Chapter 10 Rotation of Rigid Body

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)



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.




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.




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




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.




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




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




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




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



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)




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),




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




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




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.




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




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.




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.




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 .




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



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

Chapter 10 Rotation of Rigid Body

Documents

Transcript of Chapter 10 Rotation of Rigid Body