Physics 1 NTC Angular Motion General Notes When a rigid object rotates about a fixed axis in a given time interval every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration ID: 560334
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Slide1
Lecture 7:Rotational motion
Physics 1, NTCSlide2
Angular Motion, General Notes
When a rigid object rotates about a fixed axis in a given time interval, every portion on the object rotates through the same angle in a given time interval and has the same angular speed and the same angular acceleration.
So
q, w, a all characterize the motion of the entire rigid object as well as the individual particles in the object.
Section 10.1Slide3
Angular
Position
As the particle moves, the only coordinate that changes is
qAs the particle moves through q
, it moves though an arc length s.
The arc length and r are related:
s = q r
Section 10.1Slide4
Conversions
Comparing degrees and radians
Converting from degrees to radians
Section 10.1Slide5
Directions
Strictly speaking, the speed and acceleration (
w
, a
)
are the magnitudes of the velocity and acceleration vectors.
The directions are actually given by the right-hand rule.
Section 10.1Slide6
Relationship Between Angular and Linear Quantities
Every point on the rotating object has the same angular motion.
Every point on the rotating object does
not have the same linear motion.Displacementss =
θ
r
Speedsv =
ω r Accelerationsa = α r
Section 10.3Slide7
Acceleration Comparison
The tangential acceleration is the derivative of the tangential velocity.
Centripetal acceleration
Section 10.3Slide8
Rotational Kinetic Energy
The total rotational kinetic energy of the rigid object is the sum of the energies of all its particles.
I
is called the moment of inertia.
Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object.
The units of rotational kinetic energy are Joules (J).
Section 10.4Slide9
Moment of Inertia
Defined by
dimensions = ML2 or unit of kg.m
2.
Mass = inherent property, but the moment of inertia depends on the choice of rotational axis.
Moment of inertia is a measure of the resistance of an object to changes in its rotational motion, similar to mass being a measure of an object’s resistance to changes in its translational motion.
The moment of inertia depends on the mass and how the mass is distributed around the rotational axis.
Section 10.5Slide10
Moment of Inertia, cont
For a continuous rigid object, imagine the object to be divided into many small elements, each having a mass of
Δ
mi.
If
r
is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known.
Section 10.5Slide11
Moments of Inertia of Various Rigid Objects
Section 10.5Slide12
Moment of Inertia of a Uniform Rigid Rod
The shaded area has a mass
dm
= l dxThen the moment of inertia is
Section 10.5Slide13
Moment of Inertia of a Uniform Solid Cylinder
Divide the cylinder into concentric shells with radius
r
, thickness dr and length L.dm = r dV = 2
p(r
Lr) dr
Then for I
Section 10.5Slide14
9.4
Newton’s Second Law for Rotational Motion About a Fixed Axis
Example 9
The Moment of
Inertia
Depends on Where
the Axis Is.
Two particles each have mass and are fixed at theends of a thin rigid rod. The length of the rod is L.Find the moment of inertia when this object rotates relative to an axis that is perpendicular to the rod at(a) one end and (b) the center.Slide15
(a)Slide16
9.4
Newton’s Second Law for Rotational Motion About a Fixed Axis
(b)Slide17
Parallel-Axis Theorem
In the previous examples, the axis of rotation coincided with the axis of symmetry of the object.
For an arbitrary axis, the parallel-axis theorem often simplifies calculations.
The theorem states I = ICM +
MD
2
I is about any axis parallel to the axis through the center of mass of the object.I
CM is about the axis through the center of mass.D is the distance from the center of mass axis to the arbitrary axis.
Section 10.5Slide18
Moment of Inertia for a Rod Rotating Around One End
– Parallel Axis Theorem Example
The moment of inertia of the rod about its center is
D is ½ L
Therefore,
Section 10.5Slide19
Torque
Torque,
t
, is the tendency of a force to rotate an object about some axis. Torque is a vector, but we will deal with its magnitude here:t =
r F
sin
f = F d
F is the forcef is the angle the force makes with the horizontal d is the moment arm (or lever arm) of the forceThere is no unique value of the torque on an object.Its value depends on the choice of a rotational axis.
Section 10.6Slide20
Torque
is a vector!
The moment arm, d, is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force.
d
=
r sin Φ
The horizontal component of the force (F cos f) has no tendency to produce a rotation.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.
Section 10.6Slide21
Net Torque
The force
will tend to cause a counterclockwise rotation about O.The force will tend to cause a clockwise rotation about O.
S
t
= t
1 + t2 = F1d1 – F2d2
Section 10.6Slide22
Torque and Angular Acceleration, Extended
Consider the object consists of an infinite number of mass elements
dm
of infinitesimal size.Each mass element rotates in a circle about the origin, O.Each mass element has a tangential acceleration.
From Newton’s Second Law
dF
t = (dm
) atThe torque associated with the force and using the angular acceleration givesdt ext = r dFt = atr dm = ar 2 dmThis becomes St = Ia
Section 10.7Slide23
Torque and Angular Acceleration, Extended cont.
rigid body under a net torque.
The
result also applies when the forces have radial components.The line of action of the radial component must pass through the axis of rotation.
These components will
produce zero torque about the axis
.
Section 10.7Slide24
Falling Smokestack Example
When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground.
Each higher portion of the smokestack has a larger tangential acceleration than the points below it.
The shear force due to the tangential acceleration is greater than the smokestack can withstand.The smokestack breaks.
Section 10.7Slide25
9.3
Center of Gravity
Conceptual Example 7
Overloading a Cargo Plane
This accident occurred because the plane was overloaded toward
the rear. How did a shift in the center of gravity of the plane cause
the accident?Slide26
Summary of Useful Equations
Section 10.8Slide27
9.2
Rigid Objects in Equilibrium
Example 3
A Diving Board
A woman whose weight is 530 N is
poised at the right end of a diving board
with length 3.90 m. The board has
negligible weight and is supported bya fulcrum 1.40 m away from the leftend.Find the forces that the bolt and the fulcrum exert on the board.Slide28
9.2
Rigid Objects in EquilibriumSlide29
9.2
Rigid Objects in EquilibriumSlide30
9.2
Rigid Objects in Equilibrium
Example 5
Bodybuilding
The arm is horizontal and weighs 31.0 N. The deltoid muscle can supply
1840 N of force. What is the weight of the heaviest dumbell he can hold?Slide31
9.2
Rigid Objects in EquilibriumSlide32
9.2
Rigid Objects in EquilibriumSlide33
9.3
Center of Gravity
DEFINITION OF CENTER OF GRAVITY
The center of gravity of a rigid
body is the point at which
its weight can be considered
to act when the torque due
to the weight is being calculated.Slide34
9.3
Center of Gravity
When an object has a symmetrical shape and its weight is distributed
uniformly, the center of gravity lies at its geometrical center.Slide35
9.3
Center of GravitySlide36
9.3
Center of Gravity
Example 6
The Center of Gravity of an Arm
The horizontal arm is composed
of three parts: the upper arm (17 N),
the lower arm (11 N), and the hand
(4.2 N).Find the center of gravity of thearm relative to the shoulder joint.Slide37
9.3
Center of GravitySlide38
Energy in an Atwood Machine, Example
The system containing the two blocks, the pulley, and the Earth is an isolated system in terms of energy with no non-conservative forces acting.
The mechanical energy of the system is conserved.
The blocks undergo changes in translational kinetic energy and gravitational potential energy.The pulley undergoes a change in rotational kinetic energy
.
Find
?
Section 10.8Slide39
Rolling Object
The red curve shows the path moved by a point on the rim of the object.
This path is called a
cycloid.The green line shows the path of the center of mass of the object.
In
pure rolling motion
, an object rolls without slipping.
In such a case, there is a simple relationship between its rotational and translational motions.
Section 10.9Slide40
The tangential speed of a
point on the outer edge of
the tire is equal to the speed
of the car over the ground.Slide41
Pure Rolling Motion, Object’s Center of Mass
The translational speed of the center of mass is
The linear acceleration of the center of mass is
Section 10.9Slide42
Rolling Motion Cont.
Rolling motion can be modeled as a combination of pure translational motion and pure rotational motion.
The contact point between the surface and the cylinder has a translational speed of zero (c).
Section 10.9Slide43
Total Kinetic Energy of a Rolling Object
The total kinetic energy of a rolling object is the sum of the translational energy of its center of mass and the rotational kinetic energy about its center of mass.
K
= ½ ICM
w
2
+ ½ MvCM2
The ½ ICMw2 represents the rotational kinetic energy of the cylinder about its center of mass.The ½ Mv2 represents the translational kinetic energy of the cylinder about its center of mass.
Section 10.9Slide44
Total Kinetic Energy, Example
Accelerated rolling motion
is possible only if friction is present between the sphere and the incline.
The friction produces the net torque
required for rotation.
No loss of mechanical energy occurs because the contact point is at rest relative to the surface at any instant.In reality,
rolling friction causes mechanical energy to transform to internal energy.Rolling friction is due to deformations of the surface and the rolling object.
Section 10.9
Find
and
at the bottom of the incline?
Slide45
9.5
Rotational Work and Energy
Example 13
Rolling Cylinders
A thin-walled hollow cylinder (mass =
m
h
, radius = rh) anda solid cylinder (mass = ms, radius = rs) start from rest atthe top of an incline.Determine which cylinder has the greatest translationalspeed upon reaching the bottom.Slide46
9.5
Rotational Work and Energy
ENERGY CONSERVATIONSlide47
9.5
Rotational Work and Energy
The cylinder with the smaller moment
of inertia will have a greater final translational
speed.Slide48
Lecture 8: Angular MomentumSlide49
Vector Product (cross product)
The quantity
AB
sin q is equal to the area of the parallelogram formed
by
The direction of
is
perpendicular to the plane formed by
.
The best way to determine this direction is to use the right-hand rule.
Section 11.1Slide50
Properties of the Vector Product
The vector product is
not
commutative. The order in which the vectors are multiplied is important.If A
//
B (parallel q
= 0o or 180o), thenTherefore If A is perpendicular to B, then The vector product obeys the distributive law.
Section 11.1Slide51
Final Properties of the Vector Product
The derivative of the cross product with respect to some variable such as
t
iswhere it is important to preserve the multiplicative order of the vectors.
Section 11.1Slide52
Vector Products of Unit Vectors
Section 11.1Slide53
Signs in Cross Products
Signs are interchangeable in cross products
and
Section 11.1Slide54
Using Determinants
The cross product can be expressed as
Expanding the determinants gives
Section 11.1Slide55
Vector Product Example
Given
Find
Result
Section 11.1Slide56
Torque Vector Example
Given the force and location
Find the torque produced
Section 11.1Slide57
Angular Momentum
Consider a particle of mass
m
located at the vector position r and moving with linear momentum p
.
Find the net torque.
This looks very similar to the equation for the net force in terms of the linear momentum since the torque plays the same role in rotational motion that force plays in translational motion.
Section 11.2Slide58
Angular Momentum, cont
The instantaneous angular momentum of a particle relative to the origin
O
is defined as the cross product of the particle’s instantaneous position vector and its instantaneous linear momentum.
Section 11.2Slide59
Torque and Angular Momentum
The torque is related to the angular momentum.
Similar to the way force is related to linear momentum.
The torque acting on a particle is equal to the time rate of change of the particle’s angular momentum.
This is the rotational analog of Newton’s Second Law .
must be measured about the same origin.
This is valid for any origin fixed in an inertial frame.
Section 11.2Slide60
Angular
Momentum
The SI units of angular momentum are (kg
.m2)/ s.Both the magnitude and direction of the angular momentum depend on the choice of origin.The magnitude is
L
=
mvr sin f
f is the angle between and . The direction of L is perpendicular to the plane formed by r and p .
Section 11.2Slide61
Angular Momentum of a Particle, Example
The vector
is
pointed out of the diagram.The magnitude is L = mvr
sin 90
o
= mvrsin 90o is used since
v is perpendicular to r.A particle in uniform circular motion has a constant angular momentum about an axis through the center of its path.
Section 11.2Slide62
Angular Momentum of a Rotating Rigid Object
The rigid object is a non-deformable system.
Each particle of the object rotates in the
xy plane about the z axis with an angular speed of
w
The angular momentum of an individual particle is
Li
= mi ri2 wL and are directed along the z axis.
Section 11.3Slide63
9.6
Angular Momentum
DEFINITION OF ANGULAR MOMENTUM
The angular momentum
L
of a body rotating about a
fixed axis is the product of the body’s moment of
inertia and its angular velocity with respect to thataxis: Requirement: The angular speed mustbe expressed in rad/s.
SI Unit of Angular Momentum:
kg
·m
2
/sSlide64
9.6
Angular Momentum
PRINCIPLE OF CONSERVATION OFANGULAR MOMENTUM
The angular momentum of a system remains constant (is
conserved) if the net external torque acting on the system
is zero.Slide65
9.6
Angular Momentum
Conceptual Example 14
A Spinning Skater
An ice skater is spinning with both
arms and a leg outstretched. She
pulls her arms and leg inward and
her spinning motion changesdramatically.Use the principle of conservationof angular momentum to explainhow and why her spinning motionchanges.Slide66
9.6
Angular Momentum
Example 15
A Satellite in an Elliptical Orbit
An artificial satellite is placed in an
elliptical orbit about the earth. Its point
of closest approach is 8.37x10
6mfrom the center of the earth, andits point of greatest distance is 25.1x106m from the center ofthe earth.The speed of the satellite at the perigee is 8450 m/s. Find the speedat the apogee.Slide67
9.6
Angular Momentum
angular momentum conservationSlide68
9.6
Angular Momentum