Upload
lozzzzz
View
84
Download
5
Embed Size (px)
DESCRIPTION
guc physics
Citation preview
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 1/22
Dr. Aladdin. M Abdul-latif
Rotation of Rigid Bodies
And
Angular MomentumIn mechanics , a rigid body is defined as a system of particles whose relative
distances remain constant. This means that the body does not change its shape
during its motion.
A rigid body undergoes translation when all of its constituent particles are moving
on parallel trajectories and it undergoes rotation when all of its particles are moving
on circular trajectories about an axis of rotation. It can also undergo simultaneous
translation and rotation.
In rotation about a fixed axis, the rigid body is rotating about
a stationary axis and has no translational motion. Every point
of the rigid body undergoes circular motion about the axis of
rotation. The angular position of the body describes its
orientation relative to a reference line.
In Fig.(1) A compact disc rotating about a fixed axis through
O perpendicular to the plane of the figure. (a) In order to
define angular position for the disc, a fixed reference line ischosen. A particle at P is located at a distance r from the
rotation axis at O. (b) As the disc rotates, point P moves
through an arc length s on a circular path of radius r.
Angular Displacement.
As the body rotates, the angular position change from some initial angle i at initial
time ti to some final angle f at a final time t f, and a quantity called the angular
displacement
. = f i
measures how far the body has rotated from its
reference line in a time interval t = t f - ti, . The angular
displacement is measured in degrees, rads or in
revolutions.
(rad) =
180
(degree)
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 2/22
Dr. Aladdin. M Abdul-latif 2
Average angular speed ( )
It is defined as the ratio of the angular displacement, , to the time interval t in
which the angular displacement occurs.
=t
The instantaneous angular speed of the body ( )
It is the limit of the average angular speed , / t as the time interval t approaches
zero
=
t t
0lim =
dt
d
The angular velocity ()
A vector quantity (vector ), the angular velocity , can also be defined, its magnitude
is the angular speed , and its direction is that of the
axis of rotation. The sense of rotation is determined
using the "right-hand rule." For example, if the axis of
rotation of the rigid body is vertical, the fingers of the
right hand curl in the direction specifying the sense of
the rotation, i.e. either clockwise or counterclockwise,
and the extended thumb points in the direction of the
axis of rotation, i.e. vertically. When the angular speed
varies or when the orientation of the rotation axis
changes, the angular velocity also changes.
Whenever the magnitude or direction of the angular
velocity changes, the rigid body acquires an angular acceleration.
Average angular acceleration: (symbol = )
It is the ratio of the change in angular speed to the time, t , during which the
change in the angular speed occurs.
.
also a vector quantity, which is defined as the rate of change of the angular velocity
with time, d w/dt. It is expressed in radians per second squared.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 3/22
Dr. Aladdin. M Abdul-latif 3
Instantaneous angular acceleration: The limit of the ratio, / t (the average
angular acceleration), as the time interval t approaches zero.
=
t
t
0lim =
dt
d
Angular acceleration about a given point is a vector quantity that has the same the
same direction of the angular velocity .
Rotational kinematics
The simplest accelerated rotational motion is the motion with constant acceleration
. To obtain the kinematic equations for uniformly accelerated rotational motion we
do the followings
Starting with the equationdt
d , and taking ti = 0 and t f = t,
d = dt
Integrating the above equation gives
t
In this equation is the angular speed at t=0, and is the angular speed at t=t.
Using the above equation in the equation
=dt
d
and integrating once more we get
t ½ t
Substituting for t from equation into equation we get
Also we can use the average angular velocity = ½ ) for uniformly
accelerated motion to get the
displacement as
= ½ )t (****)
These equation are of the same
mathematical form of the
kinematic equation for
uniformly accelerated linear
motion.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 4/22
Dr. Aladdin. M Abdul-latif 4
Relationship between angular and linear quantities.
When a rigid body rotates about a fixed axis, every point of the rigid body
undergoes circular motion about the axis of rotation. In the given figure The point P
moves about the axis of rotation passing by the point o in a circle of radius r . At anytime the angle is related to the arc length by
S = r
The point P has a tangential velocity v that is always tangent to
the circular path of radius r. To get on the instantaneous linear
speed v we take the derivative of the above equation, noting
that the radius r is constant
(ds/dt) is the linear tangential speed v and d/dt is the angular speed . Thus
v = r
The above equation gives the relation between linear and angular speed
The tangential acceleration of the particle is obtained by differentiating the
tangential speed v with respect to time.
at = r dt d r
dt dv
this equation gives the relationship between the tangential acceleration at and the
angular acceleration .
In addition to the tangential acceleration at the point has a
centripetal acceleration ( ac = v /r = r ) so that the net
acceleration anet is given by
anet = 22
ct aa
LINEAR Connection ROTATIONAL
Position
x,y,s,d
SI: <meters>
1 m = 100 cm = .001 km
s = r Angle
,
SI: <radians>
2
rad = 360
o
= 1 Rev
dt
d r
dt
ds
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 5/22
Dr. Aladdin. M Abdul-latif 5
Velocity
SI: <m/s>
vT = r Angular velocity
SI: <rad/s>
Acceleration
SI: <m/s2>
Tangential
Acceleration
a T = r Angular Acceleration
SI: <rad/s2>
Rotational Kinetic energy
A rotating rigid body always has kinetic energy , even when it has no translational
motion. If we consider a point particle of mass mi on the rotating object wit a linear
velocity vi Its kinetic energy is given by
K = ½ mivi2
Since vi
= r i
we get for the rotational kinetic energy K = ½ (miri2) 2
The quantity miri2 is called the moment of inertia of the ith
particle and is given the symbol I.
The total kinetic energy of the rotating rigid object is the
sum of the kinetic energies of the individual particles
K = Σ ½ (miri2) 2
The quantity
Σ(mi
ri2
) is the moment of inertia of the object.
I= Σ(miri2)
For a body which is considered as a continuous distribution of particles , the moment
of inertia is calculated by the integral
I = ∫ r2dm
The moment of inertia I of a body about an axis is a measure of its rotational inertia.
The higher the value of I, the more difficult it is to change the state of the body's
rotational motion.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 6/22
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 7/22
Dr. Aladdin. M Abdul-latif 7
Example 3Moment of Inertia of a Uniform Rigid RodThe shaded area has a mass dm = dx. Then the moment of inertia is
Divide the cylinder into concentric shells with radius r, thickness dr and length Ldm = dV = 2 Lr dr
Then for I
/ 22 2
/ 2
21
12
L
y L
M I r dm x dx
L
I ML
Moment of Inertia of a Uniform Solid Cylinder
2 2
2
2
1
2
z
z
I r dm r Lr dr
I MR
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 8/22
Dr. Aladdin. M Abdul-latif 8
The parallel axis theoremAccording to the parallel axis theorem; the moment of inertia I cm of an object of mass
M about an axis passing through its center of mass is related to the moment of inertia
I o of that object about an axis through o which is parallel to the original one but
displaced from it by a distance D by
I P = I cm + MD 2
Example
Moment of Inertia for a Rod Rotating
Around One End
The moment of inertia of the rod about its center is
D is ½ L
Therefore,
D
21
12CM
I ML
2
CM
2
2 21 112 2 3
I I MD
LI ML M ML
D
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 9/22
Dr. Aladdin. M Abdul-latif 9
Torque
Torque, , is the tendency of a force to rotate an object about some axis. Angular
acceleration of an object about a given axis is the result of a torque about this axis
applied to the object. A torque () is a vector quantity but we will consider only itsmagnitude here. The magnitude of the torque is called the moment and it is defined
as the product of the moment arm which is the perpendicular distance from the
center of rotation, or pivot, to the line of action of the force and magnitude of the
force vector.
= r F sin = dFF is the force
is the angle the force makes with the position vector r
d is the moment arm (or lever arm) of the force
The moment arm, d , is the perpendicular distance from the axis of rotation to a linedrawn along the direction of the force
d = r sin
Thus, the greater the force applied or the
longer the moment arm, the larger the
resulting torque
We can see from the opposite figure that, the
moment of a force is the product of the
distance r and the component of the force(F sin ) perpendicular to r
The units of torque
The SI units of torque are N.m. Although torque is a force multiplied by a distance, it
is very different from work and energy. The units for torque are reported in N.m and
not changed to Joules
Torque direction
Torque will have direction. If the turning tendency of the force is counterclockwise,
the torque will be positive. If the turning tendency is clockwise, the torque will be
negative
Net Torque
The force 1
F
will tend to cause a counterclockwise
rotation about O The force2
F
will tend to cause a
clockwise rotation about O
=
= F 1d1 – F 2d2
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 10/22
Dr. Aladdin. M Abdul-latif 1
The Torque is defined as the cross product of the position vector r and the force
vector F. The moment is equivalent to the magnitude of the torque vector .
= r x F.
The torque is the vector (or cross) product of the position vector and the force vector F
For r = xi + y j + zk , and F= Fxi + Fy j + Fzk , the torque is given by :
= r x F =
z y x F F F
z y x
k ji
= i z y F F
z y-j
z x F F
z x+ k
y x F F
y x
= i (yFz-zFy) - j(xFz-zFx) + k(xFy -yFx)
The torque vector lies in a direction perpendicular to the plane
formed by the position vector r and the force vector F. Example
Given the force and location
Find the torque produced
Solution
Torque and Angular Acceleration
Consider a particle of mass m rotating in a circle of radius r under
the influence of tangential force. The tangential force provides a
tangential acceleration:
F t = mat
The radial force, causes the particle to move in a circular path .
The magnitude of the torque produced by around the center of
the circle is
F t r = (mat ) rThe tangential acceleration is related to the angular acceleration
= (mat ) r = (mra) r = (mr 2) a
Since mr 2 is the moment of inertia of the particle, therefore
= I
The torque is directly proportional to the angular acceleration and the constant ofproportionality is the moment of inertia
ˆ ˆ(2.00 3.00 )
ˆ ˆ(4.00 5.00 )
N
m
F i j
r i j
ˆ ˆ ˆ ˆ [(4.00 5.00 )N] [(2.00 3.00 )m]
ˆ ˆ ˆ ˆ[(4.00)(2.00) (4.00)(3.00)
ˆ ˆ ˆ ˆ(5.00)(2.00) (5.00)(3.00)
ˆ2.0 N m
r F i j i j
i i i j
j i i j
k
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 11/22
Dr. Aladdin. M Abdul-latif
Newton's second law as applied to rotational motion can be stated as :
When a non zero net torque acts on an object , the object gains angular acceleration
so that
= I
Examples(1) Torque and Angular Acceleration, Wheel Example
The wheel is rotating and so we apply
=TR= I
The tension supplies the tangential force. The mass is moving in
a straight line , so apply Newton’s Second Law
F = ma = mg - T
a = R
(2) Atwood’s Machine Example
Two blocks having masses m1 and m2 are connected to each other by a light cord that
passes over two identical frictionless pulleys, each having a moment of inertia I and
radius R , as shown in Figure. Find the acceleration of each block and the tensions T 1 ,
T 2 , and T 3 in the cord. (Assume no slipping between cord and pulleys.)
Solution
In this example, we have two masses m1 and
m2 , and two pulleys and each of the pulleys
has mass and moments of inertia about their
axis of rotation.
In the opposite figure we have (a) Atwood’s
machine. (b) Free-body diagrams for the
blocks. (c) Free-body diagrams for the
pulleys, where m pg represents the
gravitational force acting on each pulley.
We shall define the downward direction as positive for m1 and upward as thepositive direction for m2. This allows us to represent the acceleration of both masses
by a single variable a and also enables us to relate a positive a to a positive
(counterclockwise) angular acceleration (of the pulleys).
Let us write Newton’s second law of motion for each block, using the free-body
diagrams for the two blocks as shown in Figure 10.21b:
m1 g -T 1 = m1a (1)
T 3 - m2 g = m2a (2)
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 12/22
Dr. Aladdin. M Abdul-latif 2
The net torque about the axle for the pulley on the left is (T 1 - T 2)R , while the net
torque for the pulley on the right is (T 2 - T 3)R. Using the relation
where
I = ½ MR2 ( for each pulley and noting that each pulley has the same angular
acceleration ), we obtain
(T 1 - T 2)R = I (3)(T 2 % T 3)R = I (4)
In addition to the above 4 equations we have the equation a = R
These equations can be solved simultaneously. Adding Equations (3) and (4) gives
(T 1 - T 3)R = 2I (5)
Adding Equations (1) and (2) gives (T 3 - T 1 ) - m1 g - m2 g = (m1 -m2)a
T 1 - T 3 = (m1 -m2) g - (m1 - m2)a (6)
Substituting Equation (6) into Equation (5), we have
[(m1 - m2) g - (m1 -m2)a]R = 2I a
Because = a / R , this expression can be simplified to
(m1 - m2) g - (m1 - m2)a= 2I a / R 2
a
Note that if m1 > m2 , the acceleration is positive; this means that the left block
accelerates downward, the right block accelerates upward, and both pulleys
accelerate counterclockwise. If m1 < m2 , the acceleration is negative and the motions
are reversed. If m1 = m2 , no acceleration occurs at all.
The expression for a can be substituted into Equations (1) and (2) to give T 1 and T 3.
From Equation (1),
T 1 = m1 g - m1a = m1( g - a)
T 1
T 1
Similarly, from Equation (2),
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 13/22
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 14/22
Dr. Aladdin. M Abdul-latif 4
The angular speed changes from i to f . The work kinetic energy theorem for
rotational motion states that :
The net work done by external forces in rotating a symmetric rigid object about a
fixed axis equals the change in the object’s rotational energy.
In general the net work done by external forces on an object is the change in its total
kinetic energy, which is the sum of the translational and rotational kinetic energies.
For example, when a pitcher throws a baseball, the work done by the pitcher’s hands
appears as kinetic energy associated with the ball moving through space as well as
rotational kinetic energy associated with the spinning of the ball.
Exdample
A uniform rod of length L and mass M is free to rotate on a frictionless pin passing
through one end (Fig 10.24). The rod is released from rest in the horizontal position.
(A) What is its angular speed when it reaches its lowest position?
Solution To conceptualize this problem, consider Figure 10.24 and imagine the rod
rotating downward through a
A uniform rigid rod pivoted at O rotatesin a vertical plane under the action of the gravitational force
To analyze the problem, we consider the mechanicalenergy of the system of the rod and the Earth. We
choose the configuration in which the rod is hanging
straight down as the reference configuration for
gravitational potential energy and assign a value of
zero for this configuration.
When the rod is in the horizontal position, it has no rotational kinetic energy. The
potential energy of the system in this configuration relative to the reference
configuration is MgL/2 because the center of mass of the rod is at a height L/2 higher
than its position in the reference configuration.
When the rod reaches its lowest position, the energy is entirely rotational kinetic
energy ½ I2 , where I is the moment of inertia about the pivot, and the potential
energy of the system is zero. Because I = ⅓ ML2 and because the system is isolated
with no nonconservative forces acting, we apply conservation of mechanical energy
for the system:
K f - U f = K i - U i
½ I
2 + 0 = ½ (⅓ ML2) 2 = 0 + ½ MgL
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 15/22
Dr. Aladdin. M Abdul-latif 5
ROLLING MOTION OF A RIGID OBJECT
Rolling motion is the rotation of a rigid object about a moving axis. The simplest rolling
motion is the motion of objects having high degree of symmetry, such as a cylinder, a sphre,
or a hoop.
R
S=R
Figure ( ) In Pure rolling motion, as the cylinder rotatesthrough an angle q its cdenter of mass moves a
linear distance S= R
If a cylinder rolls without slipping on a flat surface (such a motion is called pure rolling
motion) a simple relationship exists between its rotational and translational motions.
The linear displacement of the center of mass
As the cylinder rotats through an angle , its center of mass moves a distance S so that S is
given by
S= R
The linear speed of the center of mass
The linear speed of the center of mass for pure rolling motion is given byvCM =
R
dt
d R
dt
ds (2)
1Where is the angular velocity of the cylindder. The above equation holds whenever a
cylinder or a spher rolss without slipping and is the condition for pure rolling motion.
The acceleration of the center of massThe acceleration of the center of mass in a pure rolling motion is given by
aCM = R dt
d =R
Where is the angular acceleration of the cylinder
P’ vp’
The linear velocity of a point on the rolling objectAs the cylinder rolls the point of contact P between the R
cylinder and the surface on which it rolls is at rest relative CM vCM
to the surface. in the figure. R
All points on the rolling cylinder have the same angular
speed but different linear velocities v.
The magnitude of the linear velocity v of a point on the rolling object depends on the distance
between this point and the the point P. vP’ = 2R while vCM=R .
P
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 16/22
Dr. Aladdin. M Abdul-latif 6
The kinetic energy of the rolling objectThe linetic energy is given by
K =2
1 Ip
p is the moment of inertia of the object about an axis passing through the point P.
Applying the parallel axis theorem
Ip =ICM + MR 2 (5)
In this equation M is the mass of the rolling object and ICM is the moment of inertia about a
parallel axis through the center of mass. Using equation (5) in equation (4) we obtain
K =2
1 ICM
2
1 MR
2
and using the relation vCM=R we get
K =2
1 ICM
2
1 M VCM
2
This equation contains two terms
(a) the first term (2
1 ICM
) is interpreted as the rotational enrgy of the rolling object
about an axis through its center of mass.
(b) The second term is2
1 M VCM
2 is interpreted as the translational thekinetic energy of
the center of mass.
Example 1
A solid cylinderical disc of mass M =1.4 Kg and whose radius is R= 8.5 cm, rolls across ahorizontal table at a speed v of 15 cm/s. (a) what is the instantaneous velocity of the top of
this disk
Yhe speed of the rolling object is the speed of its center of mass
VCM =15 cm/s
Vtop= 2VCM= 2 x 15=30.0 cm/s
(b) What is the angular speed of gthe rolling disk
= R
V CM =cm
scm
5.8
/15= 1.8 rad/s= 0.28 rev/s
© what is the kinetic energy K of the rolling disk
K =2
1 ICM
2
1 M VCM2
=2
1 (
2
1MR
2) (VCM/R)
2
2
1 M VCM
2
4
3V
=4
3 (1.4 kg)(.15m/s)2 = 0,24 J
(c) What fraction of this energy is translational and is rotational.
let this fraction be f
f =2
4
3
2
2
1
CM
CM
MV
MV = 0.67 = 67%
The remaining 33% is associated with the rotation about an axis through the center of
mass.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 17/22
Dr. Aladdin. M Abdul-latif 7
Exercise 1Repeate example1 for (a) a spher of radius R and moment of inertia (2/5 MR 2)
(c) a hoop of radius R and moment of inertia (MR 2)
Angular momentum
As the linear momentum P is defined as the product of the mass m and linearvelocity v [P= mv] , the angular momentum L of an object rotating about a fixed axis
with a fixed angular velocity is defined as the product of the moment of inertia I
and the angular velocity .
L =
L is a vector quantity that has the same direction of . and the direction of is
parallel to the fixed axis.
Note that in some cases the angular momentum has parallel and non parallel
components to the axis of rotation.
Newton's Second Law as Applied to Accelerated Rotational MotionAs the net force acting on an object of mass m is related to the linear acceleration a
by
F = ma , and to the rate of change of linear momentum dP/dt by
F = dP/dt , the
net torque acting on a body of a fixed moment of inertia I is related to the angular
acceleration , according to Newton's second law, by
Since = d/dt, then = I (d /dt) = d(I )/dt = dL/dt ,so the torque is relate to the rate
of change of the angular momentum dL/dt by
=dt
dL
This equation tells us that as the net external force equals the time rate of change of
the linear momentum, the net external torque on an object equals the time rate of
change of the angular momentum.
This equation is valid in both cases of I changing or constant.
Relationship between Angular Momentum and Linear Momentum. Consider a particle of mass m located at the vector position r
and moving with linear momentum P as in the opposite
figure. Since = r x F , and since F=dt
P d
therefore
= r xdt
P d
= P xr dt
d
=dt
Ld
Comparing this equation to dt Ld /
, we get
P xr L
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 18/22
Dr. Aladdin. M Abdul-latif 8
The instantaneous angular momentum L of a particle relative to the origin O is
defined by the cross product of the particle’s instantaneous position vector r and its
instantaneous linear momentum p. That is L is a vector quantity whose direction is
perpendicular to both r and P.
The magnitude of the angular momentum is L= rP sin = r m v sin for = 90, sin 90
= 1, therefore
L= rP = r m v
Using v= r in the above equation we easily get
L= mr 2 = I .
Angular Momentum of a Particle in a circular orbit
The linear momentum vector of a particle of mass m rotating
in a circular path of radius r (shown in the figure) is vm P
.
The angular momentum vector P xr L
is pointed out of the
diagram. The magnitude of this angular momentum vector is
L = mvr sin 90o = mvr sin 90o. = mvr
We have used = 90o because the velocity vector v
is perpendicular to r
.
For a particle in uniform circular motion both of v and r are constants, therefore the
particle has a constant angular momentum about an axis through the center of its
path
Angular Momentum of a System of Particles
The total angular momentum of a system of particles is defined as the vector sum of
the angular momenta of the individual particles
ntot L L L L ......21
Differentiating with respect to time
i
i
i
ito t
dt
dL
dt
dL
The torques acting on the particles of the system are those associated with internal
forces between particles and those associated with external forces. However, the net
torque associated with all internal forces is zero because the torques of each action
reaction pair cancels.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 19/22
Dr. Aladdin. M Abdul-latif 9
Therefore, The net external torque acting on a system about some axis passing
through an origin in an inertial frame equals the time rate of change of the total
angular momentum of the system about that origin
Although we do not prove it here, the following statement is an important theoremconcerning the angular momentum of a system relative to the system’s center of m
The resultant torque acting on a system about an axis through the center of mass
equals the time rate of change of angular momentum of the system regardless of
the motion of the center of mass.
This theorem applies even if the center of mass is accelerating, provided ! and L are
evaluated relative to the center of mass.
ExampleTwo Connected Objects
A sphere of mass m1 and a block of mass m2 are connected by a light cord that passes
over a pulley, as shown in the opposite Figure. The
radius of the pulley is R , and the mass of the rim is M.
The spokes of the pulley have negligible mass. The
block slides on a frictionless, horizontal surface. Find an
expression for the linear acceleration of the two objects,
using the concepts of angular momentum and torque.
Solution
We need to determine the angular momentum of the
system, which consists of the two objects and the pulley. We will calculate the
angular momentum about an axis that coincides with the axle of the pulley.
The angular momentum of the system includes that of two objects moving
translationally (the sphere and the block) and one object undergoing pure rotation
(the pulley).
At any instant of time, the sphere and the block have a common speed v , so the
angular momentum of the sphere is m1vR , and that of the block is m2vR. At the same
instant, all points on the rim of the pulley also move with speed v , so the angular
momentum of the pulley is MvR. Hence, the total angular momentum of the system
is
L = m1vR + m2vR + MvR + (m1 + m2 + M )vR
Now let us evaluate the total external torque acting on the system about the pulley
axle.
The force exerted by the axle on the pulley has zero torque because it has a
moment arm of zero.
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 20/22
Dr. Aladdin. M Abdul-latif 21
The normal force acting on the block is balanced by the gravitational force m2g
, and both are acting along the same line but in opposite directions, so these
forces do not contribute to the torque.
The gravitational force m1g acting on the sphere produces a torque about the
axle equal in magnitude to m1 gR, where R is the moment arm of the forceabout the axle.
The total external torque about the pulley axle is m1 gR, therefore
= m1 gR.
Using this result, together with Equation
=dt
dL
m1 gR = dt
d
(m1 + m2 + M)vR
m1 gR = (m1 + m2 + M) Rdt
dv =(m1 + m2 + M) Ra
M)m(m
mg a
21
The tension forces that the cord exerts on the objects are not included in calculating
the net torque about the axle because these forces are internal to the system, and we
analyzed the system as a whole.
Note that
Only external torques contributes to the change in the system’s angular momentum.
Angular Momentum of a Rotating Rigid ObjectConsider a rigid object rotating about the z axis of a
coordinate system, as shown in Figure and let us determine
the angular momentum of this object.
Each particle of the object rotates in the xy plane about the zaxis with an angular speed . The magnitude of the angular
momentum of a particle of mass mi about the z axis is miviri.
Because vi = ri i , we can express the magnitude of the
angular momentum of this particle as
Li = miri2 I 2
The vector Li is directed along the z axis, as is the vector .
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 21/22
Dr. Aladdin. M Abdul-latif 2
The total angular momentum of the object about the z-axis is the sum of the angular
momenta of the particles
Lz= Li = miri2 I 2 = ( miri2) I 2
Lz=I 2 Where we have replaced ( miri
2) by the moment of inertia I of the object about the z
axis.
Now let us differentiate the above Equation with respect to time, noting that I
is constant for a rigid object:
I dt
d I
dt
dL z
where is the angular acceleration relative to the axis of rotation.
Because dLz/dt is equal to the net external torque, we can express the above Equation
as
ext
That is, the net external torque acting on a rigid object rotating about a fixed axis
equals the moment of inertia about the rotation axis multiplied by the object’s
angular acceleration relative to that axis.
This result is the same as the result which was derived using a force approach, but
we get this result here by using the concept of angular momentum.
This equation is also valid for a rigid object rotating about a moving axis provided
that the moving axis (1) passes through the center of mass and (2) is a symmetry axis
Example
Bowling Ball
Estimate the magnitude of the angular momentum of A
typical bowling ball that has a mass of 6.0 kg and a radius of
12 cm and spinning at 10 rev/s, as shown in Figure.
Solution
The ball is a uniform solid sphere. The moment of inertia of
a solid sphere about an axis through its center is,
I= 2
5
2 MR = 22 .035.0)12.0)(0.6(
5
2mkg
Therefore, the magnitude of the angular momentum is
Lz = I = (0.035 kg .m2)(10 rev/s)(2 rad/rev)=2.2kg.m2/s
7/21/2019 Chapter 10 Rotation of Rigid Body
http://slidepdf.com/reader/full/chapter-10-rotation-of-rigid-body 22/22
Dr Aladdin M Abdul latif 22
Conservation of angular momentum
Since = dL/dt , then the total angular momentum of a system [L] will be constant if
the net torque acting on this system is zero
.
The law of conservation of angular momentum states that the total angular
momentum of a system remains constant with time if the net external torque acting
on this system is zero. i.e if = dL/dt =0, the angular momentum L= constant
Li =L f or I ii = I f f