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# Rotational Motion

advertisement ```Rotation of Rigid Bodies
and
Dynamics of Rotational Motion
Mark Anthony C. Burgonio
MSc in Applied Physics, Major in Medical Physics (cand.)
Physics Instructor, PUP College of Science
Outline

Translation and Rotation

Rigid Body and Its Moment of Inertia

Rotational Quantities

Newton’s Law for Rotation

Conservation Laws for Rotation
Rigid bodies
• A particle is an idealized body that occupies
only a single point in space, has mass, and has
no internal structure.
• A rigid body is a collection of particles linked
by a light rigid framework. In other words, it is
an idealization of a solid body in which
deformation is neglected. (Assumptions: perfectly
definite and unchanging shape and size)
Translation and Rotation
•
Translational Motion: linear
motion (i.e. rectilinear and
curvilinear motions)
•
Rotational Motion: rotary
motion about an axis
•
Rolling motion: combined
rotation and translation
Sense and direction of rotation
•
Use right hand rule to determine the direction of rotation.
Angle
Translational and Rotational Quantities
Translation
Quantity
Equation
Rotational
Quantity
Equation
Position
Displacement
𝑟 = 𝑟(𝑡)
∆𝑟 = 𝑟2 − 𝑟1
Angular Position
Angular
Displacement
Angular Velocity
𝜃 = 𝜃(𝑡)
∆𝜃 = 𝜃2 − 𝜃1
Velocity
Acceleration
𝑑𝑟
𝑣=
𝑑𝑡
𝑑𝑣
𝑎=
𝑑𝑡
Angular
Acceleration
𝑑𝜃
𝜔=
𝑑𝑡
𝑑𝜔
𝛼=
𝑑𝑡
𝝎
Translational and Rotational Quantities
Quantity
Linear speed, 𝑣
Radial acceleration, 𝑎𝑟𝑎𝑑
Tangential acceleration,
𝑎𝑡𝑎𝑛
Relation
𝑑𝜃
𝑣=𝑟
= 𝑟𝜔
𝑑𝑡
𝑣2
𝑎𝑟𝑎𝑑 =
= 𝑟𝜔2
𝑟
𝑑𝑣 𝑟𝑑𝜔
𝑎𝑡𝑎𝑛 =
=
= 𝑟𝛼
𝑑𝑡
𝑑𝑡
The non-uniform acceleration is:
𝑎 = 𝑟𝛼 𝑡 − 𝑟𝜔2 𝑛
where: 𝑡 = tangential unit vector and 𝑛 = normal unit vector
Example
2. A centrifuge used to accustom astronaut
trainees to high accelerations has radius r of
the circle traveled by an astronaut is 15 m.
(a) At what constant rotational speed must
the centrifuge rotate if the astronaut is to
have a translational acceleration of
magnitude 11g?
(b) What is the tangential acceleration of the astronaut if the centrifuge
accelerates at a constant rate from rest to the rotational speed found in part (a)
in 120 s?
Moment of inertia
For discrete mass distribution: For continuous mass
distribution:
𝑛
𝑚𝑖 𝑟𝑖2
𝐼=
𝐼=
𝑟 2 𝑑𝑚
𝑖=1
where:
where:
• 𝑚𝑖 is the mass of the nth object
• 𝑑𝑚 is the infinitesimal mass
• 𝑟𝑖 is the distance of the nth
• 𝑟 is the distance of dm from the
object from the axis of rotation
axis of rotation
Moment of inertia: various bodies
Example
1. Three identical balls, with masses of M, 2M, and 3M, are
fastened to a massless rod of length L as shown.
Determine the rotational inertia about the left end of the
rod.
Translation
Quantity
Equation
𝑝 = 𝑚𝑣
Momentum
Net Force
Impulse
Work done by
Force
𝑑𝑝
𝐹=
= 𝑚𝑎
𝑑𝑡
𝑗 = ∆𝑝 =
𝑊=
𝐹𝑛𝑒𝑡 𝑑𝑡
𝐹 ∙ 𝑑𝑟
Rotational Quantity
Angular Momentum
𝐿 = 𝐼𝜔
Net Torque
Angular Impulse
Work done by Torque
Power
℘=𝐹∙𝑣
Kinetic energy
Kinetic energy
𝑝2
1
2
=
= 𝑚𝑣
2𝑚 2
𝑇𝑡𝑟𝑎𝑛𝑠
Equation
𝑑𝐿
𝜏=
= 𝐼𝛼
𝑑𝑡
𝑗 = ∆𝐿 =
𝑊=
𝜏𝑛𝑒𝑡 𝑑𝑡
𝜏 ∙ 𝑑𝜃
℘=𝜏∙𝜔
Power
𝑇𝑟𝑜𝑡
𝐿2 1 2
=
= 𝐼𝜔
2𝐼 2
Examples
3. A 25.0-kg boy travels around a 0.50-m radius circle with
an angular speed of 12 rad/s. Determine (a) the
magnitude of its angular momentum, and (b) the
rotational kinetic energy about the center of the circle.
4. A 40-g cylinder is 0.10 m in radius and 0.05 m in length.
A string is wound around it and a force is applied to cause
its angular acceleration of 5.0 𝑟𝑎𝑑 𝑠2. Determine the net
torque acting on it.
Torque and Angular Momentum
Quantity
Operational Definition
Torque (Moment of
Force)
Angular Momentum
𝜏 ≡ 𝑟 × 𝐹 = 𝑟𝐹 sin 𝜃 𝑛
𝐿 ≡ 𝑟 × 𝑝 = 𝑟𝑝 sin 𝜃 𝑛
SI Unit:
𝑁 ∙ 𝑚 for Torque
kg ∙ 𝑚2 𝑠 or J ∙ 𝑠 for angular momentum
Application of Rotation
Stability
The center of gravity is located at the center
of mass in location where the acceleration
due to gravity is constant
Simple Machine
Example
5. Two objects are moving in the xy plane as shown.
Determine the magnitude of their total angular momentum
(about the origin O).
Conservation of Energy Extended
Example
6. You make a primitive yo-yo by wrapping a massless
string around a solid cylinder with mass M and radius R.
You hold the free end of the string stationary and release
the cylinder from rest. The string unwinds but does not
slip or stretch as the cylinder descends and rotates. (a)
Find the speed of the center of mass of the cylinder after
it has descended a distance h. (b) Find the downward
acceleration of the cylinder and the tension in the wire.
Conservation of Angular Momentum
Example
7. A wheel is rotating freely at rotational speed 800
rev/min on a shaft whose rotational inertia is
negligible. A second wheel, initially at rest and
with twice the rotational inertia of
the first, is suddenly coupled to the same shaft.
(a) What is the rotational speed of the resultant
combination of the shaft and two wheels? (b)
What fraction of the original rotational kinetic
energy is lost?
Problem Set: (Refer to UP,
9.47
9.68
10.10
10.25
10.37
10.67
10.87
th
13
ed.)
Grandfather clock
Celestial Motion
Communications
Oscillation: Periodic Motion
Molecules
Locomotion
Why things oscillate?
Newton’s first law explains why harmonic motion happens for moving
objects.
•
According to the first law: an object in motion stays in motion unless
acted by a force.
Spring-Mass System
Simple Pendulum
System in Equilibrium
In unstable systems, there are forces that act to pull the system the system away
from equilibrium when disturbed. In stable systems, there are restoring forces
Unstable systems do not result in
harmonic motion.
Stable systems result in harmonic
motion.
Oscillator: system in harmonic or
periodic motion
Spring-Mass System
Cycle and Phase
Cycle: a unit of harmonic motion (complete
back-and-forth motion)
Phase: position and direction of an oscillator
in the cycle
Oscillator B is in-phase with oscillator A.
Oscillator C is 180° out-of-phase with oscillator A
Basic Quantities: Periodic Motion
Period, T: time for one cycle
Frequency, f: number of cycles per
second
𝑓=
1
𝑇
T=
1
𝑓
SI unit: Hertz (Hz), 1 Hz = 1 cycle/s
Displacement: shortest distance from
Angular frequency, 𝝎: time rate of
equilibrium position
phase
Amplitude, A: maximum displacement
2𝜋
𝜔 = 2𝜋𝑓 =
𝑇
Simple Harmonic Motion (SHM)
Necessary Conditions:
1. It arises whenever a system vibrates around an equilibrium position.
2. It is caused by a restoring force on the oscillator.
Restoring force:
It is a force that is directed toward the equilibrium position and
proportional to the displacement of the oscillator from the equilibrium
position. It maintains the system to be in equilibrium.
Examples: Spring force in a spring-mass system; tangential component of
gravitational force in a simple pendulum.
Linear Motion vs. Harmonic Motion
SHM of mass m on the end of a spring
The harmonic motion is caused by a
restoring force linearly proportional to
the displacement from equilibrium.
𝐹𝑟𝑒𝑠𝑡𝑜𝑟𝑖𝑛𝑔 ∝ 𝑥
The force satisfying this condition is elastic force or spring force.
Hooke’s Law:
𝐹𝑠𝑝𝑟𝑖𝑛𝑔 = −𝑘𝑥
Harmonic Oscillators
Horizontal SHM
Vertical SHM
Uniform Circular Motion and SHM
•
Both are cyclic
•
The key difference is that cycles of circular
motion is 2𝜋
•
The motion of the phasor is simple harmonic
UCM
SHM
Radius
Amplitude
Center
Mean position of
oscillation
Angular speed
Angular frequency
Angular position
Phase
Initial angular position Initial phase or epoch
The system in uniform
circular motion has
projection called as
phasor.
Uniform Circular Motion and SHM
Uniform Circular Motion and SHM
Recall, the magnitude of the acceleration
of the SHO:
𝑎=
𝑣2
𝐴
=
1
2
(𝐴𝜔)
𝐴
= (𝐴𝜔)2
The projection acceleration of the phasor,
𝑎𝑥 is:
𝑎𝑥 = −𝜔2 𝐴 cos(𝜔𝑡 + 𝜙) = −𝜔2 𝑥
Comparing the projection acceleration and
the SHO’s acceleration:
𝑘
2
−𝜔 𝑥 = − 𝑥
𝑚
Then, the angular frequency is:
𝜔=
𝑘
𝑚
Also, the frequency and period are:
𝜔
1 𝑘
𝑎=
=
2𝜋 2𝜋 𝑚
1 2𝜋
𝑘
𝑇= =
= 2𝜋
𝑓
𝜔
𝑚
Example
When a family of four with 200-kg net mass step into their 1200kg
car, the car’s springs compress by 3.0 cm.
(a) What is the spring constant of the car’s springs, assuming they
act as a single spring?
(b) Determine the period and frequency of the car after hitting a
bump
Kinematics of SHM
The solution to the differential equation,
𝑑2 𝑥
𝑑𝑡 2
= −𝜔2 𝑥 is:
𝑥 = 𝐴 cos 𝜃 = 𝐴 cos(𝜔𝑡 + 𝜙)
The initial position of the SHO is:
𝑥0 = 𝑥 𝑡 = 0 = 𝐴 cos 𝜃
Figure: x-t graph at 𝜙 = 0
Kinematics of SHM
The velocity of the simple harmonic oscillator is:
𝑑𝑥
𝑣𝑥 =
= −𝜔𝐴 sin 𝜔𝑡 + 𝜙
𝑑𝑡
𝑣0𝑥 = 𝑣𝑥 (𝑡 = 0) = −𝜔𝐴 sin 𝜙
𝑣𝑚𝑎𝑥 = 𝜔𝐴
at 𝜙 = 𝜋 2
The acceleration of the SHO is:
𝑑𝑣𝑥
𝑎𝑥 =
= −𝜔2 𝐴 cos 𝜔𝑡 + 𝜙 = −𝜔2 𝑥
𝑑𝑡
𝑎𝑥 = −𝜔2 𝐴 cos 𝜙
𝑎𝑚𝑎𝑥 = ±𝜔2 𝐴
Example
A large motor in a factory causes the floor to vibrate at a frequency
of 10 Hz. The amplitude of the floor’s motion near the motor is
about 3.0 mm. Estimate the maximum acceleration of the floor near
the motor.
Phase angle and amplitude in SHM
Given the initial velocity and initial position, the phase angle is obtained
from their ratio:
𝑣0𝑥 −𝜔𝐴 sin 𝜙
=
= −𝜔 tan 𝜙
𝑥0
𝐴 cos 𝜙
𝑣0𝑥
𝜙 = arctan(−
)
𝜔𝑥0
Also from the initial velocity and initial position, the amplitude is:
2
𝑣
0𝑥 1 2
2
𝐴 = [𝑥0 + 2 ]
𝜔
Energy in SHM
Since there is no nonconservative force in the system,
the total mechanical energy of the SHO is conserved:
𝐸 = 𝐾 + 𝑈 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
1
1
𝑚𝑣𝑥2 + 𝑘𝑥 2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
2
2
Recall,
𝑥 = 𝐴 cos 𝜔𝑡 + 𝜙
𝑣𝑥 = −𝜔𝐴 sin 𝜔𝑡 + 𝜙
Then,
1
1
1
2
2
𝑚𝑣𝑥 + 𝑘𝑥 = 𝑘𝐴2
2
𝐸=
2
1
𝑘𝐴2
2
2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Simple Pendulum
A simple pendulum consists of a small bob
of mass m suspended by a light inextensible
string of length L, free to swing in a vertical
plane about the equilibrium position.
A simple coordinate, 𝜃 or 𝑥, is required to
describe completely the position of the bob
at any time
Simple Pendulum
Simple Pendulum
Free-Body Diagram:
The restoring force is the tangential component of the
gravitational force:
𝐹𝜃 = −𝑚𝑔 sin 𝜃
To be simple harmonic, 𝜃 must be small. Then,
sin 𝜃 ≈ 𝜃
Thus,
𝑚𝑔
𝐹𝜃 = −𝑚𝑔𝜃 = −
𝑥
𝐿
𝑚𝑔
The restoring force constant: 𝑘 =
𝐿
The period of simple pendulum, small amplitude:
𝑇=
2𝜋
𝜔
= 2𝜋
𝑚
𝑘
= 2𝜋
𝐿
𝑔
Example
A geologist uses a simple pendulum that has a length of
37.10 cm and a frequency of 0.8190 Hz at a particular
location on the Earth. What is the acceleration of gravity at
this location?
Torsional Pendulum
Physical Pendulum
Wave Pendulum
The pendulum wave is a
special case of multiple
simple pendulum, which
has a fixed relation between
the length of the pendulum
and its period.
Oscillators: system with motion caused by restoring force
Damping force: dissipative force
proportional to the speed of oscillator
Driving force: applied force, either
periodic or arbitrary function of time
Underdamped oscillation
Over and critically damped
Forced Oscillation
Resonance
It is the phenomenon in which a system is made to oscillate by external
force whose frequency is equal to the natural frequency of the system.
At resonance, the amplitude of the system is maximum. It is a special
case of forced oscillation.
The condition for resonance is:
𝜔 = 𝜔0
References
1. Young, H. D., Freedman, R. A., Ford, A. L., & Sears, F. W. (2016). Sears and
Zemansky's University Physics: With modern physics. San Francisco:
Pearson Addison Wesley.
2. Gibilisco, Stan. Physics DeMYSTiFieD, Second Edition. US: McGrawHill
Professional, 2010.
3. Feynman, R. P., Leighton, R. B., & Sands, M. L. (1963). The Feynman
lectures on physics. Reading, Mass: Addison-Wesley Pub. Co.
4. Serway, Raymond A. (2007). Essentials of college physics. Belmont, Calif. ;
[Toronto] :Thomson-Brooks/Cole
5. Giancoli, Douglas C. (2005) Physics: principles with applications Upper
Saddle River, N.J. : Pearson/Prentice Hall
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