds to 360o, so the number of degrees in one radian is
1 rad
=
360
= 57.3o
2
Definition of average angular velocity (  )
“The rate of change of angular displacement”
Average angular velocity =

Angle turned
Time taken
=
  2   1
=
t t 2  t1
7.1.2 Definition of instantaneous angular velocity (  )
“The limit of average angular velocity, (  av) as time interval, ( t ) approaches
zero, that is, the derivative of  with respect to t :
 = lim
 d 
=
dt
t
which is
 = instantaneous angular velocity
 = angular displacement
t = time interval
 is the small angular distance moved by a rotating wheel in the small time t .
7.2
Relationship between linear and rotational motion.
Objective :
At the end of this topic, students should be able to write and explain;
s  r
v  r
a t  r
a c  r 2
  2f
S
v
r

Figure 7.2. : Rotation of dics: the angle  is defined to be the arc length s devided
by the radius.
The picture focuses attention on a point P on the disc. This points starts out on the
reference line, so that  defines in radius only.
 (in radians) =
or
s  r
Arc length s

Radius
r
(7.2.1)
The angular coordinat  of a rigid body rotating around a fixed axis can be
positive or negative. If we choose positive angles to be measured counterlockwise
from the positive x-axis, than the angle in Figure 7.2 is positive.
If we instead choose the positive rotation direction to be clockwise, the  in the
Fig.7.2 is negative as mention before.
From the equation 7.2.1
s  r ,
ds d
 ( r )
dt dt
then,
v =r
d
dt
,
d
=
dt
v = r
which is
v = instantaneous linear speed of particle,
The farther a points is from the axis, the greater its linear speed
r = radius
 = instantaneous angular speed
- that is, the magnitude of the instantaneous angular velocity in
rad/s
- every points on the rigid object has same angular speed
We can represent the acceleration of a particle moving in a circle in terms of its
centripetal and tangential components arad and atan (Figure 7.2.2)
arad
atan
arad
Figure 7.2.2
They can be write as
a tan 
v2
r
(7.2.3)
(always points toward the axis of rotation)
=  2r
a tan 
dv
dt
and
=r
d
dt
= r
(7.2.4)
(always tangent to the circular path of the particle)
but when we combine both of equations, we get magnitude a vector of acceleration
a  a rad  a tan
2
=
2
(total linear acceleration)
r 2 4  r 2 2
= r 4  2
In the study of linear motion, we found that the simplest form of accelerated motion to
analyze is motion under constant linear acceleration. Like wise for rotational motion
about fixed axis, the simplest accelerated motion to analyze is motion under constant
angular acceleration. Therefore we next develop kinematic relationships for rotational
motion under contant angular acceleration.
A comparison of kinematic equation for rotational and linear motion under constant
acceleration.
Rotational Motion About a fixed axis
with  = constant variable =  and 
Linear Motion with a constant variables :
x and v
 =  0  t
v  v0  at
1
2
 =  0 t  ( 0   )t
1
2
x  x0 
1
(v0  v)t
2
1 2
at
2
 =  0   0 t  t 2
x  x0  v0 t 
 2  0 2  2 (   0 )
v 2  v0  2a( x  x0 )
2
Example : Relationship between angular ang linear quantities
A racing car travels on a circular track of radius 250 m. If the car moves with a constant
linear spead of 45 m/s, find
a. its angular speed
b. the magnitude and direction of its acceleration
Solution :
v 45m / s

 0.180rad / s
r
250m
v 2 (45m / s) 2
b. a r 

 8.10m / s 2
r
250m
a. v  r ;

Exercise :
A racing car travels on a circular track of radius R. If the car moves with a constant linear
with a constant linear speed , find
a. its angular speed
b. the magnitude and direction of its acceleration
7.3
Rotational motion with uniform angular acceleration
Objectives
: 1. Describe and use equations for rotational motion with constant angular
acceleration
 = ½ ( o +  )
 = ot + ½ t2
 = o + t
2 = o2 + 2
2. Relate them with their corresponding quantities in linear motion.
It is understood that straight-line motion is particulary simple when the acceleration is
constant. This also true of rotational motion about a fixed axis. When the angular
acceleration is constant, we can derive equations for angular velocity and angular
position using exactly the same procedure that we used for straight-line motion.
Table 7.3
: comparison of linear and angular motion with constant acceleration
Straight-line with constant linear
Acceleration
Fixed-axis rotational with constant angular
acceleration
a = constant
 = constant
 = o + at
 = o + t
x = xo + ot + ½ at2
 = o + ot + ½ t2
2 = o2 + 2a ( x- xo )
2 = o2 + 2  (  - o )
x – xo = ½ (  + o )t
 - o = ½ (  + o )t
Therefore, we next develop kinemetic relationships for rotational motion constant angular
acceleration. We can rewrite instantaneous angular acceleration in the form,
d =  dt
………..
7.3.1
and let  = o at to = 0, we can integrate this expression directly
 = o + t
………..
7.3.2
(  = constant )
Likewise, substituting equation 7.3.2 into equation 7.3.1 and integrating once more ( with
 = o at to = 0 ) we get
 = o + ot + ½ t
……….
7.3.3
If we eleminate t from equations 7.2.2 and 7.2.3, we get
2 = o2 + 2  (  - o )
………..
7.3.4
Notice that these kinemetic expressions for rotational motion under constant angular
acceleration are of the same from as those for linear motion under constant linear
acceleration with the substitutions
X  ,    and a  
Furthermore, the expressions are valid for both rigid-body rotation and particle motion
about a fixed axis.
Example :
A wheel rotates with a constant angular acceleration of 3.50 rad s-2. If the angular speed
of the wheel is 2.00 rad s-1 at to = 0,
a)
What angle does the wheel rotate through in 2.00 s?
b)
What is the angular speed at t = 2.00 s?
Solution :
a)
 - o = ot + ½ t2
= ( 2.00 rad s-1 )(2.00 s ) +1/2 ( 3.5 rad s-2 )( 2.00 s )2
= 11.0 rad
= 630 o
= 1.75 putaran
b)
 = o + t
= ( 2.00 rad s-1) + ( 3.50 rad s-2 )( 2.00 s )
= 9.00 rad s-1
Exercise :
Find the angle that the wheel rotates through between t = 2.00 s and 3.00 s.
Answer : 10.8 rad
7.4 Centre of mass, Centre of Gravity, Moment of inertia and Torque
7.4.1 Centre of mass (COM)
COM of a rigid body is the point where its entire mass can be considered to act
when caalculating the
xcm 
=
m1 x1  m2 x2  ..........mn xn
m1  m2  ........mn
m x
m
i
i
i
Example:
1.
For the system of masses shown as shown in figure below, find the centre
of mass.
2.0 kg
5.0 m
3.0 kg
3.0 m
3.0 m
1.0 kg
5.0 m
4.0 kg
Solution:
m3 =2.0 kg
(0,3)
(0,0)
m1=1.0 kg
m4=3.0 kg
(5,3)
(5,0)
m2=4.0 kg
Centre of mass
(Xcm , Ycm) =
m x , m y
m m
i
i
i
m x
m
i
i
i
i
i
i
i
=
(1)(1)  (4)(5)  (2)(0)  (3)(5)
1.0  4.0  2.0  3.0
=
0  20  0  15
10
i
m y
m
i
= 3.5 cm
(1)(0)  (4)(0)  (2)(3)  (3)(3)
=
1.0  4.0  2.0  3.0
=
0  0  6 9
10
= 1.5 cm
Centre of mass = (3.5,1.5)
Centre of mass
1.5 cm
3.5 cm
7.4.2
Centre of gravity (COG)
COG of a rigid body is the point where the entire weight can be considered to act.
xcg 
=
=
m1 gx1  m2 gx2  ..........mn gxn
(m1  m2  ........mn ) g
( mi xi ) g
( mi ) g
m x
m
i
i
i
For a regular-shaped object the centre of mass is the centre of gravity.
7.4.3 Moment Inertia
Moment inertia of a body is the sum of the moments inertia of each individual
element of the body.
I
= m1r1  m2 r2  ............mn rn
2
n
I   mi ri
2
2
2
i 1
which is
I = moment inertia
m = mass of the element
r = perpendicular radial distance of each particle from the axis of
rotation.
Common Moments of Inertia
Example:
Two particles of mass 5.0 kg and 7.0 kg are mounted 4.0 m apart on a light rod (whose
mass is negligible). Calculate the moment of inertia of the system
a)
when rotated about an axis passing halfway between the masses.
b)
when the system rotates about an axis located 0.5 m to the left of the 5.0 kg mass.
Solution:
a)
Both particles are the same distance 2.0 m, for the axis of rotation.
I =  mr 2
= ( 5.0)(2.0)2 + (7.0)(2.0)2
= 48 kg m2
b)
The 5.0 kg mass is now 0.5 m from the axis and the 7.0 kg mass is
4.50 m from the axis. Then
I =  mr 2
= (5.0)(0.5)2 + (7.0)(4.5)2
= 1.3 + 142
= 143.3 kg m2
7.4.4 Torque
Torque is a measure of how much a force acting on an object causes that object to rotate.
The object rotates about an axis called pivot point. The distance from the pivot point to
the point where theforce acts is called the moment art, and is denoted by r.
Torque is defined as
  r x F  rF sin 
Imagine a force F acting on some object at a
distance r from its axis of rotation. We can break
up the force into tangential (Ftan), radial (Frad)
(see Figure 1). (This is assuming a twodimensional scenario. For three dimensions -- a
more realistic, but also more complicated situation
-- we have three components of force: the
tangential component Ftan, the radial component
Frad and the z-component Fz. All components of
force are mutually perpendicular, or normal.)
From Newton's Second Law,
Ftan = m atan
Figure 1 Radial and Tangential
Components of Force, two
dimensions
However, we know that angular acceleration, ,
and the tangential acceleration atan are related by:
atan = r 
Then,
Ftan = m r 
If we multiply both sides by r (the moment arm),
the equation becomes
Ftan r = m r 2 
Note that the radial component of the force goes
through the axis of rotation, and so has no
contribution to torque. The left hand side of the
equation is torque. For a whole object, there may
Figure 2 Radial, Tangential and zComponents of Force, three
dimensions
be many torques. So the sum of the torques is
equal to the moment of inertia (of a particle mass,
which is the assumption in this derivation),
I = m r2 multiplied by the angular acceleration, .
If we make an analogy between translational and rotational motion, then this relation
between torque and angular acceleration is analogous to the Newton's Second Law.
Namely, taking torque to be analogous to force, moment of inertia analogous to mass,
and angular acceleration analogous to acceleration, then we have an equation very much
like the Second Law.
Example
A cable is wrapped around a uniform, solid cylinder of radius 'R' and mass 'M'. The
cylinder rotates about its axis, and the cable unwinds without stretching or pulling. If the
cable is pulled with a force of 'F' Newtons, what is its acceleration?
Hints
What is the moment of inertia for a uniform, solid cylinder, with the axis through its
center?
What is the torque exerted?
What is the relationship between acceleration of the cable, a, and the angular
acceleration, ?
Solution
Drawing a diagram will aid us in solving this problem; refer to Figure 1.
For a uniform, solid cylinder of radius R and mass M,
the moment of inertia is:
The torque exerted by a force F is found to be:
=RF
since the force is perpendicular to the moment arm.
(That is,  = 90o, so sin(90)=1.)
We also learned in this section that
=I
Solving for , we get:
Figure 1 Diagram of the cable
unwinding from a cylinder.
7.5
Rotational Kinetic Energy and Power
Objective:
a)
b)
Distinguish between pure translational and pure rotational motions of a
rigid body
State the condition for rolling without slipping
7.5.1 Rotational Work:
It is necessary to apply torque to mass of a body which is initially at rest to start
rotating about a fixed axis. For a single force F acting tangentially along an arc
length s: is given by
W = Fs
Thus for a single torque acting though an angle of rotational , work is given by
W = 
where  = Torque
 = angular displacement
7.5.2 Rolling Without Slipping:


If an object rolls without slipping, then the bottom of the rolling object (at the
point of contact) must be momentarily at rest relative to a fixed observer.
Rolling without slipping can be thought of as the motion of the center of mass
(fixed observer) plus rotational motion about its center of mass (observer moving
with the object).
i)
ii)
iii)

in pure translational motion – all the particles of an object have the same
instantaneous velocity
For pure rotational motion – all the particles of an object have the same
instantaneous angular velocity
Rolling is a combination of translational + rotational motion
In the case for rolling without slipping, the distance, the velocity, and the
acceleration of the center of mass is directly related to the angle of rotation, the
angular velocity, and the angular acceleration about the center of mass.
7.5.3 Rotational Kinetic energy
The kinetic energy of a body undergoing translational motion, K.Etrans is given
by the quantity
K.Etrans = ½ mv2
By analogy the K.E of rotational kinetic energy
K.Erot =
=
½ I2
½ (mR2)( 2)
Since  mR2 is moment inertia, I,
So the K.Erot = ½ I2
The moment inertia about the fixed axis is given by the parallel axis theorem
I = Icm + mR2
When R is the radius of the cyclinder.
Then, K= ½ I2
= ½ ( Icm + mR2 )2
= ½ Icm 2 + ½ m R2 2
Since there is no slipping,
cm = R
And
K = ½ Icm 2 + ½ m cm2
Where cm is the linear velocity of the centre mass.
Icm is the moment of inertia about axis through the centre of mass
 is the angular velocity
m is the total of mass
Thus the total kinetic energy of a rolling without slipping object is the sum of :
i.
the translational kinetic energy of the centre of mass and
ii.
the rotational kinetic energy relative to harizontal axis through the centre
of mass
7.5.4 The work energy theorem and kinetic energy
The relationship between the net ratational work and change in rotational kinetic
energy can be derived as follows,
=
=I
For a constant angular acceleration
2 = o2 + 2  
2 = 2 - o2
 = (2 - o2) / 2
 = I (2- o2) / 2
 = ½ I2 – ½ Io2
therefore
Thus the net rotational work is equal to the change in the rotational kinetic
energy.
Example:
Find the rotational kinetic energy of the earth due to its daily rotation on its axis.
Assume it to be a uniform sphere, m = 5.89 x 1024 kg, r = 6.37 x 106 m.
Solution:
For a uniform sphere, I = 2/5 mr2
= (2/5)(5.98 x 1024 kg)(6.37 x 106 m)2
= 9.71 x 1037 kgm2
The angular relocity of the earth is
W = (1 rev/day)(1/86,400 day/s)(2 rad/rev) = 7.27 x 10-5 rad/s
K. Er = ½ I2
= ½ (9.71 x 1037 kg.m2)(7.27 x 10-5 rad/s)2
= 2.56 x 1029 J
7.5.5 Rotational power
Rotational power, P is the rate of doing rotational work and given by
P = W/t
= (/t)
= 
Example : Rotational kinetic energy
The center of mass of a pitched baseball (radius = 3.8 cm) moves at 38 m/s . The ball
spins about an axis through its center of mass with an angular speed of 125 rad/s, find
a. Calculate the ratio of the rotational energy to the translational kinetic energy
Treat the ball as uniform sphere
b. The total energy
Solution :
a.
b.
 = 38 m/s
 = 125 rad/s
12 2 2
1 2
 mr 
I
25

2
RATIO 

1
1
mv 2
mv 2
2
2
2
0.382 1252 1
RATIO  5

160
382
1
1
mv 2  I 2
2
2
1
12
2
2
2
 m38   m0.3 125 
2
25

 1003.25m Joule
K
7.6 Angular Momentum
Definition: Angular Momentum
The angular momentum, L of a rigid body with moment of inertia I rotating with
angular velocity is:
(14)
L=I .
This is the rotational analogue of linear momentum.
Note: The units of angular momentum are kg m 2/s.
The dynamical torque equation can be written in terms of angular momemtum:
= I
=
I
=
=
(15)
.
This is the rotational analogue of Newton's second law:
Idea: Conservation of Angular Momentum
In the absence of external torque (
body is conserved
L = 0 , or
=
.
= 0) , the angular momentum of a rotating rigid
(16)
Ii
= If
.
For systems that consist of many rigid bodies and/or particles, the total angular
momentum about any axis is the sum of the individual angular momenta. The
conservation of angular moment also applies to such systems. In the absence of
external forces acting on the system, the total angular momentum of the system remains
constant.
Note:


Angular momentum and torque are really vector quantities. Their direction is
always along the axis of rotation. For two dimensional motion they always point
either out of the page (if they are positive) or into the page (if they are negative).
Thus we don't need to explicitly consider their vector properties. We need only
insure that we have the correct sign.
Table 8.1 gives the rotational analogues of some linear quantities.
Table 8.1: Linear vs. Angular Quantities
Linear Stuff
Angular Stuff
Quantity
Units
a
m/s 2
m
kg
p = mv
kg m/s
KEt = mv 2
F = ma
Quantity
Units
rads / s 2
I
L=I
J
kg m 2
kg m 2 /s
J
KEr = I
N
=I
N m
Example : Angular momentum
At a certain instant the position of a stone in a sling is given r = (1.7 i) m. The linear
momentum p of the stone is (12 j ) kg m/s. Calculate its angular momentum.
Solution :
L=rxp
= ( 1.7 i ) x ( 12 j )
= 20.4 (k) kgm2/s
Example : Angular momentum
A light rigid rod 1.00 m in length rotates in the xy plane about a pivot through the rod’s
center. Two particles of masses 4.00 kg and 3.00 kg are connected to its ends. Determine
the angular momentum of the system about the origin at the instant the speed of each
particle is 5.00 m/s.
V
3.00 kg
1.00 m
4.00 kg
V
L   m i v i ri
= (4 +3)(5)(0.5)
= 17.5 kgm2/s