1. Fundamentals of Physics School of Physical Science and Technology
Mechanics
(Bilingual Teaching)
谢 丽 张昆实
School of Physical Science and Technology
Yangtze University

2. Rolling, Torque, and Angular Moment
Chapter 12
Rolling, Torque, and Angular Moment
12-1 Rolling
12-2 The Kinetic Energy of Rolling
12-3 The Forces of Rolling
12-4 The Yo-Yo
12-5 Toque Revisited
Lee Raymond, Chairman & CEO
Rex Tillerson, President
$600M per year
Most in-house
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3. Rolling, Torque, and Angular Momentum
Chapter 12
Rolling, Torque, and Angular Momentum
12-7 Newton’s Second Law in Angular form
12-6 Angular Momentum
12-8 The Angular Momentum of a System
of Particles
12-9 The Angular Momentum of a Rigid
Body Rotating About a Fixed Axis
12-10 Conservation of Angular Momentum
Lee Raymond, Chairman & CEO
Rex Tillerson, President
$600M per year
Most in-house
1/3 E&P, 1/3 Refining, 1/3 Petrochemicals
Over $200M for E&P -- more than any competitor

4. 12-1 Rolling Rolling as Rotation and Translation Combined
When a bicycle moves along a straight track, the center
of each wheel moves forward in pure translation.
In what follows, we analyze the motion of a rolling wheel first by viewing it as a combination of pure translation and pure rotation, and then by viewing it as rotation alone.

5. 12-1 Rolling
The wheel of a bicycle is rolling smoothly (without sliding) along a street
center of the mass
the point P
During a time interval t both O and P move forward
by a distance s. The wheel rotates through an angle
about the center of the wheel, with the point of
the wheel that was touching the street at the begin-
ning of t moving through the arc length s, so we get
(12-1)
Thus, differentiating Eq 12-1
with respect to time (with R
held constant) gives us
(12-2)
(smooth rolling motion)

6. 12-1 Rolling = +
The combination of Figs (a) and (b) yields the actual rolling motion of the wheel, Fig. (c). The portion of the wheel at the bottom (at point P) is stationary and the portion of the wheel at the top (at point T ) is moving at , faster than any other portion of the wheel.
The motion of any round body rolling smoothly over a surface can
be separated into purely rotational and purely translational motions

7. 12-1 Rolling +
Rotation
Translation =
Real Motion +

8. 12-1 Rolling Rolling as Pure Rotation
Another way to look at the rolling motion of the wheel
We consider the rolling motion to be pure
rotation about an axis passing through
point P in Fig 12-3(c) and perpendicular to
the plane of the figure. The vectors i n
Fig.12-5 then represent the instantaneous
velocities of points on the rolling wheel.
What is the angular speed of the rolling wheel about the new axis ?
The same angular speed as that the wheel rotates about its “ COM ” axis !
Check this answer by calculating
the lineal speed at the top point T,
and the point O and P
We get the same results !

9. 12-2 The Kinetic Energy of Rolling
If we view the rolling as pure rotation about an axis through
P in Fig. 12-5, then from Eq we have
(12-3)
where is the angular speed of the wheel and is
the rotational inertia of the wheel about the axis through P.
From the parallel-axis theorem of Eq , we have
(12-4)

10. 12-2 The Kinetic Energy of Rolling
where M is the mass of the wheel, is its rotational
inertia about an axis through its center of mass. Substi-
tuting Eq.12-4 into Eq.12-3:
(12-4)
(12-3)
And using the relation Eq.12-2 yields
Konig’s theorom
漆安慎力学 P240 (7.5.6)
(12-5)
A rolling object has two type of kinetic energy :
a rotational kinetic energy due to its rotation about its center of mass and a translational kinetic energy due to translation of its center of mass.

11. 12-3 The Forces of Rolling Friction and Rolling
A wheel is being made to rotate faster while rolling along a flat surface, The
faster rotation tends to make the bottom
of the wheel slide to the left at point P.
Tendency of slide
A frictional force at P, directed to the right , opposes this tendency to slide.
If the wheel does not slide, that frictional force is a static frictional force , the motion is smooth rolling.
By differentiating Eq with respect to time we have
(12-2)
(12-6)
(smooth rolling motion)

12. 12-3 The Forces of Rolling Friction and Rolling
(12-6)
Eq holds only for
smooth rolling motion.
Tendency
of slide
If the wheel were made to rotate slower, The directions of the cente-of-mass acceleration and the frictional force at point P would now be to the left !

13. 12-3 The Forces of Rolling Rolling down a Ramp
A round uniform body of radius R rolls
smoothly down a ramp at angle , along an axis.
★ The gravitational force , acting at the
COM of the body. Its component along the
ramp is
★ A normal force , acting at point P but
is shifted to the COM of the body.
The body would slide down the ramp.
Thus, the frictional force opposing the
sliding must be up the ramp.
★ A static frictional force , acting at point
P and is directed up the ramp.
We can write Newton’s second law for components along the x axis as
(12-7)

14. 12-3 The Forces of Rolling Rolling down a Ramp
★ Next, check all the external torques
acting on the rolling body :
★ the torque of the frictional force :
( counterclockwise )
★ the torque of the gravitational
force :
( zero moment arm )
★ the torque of the normal force :
( zero moment arm )
★ Apply Newton’s second law in
angular form to the body’s rotation
about its center of mass.
(12-8)

15. 12-3 The Forces of Rolling
Because the body is rolling smoothly, we can use Eq but we must be cautious because here is negative ( in the negative direction of x axis ) and is positive (counterclockwise).
Thus substitute for
Then, solving for , we obtain
(12-9)
Substituting the right side of Eq for in Eq. 12-7, we then find
(12-10)
We can use this equation to find the linear acceleration of
any body rolling along an incline of angle with horizontal.

16. 12-4 The Yo-Yo We change the notation in Eq. 12-10 and set
If a yo-yo rolls down its string for a distance h, it lose potential energy in amount
mgh but gains kinetic energy in both translational and
rotational forms. As it climb back up, it loses kinetic
energy and regains potential energy. R R0
Instead of rolling down a ramp at angle with the
horizontal, The Yo-Yo rolls down a string at angle
with horizontal.
Instead of rolling on its outer surface at radius R, the
Yo-Yo rolls on an axle of radius R0 (Fig.12-8a).
(a)
Instead of being slowed by friction force , the Yo-Yo is
slowed by the force on it from the string (Fig12-8b). R0
We change the notation in Eq and set
to write the linear acceleration as
(b)
(12-13)

17. 12-5 Toque Revisited o A O P *
We now expand the definition of torque to apply it into an individual particle that moves along any path relative to a fixed point (rather than
a fixed axis). The path need no longer a circle, and we must write the torque as a vector that may have any direction.
a particle is at point A, is its position vector a single force acts on the particle. A o P
★ The torque acts on the particle relative
to the fixed point O is a vector quantity
defined as * O
(12-14)
We use the right-hand rule for vector products, sweeping the fingers of the right hand from (the first vector in the product) into (the second vector). The outstretched right thumb then gives the direction of

18. 12-5 Toque Revisited o A or To determine the magnitude of ,
we apply the general result of Eq.3-27
, finding A o
(12-15)
We can rewrite the Eq as
(12-17)
(12-16) or

19. 12-6 Angular Momentum
A particle of mass m passes through point A with linear momentum
is the position vector of the particle.
right-hand-rule
The angular momentum of the particle with
respect to the origin O is a vector quantity
(12-18)
The SI unit of is (kg.m2/s) or (J. s)
Using the right-hand rule we can determine the direction of the angular momentum .
The magnitude of
(12-19)

20. 12-6 Angular Momentum or Note:
As the particle moves relative to O in the direction
of its momentum position vector
rotates around O. Note carefully that to have
angular momentum about O, the particle does
not Itself have to rotate around O !
right-hand-rule
(12-21)
rewrite the Eq as
(12-20) or
(12-20)
Note:
1. Angular momentum has meaning only with respect to a specified origin !
2. The direction of the angular momentum vector is always perpendicular to the plane formed by the position and linear momentum vectors !

21. 12-7 Newton’s Second Law in Angular form
Newton’s second law written in the form
(single particle)
(12-22)
We have seen enough of the parallelism between linear and angular quanties. Guided by Eq.12-22, we guess the relation between torque and angular momentum must be
(single particle)
(12-23)
★ The (vector) sum of all the torques acting on a particle is equel to the time rate of change of the angular momentum of that particle.
The torques and the angular momentum must be defined with the same origin !
Note:

22. 12-7 Newton’s Second Law in Angular form
Prove of Eq.12-23
(12-23)
We know the definition of the
angular momentum of a particle:
Differentiating each side with respect to time t yields
(12-24)
We now use Newton’s Second law, obtaining
(12-23)
(12-25)

23. 12-8 The Angular Momentum of a System of Particles
The total angular momentum of a system of particles is the (vector)
sum of the angular momenta of the individual particles:
(12-26)
By taking the time derivative of Eq The change in can be found
(12-27)
(12-28)
The only torques that can change the total angular momentum
of the system are the external torques acting on the system.
The sum of the internal torques due to the interaction between
the particles is zero !

24. 12-8 The Angular Momentum of a System of Particles
Let represent the net external torque. Then Eq becomes
(12-29)
(system of particles)
This is the Newton’s second law for rotation in angular form.
The net extenal torque acting on a system of particles is equal to the time rate of change of the system’s total angular momentum
caution
1. Torques and system’s angular momentum must be
measured relative to the same origin.
2. If the center of mass of the system is not accelerating relative to an inertia frame, that origin can be any point ; if the center of mass of the system is accelerating, the origin can be only at that center of mass .

25. 12-9 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis
Fig.12-14(a) : a rigid body rotates about the axis with angular speed .
A mass element , of mass , within the body is located by the position
vector It moves about the axis in a circle with radius and has
the linear momentum
The magnitude of the angular momentum of this mass
element, with respect to O is
The angular momentum vector for the mass element is
Shown in Fig.12-14(b); its direction must be perpendicular
to those of and The component of is

26. 12-9 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis
(12-30)
Because , we may write
The quantity is the rotational inertia of the body about the
fixed Axis .Thus Eq reduces to
(12-31)
( rigid body, fixed axis )

27. 12-9 The Angular Momentum of a Rigid Body Rotating About a Fixed Axis

28. 12-10 Conservation of Angular Momentum
The law of conservation of angular momentum
If no net external torque acts on the system , become
= a constant
(12-32)
( isolated system ) or
net angular momentum
at some later time
at some initial time =
(12-33)
( isolated system )
★ The law of conservation of angular momentum
If the net external torque acting on a system
is zero, the angular momentum of the system
remains constant, no matter what changes take
place within the system.

29. 12-10 Conservation of Angular Momentum
The law of conservation of angular momentum
= a constant
(12-32)
(12-33)
( isolated system ) or
Depending on the torques acting on a system, the angular momentum of the system might be conserved in only one or two directions:
★ If the component of the net external torque on a system
along a certain axis ( ) is zero, then the component of the
angular momentum of the system along that axis ( ) can-
not change, no matter what changes take place within the
system.
Substituting into ,we write this conservation law as
(12-34)

30. 12-10 Conservation of Angular Momentum
The law of conservation of angular momentum
= a constant
(12-32)
(12-33)
( isolated system ) or
(12-34)
Like the other two conservation laws that we have studied, The law of conservation of angular momentum holds beyond the limitations of Newtonian mechanics. It holds for particles whose speeds approach that of light (where the theory of relativity reigns) and it remains true in the world of subatomic particles (where quantum mechanics reigns). No exceptions to it have ever been found !

31. 12-10 Conservation of Angular Momentum
Examples
1. The spinning volunteer
A student sits on a stool which may rotate around a vertical axis ( ignore friction). At the beginning, the student stretches out his both arms with each hand holding a dumbbell,
and the student and the stool rotate together with a certain angular velocity. Since there is no external torque on the horizontal plane, the sum of the angular momentum of the person and the stool should remain unchanged, therefore, when the person drops both arms to reduce the moment of inertia the rotational angular velocity of the person and the stool is increased.

32. 12-10 Conservation of Angular Momentum
2. The springboard diver
The diver leaves the high-diving board with outstretched arms and legs and some initial angular velocity about his centre of gravity.
His angular momentum ( ) remains constant since no external torques act on him (gravity exerts no torque about his centre of gravity). To make a somersault he must increase his angular velocity.
He does this by pulling in his legs and arms so that decreases and therefore increases By extending his arms and legs again, the diver can enter the water with little splash.

33. 12-10 Conservation of Angular Momentum
3. Spacecraft orientation
The spacecraft + fiywheel form an isolated system. Therefore, if the system's total angular momentum
is zero, it must remain zero.
To change the orientation of the spacecraft, the flywheel is made to rotate. The spacecraft will start
to rotate in the opposite sense to maintain the system's angular momentum at zero.
When the flywheel is then brought
to rest, the spacecraft will also stop rotating but will have changed its orientation.

34. 12-10 Conservation of Angular Momentum
4. The incredible shrinking star
before shrinking
When the nuclear fire in the core of a star bums low, the star may eventually begin to collapse.
before shrinking
The collapse may go so far as to reduce the radius of the star from something like that of the Sun to the incredibly small value of a few kilometers. The star then becomes a neutron star.
During this shrinking process, the star is an isolated system and its angular momentum cannot change. Because its rotationaI inertia is greatly reduced, its angular speed is correspondingly greatly increased, to as much as 600 to 800 revolutions per second.
The neutron star
after shrinking