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•We know that a net force applied to a body gives that
body an acceleration. But what does it take to give a
body an angular acceleration?
•The physical quantity that make things turn is called
torque. The net torque acting on a rigid body determines
its angular acceleration, in the same way that the net
force on a body determines its linear acceleration.
Ch 10 – learning goals
1. What is meant by the torque produced by a force.
2. How the net torque on a body affects the rotational motion of
the body.
3. How to analyze the motion of a body that both rotates and
moves as a whole through space.
4. How to solve problems that involve work and power for
rotating bodies.
5. What is meant by the angular momentum of a particle or of a
rigid body.
6. How does the angular momentum of a system changes with
time.
The quantitative measure of
the tendency of a force to
change a body’s rotational
motion is called torque;
Fa applies a torque about
point O to the wrench.
Fb applies a greater torque
about O,
and Fc applies zero torque
about O.
• The tendency of a force F to cause a rotation about O depends
on
– its magnitude F
– the perpendicular distance l1 between point O and the
line of action of the force. We call the distance l1 the
lever arm of force F1about O.
• We define the torque of the force F1 with respect to O as the
product F1l1. we use the Greek letter τ (tau) for torque.
τ = Fl
F: force
l: the perpendicular distance l1 between point O
and the line of action of the force
CAUTION: Torque is always
measured about a point
The direction of torque
• counterclockwise torques
are positive and clockwise
torques are negative.
The units of torque
• The SI unit of torque is the Newton-meter.
• Torque is not work or energy, and torque
should be expressed in Newton-meters,
not joules.
If φ is angle between force F and distance r
τ = F∙(r∙sin)
r∙sin - perpendicular distance
τ = r∙(F∙sin)
F∙sin - perpendicular force
When a force F acts at a point having a position vector r
with respect to an origin O, the torque of the force with
respect to O is the vector quantity
Magnitude:
The direction of torque is perpendicular to both r
and F. The torque vector is directed along the axis
of rotation, with a sense given by the right-hand
rule.
A dot ●
means pointing out of the screen
A cross × means pointing into the screen
example
• Rank the design scenarios (A through C) on the basis of
the tension in the supporting cable from largest to
smallest. In scenarios A, and C, the cable is attached
halfway between the midpoint and end of the pole. In B,
the cable is attached to the end of the pole.
60o
sign
A
45o
30o
sign
sign
B
A, C, B
C
example
•
1.
2.
3.
4.
5.
If Anya decides to make the star twice as massive, and
not change the length of any crossbar or the location of
any object, what does she have to do with the mass of
the smiley face to keep the mobile in perfect balance?
Note that she may have to change masses of other
objects to keep the entire structure balanced.
make it eight times more massive
make it four times more massive
make it twice as massive
Nothing
impossible to tell
• Newton’s second law for
the tangential component
is:
since
We write an equation like this for every particle in the body
and then add all these equations:
Just as Newton’s second law says that the net force on a
particle equals the particle’s mass times its acceleration, the
equation says that the net torque on a rigid body equals
the body’s moment of inertia about the rotation axis
times its angular acceleration.
4 things to note in Eq.
1. The equation is valid only for rigid bodies.
2. Since we used atan = r∙αz, αz must be measured in rad/s2.
3. Since all the internal torques add to zero, so the sum ∑τ
in Eq. ∑τ = Iα includes only the torques of the external
forces.
4. Often, an important external force acting on body is its
weight. We assume that all the weight is concentrated at
the center of mass of the body to get the correct torque
(about any specified axis).
Consider the situation on
the diagram, find the
acceleration of the block
of mass m.
•
The figure shows a glider of mass m1 that can slide without friction
on horizontal air tract. It is attached to an object of mass m2 by a
massless string. The pulley has radius R and moment of inertia I
about it axis of rotation. When released, the hanging object
accelerates downward, the glider accelerates to the right, and the
string turns the pulley without slipping or stretching. Rank the
magnitudes of the following forces that acting during the motion, in
order from largest to smallest magnitude.
1. The tension force (magnitude T1) in the horizontal part of the string;
2. The tension force (magnitude T2) in the vertical part of the string;
3. The weight m2g of the hanging object.
example
• Find the magnitude of the angular
acceleration α of the swing bar.
• When a rigid body rotate about a moving axis, the motion of
the body is combined translation and rotation. We need
to combine:
• Translational motion of the center of mass
• Rotation about an axis through the center of mass.
• The kinetic energy of a rigid body that has
both translational and rotational motions is the
sum of a part ½ Mvcm2 associated with motion
of the center of mass and a part ½ Icmω2
associated with rotation about an axis through
the center of mass.
The point on the wheel that contacts the surface must be
instantaneously at rest so that it does not slip. Hence the velocity v1’
of the point of contact relative to the center of mass must have the
same magnitude but opposite direction as the center-of-mass
velocity vcm. If the radius of the wheel is R and its angular speed
about the center of mass is ω, then the magnitude of v1’ is R∙ω;
hence we must have
vcm = Rω (condition for rolling without slipping)
• Note that the relationship vcm = Rω holds only if there is
rolling without slipping.
– When a drag racer first starts to move, the rear tires
are spinning very fast even though the racer is hardly
moving, so Rω is greater than vcm.
– If a driver applies the brakes too heavily so that the
car skids, the tires will spin hardly at all and Rω is
less than vcm.
• If a rigid body changes height as it moves, we must also
consider gravitational potential energy. U = Mgycm
• The speed doesn’t depend on either the mass M of the
body or its radius R. All uniform solid cylinders have the
same speed at the bottom, even if their masses and radii
are different, because they have the same c. All solid
spheres also have the same speed, an so on.
• The smaller the value of c, the faster the body is moving
at the bottom (and at any point on the way down).
• Small-c bodies always beat large-c bodies because they
have less of their kinetic energy tied up in rotation and
have more available for translation.
• Reading the values of c from the reference sheet, the
order of finish is as follows:
– Solid sphere
– Solid cylinder,
– Thin-walled hollow sphere
– Thin-walled hollow cylinder
When a rigid body with total mass M moves, its motion can be
described by combining translational motion and rotational motion
In translation, the acceleration acm of the center of mass is the
same as that of a point mass M acted on by all the external forces
on the actual body:
The rotational motion about the center of mass is described by the
rotational analog of Newton’s 2nd law:
Note: when we learned this equation, we assumed that the
axis of rotation was stationary. But in fact, this equation is
valid even when the axis of rotation moves, provided the
following two conditions are met:
1. The axis through the center of mass must be an axis of
symmetry.
2. The axis must not change direction.
• The axle of a bicycle wheel passes through the
wheel’s center of mass and axis of symmetry. Hence
the rotation of the wheel is described by ∑τz = Icmαz
provided the bicycle doesn’t turn or tilt to one side
(which would change the orientation of the axle).
When a perfectly rigid sphere is rolling down a perfectly rigid
incline, there is no sliding at the point of contact, so friction does
no work. However, in reality, when a not so perfectly rigid sphere
rolling down a not so perfectly rigid incline, there are some
deformations at the points of contact. As a result, there is rolling
friction.
Often the rolling body and the surface are rigid enough that
rolling friction can be ignored.
•
Suppose the solid cylinder used as a yo-yo in
example 10.6 is replaced by a hollow cylinder
of the same mass and radius
1. Will the acceleration of the yo-yo
a. Increase
b. Decrease,
c. Remain the same?
2. Will the string tension
a. Increase,
b. Decrease,
c. Remain the same?
example
•
•
•
Two uniform identical solid spherical balls each of mass M,
radius r and moment of inertial about its center 2/5MR2, are
released from rest from the same height h above the
horizontal ground Ball A falls straight down, while ball B
rolls down the distance x along the inclined plane without
slipping.
If the velocity of ball A as it hits the ground is VA, what is
the velocity VB of ball b as it reaches the ground?
In terms of acceleration due to earth’s gravity g, the
acceleration of ball B along the inclined plane would be
A
B
h
x
30o
horizontal
The work dW done by the force Ftan
while a point on the rim moves a
distance ds is dW = Ftan∙ds. If dθ is
measured in radians, then ds = R∙dθ
• The total work W done by the torque
during an angular displacement from θ1 to
θ2 is
• If the torque remains constant while the
angle changes by a finite amount ∆θ = θ1
– θ2
The work done by a constant torque is the product of torque
and the angular displacement. If torque is expressed in
Newton-meters and angular displacement in radian, the work
is in joules.
Only the tangent component of force does work, other
components do no work.
When a torque does work on a rotating rigid body, the
kinetic energy changes by an amount equal to the work
done.
Wtot = ½ I∙ω22 – ½ I∙ω12
•Power is the rate of doing work.
P = dW/dt = τz(dθ/dt) = τzω
When a torque acts on a body the rotates with angular
velocity ωz, its power is the product of τz and ωz. This is
the analog of the relationship P = F∙v
•
1.
2.
3.
You apply equal torques to two different cylinders,
one of which has a moment of inertial twice as large
as the other cylinder. Each cylinder is initially at rest
after one complete rotation, which cylinder has the
greater kinetic energy?
The cylinder with the larger moment of inertia;
The cylinder with the smaller moment of inertia;
Both cylinders have the same kinetic energy.
• Every rotational quantity that we have encountered so
far is the analog of some quantity in the translational
motion of a particle.
∑F = ma
K = ½ mv2
Wtot = ½ mv22 - ½ mv12
P = Fv
• The analog of momentum of a particle is angular
momentum, a vector quantity denoted as L.
L=I∙ω
The value of L depends on the choice of origin O, since it
involves the particle’s position vector relative to O. the units
of angular momentum are kg∙m2/s.
• If a particle moves in the xy-plane counterclockwise, then the
angular momentum vector L is perpendicular to the xy-plane.
The right-hand rule for vector products shows that is direction
is along the +z-axis.
If at moment t the position of the
particle is r and its linear
momentum is p = m∙v, then the
magnitude of angular momentum
is:
or
L = (mvsinΦ)r
L = mv(rsinΦ) = mvl
Where l is the perpendicular
distance from the line of v to O.
this distance plays the role of
“lever arm” for the momentum
vector.
• When a net force F acts on a particle, its velocity and
momentum changes:
= Fnet
• The rate of change dp/dt of the linear momentum of a
particle equals the net force acting on it.
• Similarly, when the torque of the net force acting on a
particle, its angular velocity and angular momentum
changes:
• The rate of change of angular momentum of a
particle equals the torque of the net force acting on
it.
The derivation of
If we take the time derivative of
the rule for the derivative of a product:
, using
Consider a rigid body rotating about the
z-axis with angular speed ω. First let’s
think about a thin slice of the body lying
in the xy-plane.
Each particle moves in a circle centered
at the origin, and at each instant its
velocity vi is perpendicular to its position
vector ri. Hence Φ = 90o for every
particle. A particle with mass mi at a
distance ri from O has a speed vi = riω.
The magnitude of its angular
momentum is:
Li = (mivisinΦ)ri = (miriω)ri = miri2ω
The direction of each particle's angular momentum, as given by
the right-hand rule for the vector product, is along the +z-axis.
• The total angular momentum of the slice of the body lying in the
xy-plane is the sum ∑Li of the angular momenta Li of the
particles:
Where I is the moment of inertia of the slice about the z-axis.
We can do this same calculation
for the other slides of the body, all
parallel to the xy-plane. Since the
rigid body is rotating the axis of
symmetry, their angular
momentum vector sum L1 + L2
also lies along the symmetry, and
its magnitude is L = Iω.
Since ω and L have the same
direction, we have a vector
relationship: L = Iω
• We know that the rate of change of angular momentum of a
particle equals to the torque of the net force acting on the
particle.
•For any system of particles (including both rigid and non-rigid
bodies), the rate of change of the total angular momentum equals
the sum of the torques of all forces acting on all the particles. The
torques of the internal forces add to zero. So the sum of the torques
includes only the torques of the external forces. If the total angular
momentum of the system of particles is L and the sum of the
external torques is ∑τ, then
If the system of particles is a rigid body rotating about a symmetry
axis (z-axis), the Lz = Iωz and I is constant. If this axis has a fixed
direction in space, then the vectors L and ω change only in
magnitude, not in direction. dLz/dt = Idωz/dt = Iαz ∑τ = Iαz
•
A ball is attached to one end of a piece of
string. You hold the other end of the
string and whirl the ball in a circle around
your hand.
1. If the ball moves at a constant speed, is
its linear momentum p constant? Why or
why not?
2. Is its angular momentum L constant?
why or why not?
practice
•
A uniform rod of mass M and length L has a moment of
inertia about one end I = ML2/3. It is released from rest
in horizontal direction about the fixed axis
perpendicular to the paper as shown below.
Axis
P
cm
A
1. What is the linear velocity of the center of mass, cm,
when the rod is in the vertical position?
2. What is the angular momentum of the rod about the
axis of rotation at one end?
• Like conservation of energy and of linear momentum, the
principle of conservation of angular momentum is a
universal conservation law, valid at all scales from atomic and
nuclear systems to the motions of galaxies.
• The principle follows directly from equation: ∑τ = dL/dt
• If ∑τ = 0, then dL/dt = 0, and L is constant.
When the net external torque acting on a system is zero, the
total angular momentum of the system is constant (conserved).
• A circus acrobat, a diver, and an ice skater pirouetting on the
toe of one skate all take advantage of this principle.
I1ω1z = I2ω2z
When a system has several parts, the internal forces that the parts
exert on each other cause changes in the angular momenta of the
parts, but the total angular momentum doesn’t change. The total
angular momentum of the system is constant.
•
If the polar ice caps were to completely melt
due to global warming, the melted ice would
redistribute itself over the earth. This change
would cause the length of the day (the time
needed for the earth to rotate once in its axis)
to
L = Iω
1. Increase
2. Decrease
I increases, ω decreases, days is
longer.
3. Remain the same
(hint: use angular momentum ideas. Assume
that the sun, moon, and planets exert negligibly
small torques on the earth.)
example
•
A uniform rod of mass M and length L has a moment of
inertia about one end I = ML2/3. It is released from rest in
horizontal direction about the fixed axis perpendicular to the
paper as shown below.
Axis
P
cm
A
1.
What is the linear velocity of the center of mass, cm, when
the rod is in the vertical position?
2.
What is the angular momentum of the rod about the axis of
rotation at one end?
example
•
Two equal masses, each m, are resting at the ends of
a uniform rod of length 2a and negligible mass. The
system is in equilibrium about the center C of the rod.
A piece of clay of mass m is dropped down on the
mass at the right end, hits it with velocity v as shown
below and sticks to it.
m
v
m
a
a
m
C
•
What is the ratio of the kinetic energy Ef just after the
collision to the kinetic energy Ei just before the
collision, Ef/Ei, of the system?
example
•
A skater is spinning on ice with her arms
outstretched about the vertical axis at an angular
speed of ω. When she brings her arms close to her
body, which of the following statements is correct?
A.
Her angular velocity and angular momentum remain
constant.
Her angular momentum is increased.
Her kinetic energy is increased.
Her kinetic energy is decreased.
The net torque on her about the axis of rotation increases.
B.
C.
D.
E.
example
•
A uniform diving board, 12 meters long and 20 kg in
mass, is hinged at P, which is 5 meters from the edge
of the platform. An 80 kg diver is standing at the other
end of the board.
P
5m
0
hinge
platform
1.
2.
What will be the force exerted by the hinge on the
board?
What will be the normal force on the board at the edge
of the platform?
example
M
axis
P
M
h
L
rod
•
•
A mass M slides down a smooth surface from height h
and collides inelastically with the lower end of a rod
that is free to rotate about a fixed axis at P as shown
below. The mass of the rod is also M, the length is l,
and the moment of inertial about P is ml2/3.
What is the angular velocity of the rod about the axis P
jut after the mass sticks to it?
example
•
A square metal plate 0.180 m on each side is pivoted about an
axis through point at its center and perpendicular to the plate.
Calculate the net torque about this axis due to the three forces
shown in the figure if the magnitudes of the forces are F1
= 21.0 N , F2 = 17.0 N, and F3 = 14.9 N . The plate and all
forces are in the plane of the page. Take positive torques to be
counterclockwise.
example
M1
L
M2
P
• Masses M1 and M2 are separated by a distance L. what is the
distance of the center of mass of the system at P from M1 as
shown above?
• What is the moment of inertial of the system about the center
of mass at P?
example
• A solid, uniform cylinder with mass 8.45 kg and diameter
11.0 cm is spinning with angular velocity 230 rpm on a
thin, frictionless axle that passes along the cylinder axis.
You design a simple friction-brake to stop the cylinder by
pressing the brake against the outer rim with a normal
force. The coefficient of kinetic friction between the brake
and rim is 0.334. What must the applied normal force be
to bring the cylinder to rest after it has turned through 5.15
revolutions?
example
•
A wheel with a weight of 395 N comes off a moving truck
and rolls without slipping along a highway. At the bottom
of a hill it is rotating at an angular velocity of 26.8 rad/s.
The radius of the wheel is 0.652 m and its moment of
inertia about its rotation axis is 0.800 MR2 . Friction does
work on the wheel as it rolls up the hill to a stop, at a
height of above the bottom of the hill; this work has a
magnitude of 3520 J. Calculate h. Use 9.81 m/s2 for
the acceleration due to gravity.
example
• The radius of the pulley is r and mass M is initially at height h.
The system is initially at rest and is then released at time t = 0.
Assume M>>m.
• Assuming the pulley to be massless and frictionless, what is the
angular acceleration of the pulley while M is falling?
r
M
m
h
example
•
An object of moment of inertial I is initially
at rest when torque T begins to act on it as
shown below. After t seconds,
1. what is the angular velocity of the object in
terms of T, I and t?
2. What is the kinetic energy of the object?
T
Moment of
inertial I
example
• A solid, uniform cylinder with mass 8.45 kg and diameter
11.0 cm is spinning with angular velocity 230 rpm on a
thin, frictionless axle that passes along the cylinder axis.
You design a simple friction-brake to stop the cylinder by
pressing the brake against the outer rim with a normal
force. The coefficient of kinetic friction between the brake
and rim is 0.334. What must the applied normal force be
to bring the cylinder to rest after it has turned through 5.15
revolutions?
example
•
A hoop is rolling to the right without slipping on a horizontal
floor at a steady 1.8 m/s (Vcm).
1. Find the velocity vector of each of the following points, as
viewed by a person at rest on the ground:
A. The highest point on the hoop
B. The lowest point on the hoop
C. A point on the right side of the hoop, mideway abetween
the top and the bottom
2. find the velocity vector of each of the above points, as
viewed by a person moving along with the same velocity as
the hoop.
example
•
Find the magnitude of the angular momentum of
the second hand on a clock about an axis
through the center of the clock face. The clock
hand has a length of 15.0 cm and a mass of
6.00 g. Take the second hand to be a slender
rod rotating with constant angular velocity about
one end.