Transcript Rotational Motion - My Teacher Pages

Unit 8
Center of Mass
• A point that represents the average location for the
total mass of a system
• For symmetric objects, made from uniformly
distributed material
• Center of mass = Geometric center
xcm
m1 x1  m2 x2

m1  m2
Center of Mass & Motion
• Center of mass follows a projectile path
Rotational Motion
• Focuses on pure rotational motion
• Motion that consists of rotation about
a fixed axis
• Points on a rigid object move in circular
paths around an axis of rotation
• Examples:
• Ferris Wheel; CD
Angular Position
• In the study of rotational motion, position is
described using angles (θ)
• Angular Position = The amount of rotation
from a reference point
• Counterclockwise rotation = positive angle
• Clockwise rotation = negative angle
• Units = radians
Angular Position
Arc length s
 (in radians) 

Radius
r
For a full revolution:
2 r

 2 rad
r
2 rad  360

Angular Displacement
• In the study of rotational motion, angular
displacement is also described using angles (θ)
• Change in angular position
• Δθ = θf − θi
•
•
•
•
Δθ = angular displacement
θf = final angular position
θi = initial angular position
Units = radians (rad)
• Counterclockwise rotation = Positive
• Clockwise rotation = Negative
Angular Displacement
The angle through which the
object rotates is called the
angular displacement.
     o
PROBLEM:
Adjacent Synchronous Satellites
Synchronous satellites are put into
an orbit whose radius is 4.23×107 m.
If the angular separation of the two
satellites is 2 degrees, find the
arc length that separates them.
ANSWER
Arc length s
 (in radians) 

Radius
r
 2 rad 
  0.0349 rad
2.00 deg
 360 deg 
s  r  4.23 107 m0.0349 rad
1.48 106 m (920 miles)
Angular Velocity
     o
How we describe the rate at which angular
displacement is changing…
Average Angular Velocity
Angular displaceme nt
Average angular ve locity 
Elapsed time
  o



t  to
t
SI Unit of Angular Velocity: radian per second (rad/s)
Problem
Gymnast on a High Bar
A gymnast on a high bar swings through
two revolutions in a time of 1.90 s.
Find the average angular velocity
of the gymnast.
Answer
2 rad 
  2.00 rev
 12.6 rad
 1 rev 
 12.6 rad

 6.63 rad s
1.90 s
Problem
It takes a motorcycle rider 2 seconds to make a
counterclockwise lap around a track. What is his
average angular velocity?
  o



t  to
t
2 rad
 
  rad s
2s
Angular Acceleration
Changing angular velocity means that an angular acceleration is
occurring.
DEFINITION OF AVERAGE ANGULAR ACCELERATION
Change in angular ve locity
Average angular accelerati on 
Elapsed time
 
  o
t  to


t
SI Unit of Angular acceleration: radian per second squared (rad/s2)
Problem
A Jet Revving Its Engines
As seen from the front of the
engine, the fan blades are
rotating with an angular
speed of -110 rad/s. As the
plane takes off, the angular
velocity of the blades reaches
-330 rad/s in a time of 14 s.
Find the angular acceleration, assuming it to
be constant.
Answer
 

  o
t  to


t

 330 rad s    110 rad s 

 16 rad
14 s
s
2
Problem
Over the course of 1 hour, what is
(a) The angular displacement
(b) The angular velocity and
(c) The angular acceleration of the minute
hand?
Answer
VARIABLES:
Elapsed Time
Δt = 1 h  3600 s
Angular Displacement Δθ
Angular Velocity
ω
Angular Acceleration
α
STEP-BY-STEP SOLUTION
* Angular Displacement
Δθ = -2πrad (Minute hand travels clockwise one revolution)
Δt = 1 h  3600 s
* Angular Velocity
ω = Δθ/Δt = -2πrad/3600s = -((2)(3.14))/3600 = -1.75 x 10-3 rad/s
* Since the angular velocity is constant, angular acceleration is zero
Problem
At a particular instant, a potter's wheel
rotates clockwise at 12.0 rad/s; 2.50 seconds
later, it rotates at 8.50 rad/s clockwise. Find
its average angular acceleration during the
elapsed time.
Answer
Average Angular Acceleration:
α = Δω / Δt
= (-8.5 rad/s) – (-12 rad/s) / 2.5 s
= 1.4 rad/s2
Did NOT do slides
beyond this point
Angular Velocity Vector
Angular Velocity Vector:
* Parallel to the axis of rotation
* Magnitude of the angular velocity vector is proportional
to the angular speed
- Faster the object rotates  longer the vector
* Direction of the angular velocity vector determined by the
right-hand rule
Right-Hand Rule
Right-hand Rule:
* Used to determine the direction of the angular velocity
vector
* Fingers of the right hand curl in the direction of rotation
* Thumb then points in the direction of the angular
velocity vector
Angular Velocity Vector
Right-Hand Rule: Grasp the
axis of rotation with
your right hand, so that your
fingers circle the axis
in the same sense as the
rotation.
Your extended thumb points
along the axis in the
direction of the angular
velocity.
Angular Acceleration
Vector
Angular Acceleration Vector:
* If the object is speeding up  its angular velocity vector
is increasing in magnitude  Angular acceleration vector
points in the same direction as the angular velocity vector
* If the object is slowing down  its angular velocity vector
is decreasing in magnitude  Angular acceleration vector
points in the opposite direction as the angular velocity
vector
Preview Kinetic Books: 10.15
Recall… The Kinematic Equations For Constant
Acceleration
Five kinematic variables:
1. displacement, x
2. acceleration (constant), a
3. final velocity (at time t), v
4. initial velocity, vo
5. elapsed time, t
v  vo  at
x  12 vo  v  t
v  v  2ax
2
2
o
x  vot  at
1
2
2
Rotational Kinematic Equations:
For Constant Angular Acceleration
ANGULAR ACCELERATION
ANGULAR VELOCITY
  o   t
  12 o    t
ANGULAR DISPLACEMENT
    2
2
2
o
  o t   t
1
2
2
TIME
Rotational Kinematic Equations
Reasoning Strategy
1. Make a drawing.
2. Decide which directions are to be called positive (+) and
negative (-).
3. Write down the values that are given for any of the five
kinematic variables.
4. Verify that the information contains values for at least three
of the five kinematic variables. Select the appropriate equation.
5. When the motion is divided into segments, remember that
the final angular velocity of one segment is the initial velocity
for the next.
6. Keep in mind that there may be two possible answers to a
kinematics problem.
PROBLEM: Blending with a Blender
The blades are whirling with an
angular velocity of +375 rad/s when
the “puree” button is pushed in.
When the “blend” button is pushed,
the blades accelerate and reach a
greater angular velocity after the
blades have rotated through an
angular displacement of +44.0 rad.
The angular acceleration has a
constant value of +1740 rad/s2.
Find the final angular velocity of the blades.
θ
α
ω
ωo
+44.0 rad
+1740 rad/s2
?
+375 rad/s
t
 2  o2  2
  o2  2

375 rad s 
2

 2 1740 rad s
2
44.0rad   542 rad s
Tangential Velocity
Tangential Velocity:
* Linear velocity at an instant
- Magnitude = magnitude of linear velocity
- Direction: Tangent to circle
* vT = rω
vT = tangential speed
r = distance to axis
ω = angular velocity
Direction: tangent to circle
Preview Kinetic Books: 10.11
Tangential Velocity

vT  tangentia l velocity
Tangential Velocity
s r
 
vT  
 r 
t
t
t
vT  r ( in rad/s)


t
Tangential Acceleration
Tangential Acceleration:
- Rate of change of tangential speed
- Increases with distance from center
- Direction of vector is tangent to circle
- aT = rα
aT = tangential acceleration
r = distance to axis
α = angular acceleration
Preview Kinetic Books: 10.12
Tangential Acceleration
vT  vTo r   ro 
  o
aT 

r
t
t
t
aT  r
( in rad/s )
2
  o

t
PROBLEM: A Helicopter Blade
A helicopter blade has an angular speed of 6.50 rev/s and an
angular acceleration of 1.30 rev/s2.
For point 1 on the blade, find
the magnitude of
(a) The tangential speed
(b) The tangential acceleration.
rev  2 rad 

   6.50

  40.8 rad s
s  1 rev 

vT  r   3.00 m 40.8rad s   122m s
rev  2 rad 

2
  1.30 2 
  8.17 rad s
s  1 rev 



aT  r  3.00 m 8.17 rad s  24.5 m s
2
2
Centripetal Acceleration

v
r 
ac 

 r 2
r
r
2
T
2
( in rad/s)
Problem: A Discus Thrower
Starting from rest, the thrower
accelerates the discus to a final
angular speed of +15.0 rad/s in
a time of 0.270 s before releasing it.
During the acceleration, the discus
moves in a circular arc of radius
0.810 m.
Find the magnitude of the total
acceleration.
Centripetal Acceleration & Tangential
Acceleration
ac  r  0.810 m 15.0 rad s 
2
2
 182 m s 2
ω-ωo
 15.0 rad s 
aT  r  r
 0.810 m 

t
 0.270 s 
 45.0 m s 2
a  a a 
2
T
2
c
182 m s

2 2
  45.0 m s

2 2
 187 m s 2
Rotational Motion Review
• UNIFORM CIRCULAR MOTION:
• Example- Spinning Ferris wheel or an orbiting satellite
• Object moves in a circular path and at a constant speed
• The object is accelerating, however, because the direction of
the object’s velocity is constantly changing
• Centripetal acceleration  Directed toward the center of the
circle
• Net force causing the acceleration is a centripetal force 
Also directed toward the center of the circle
• In this section, we will examine a related type of
motion  The motion of a ROTATING RIGID OBJECT
The Motion of a Rotating
Rigid Object
• A football spins as it flies through the air
• If gravity is the only force acting on the football
 Football spins around a point called its
center of mass
• As the football moves through the air, its center
of mass follows a parabolic path
Rotational Dynamics
• Study of Rotational Kinematics
• Focuses on analyzing the motion of a rotating
object
• Determining such properties as its angular
displacement, angular velocity or angular
acceleration
• Study of Rotational Dynamics
• Explores the origins of rotational motion
* Preview Kinetic Books 11.1
Torque
• Every time you open a door, turn of a water faucet, or
tighten a nut with a wrench  you are exerting a
TURNING FORCE
• Turning force produces a TORQUE
• Review…
• If you want to make an object move  Apply a force
• Forces make things accelerate
• If you want to make an object turn or rotate  Apply a
torque
• Torques produce rotation
Torque
• Torque
• A force that causes or opposes rotation
• The amount of torque depends on…
• The amount of force
• When more force is applied  there is more
torque
• The distance from the axis of rotation to the
point of force
The amount of torque depends on where and in
what direction the force is applied, as well as the
location of the axis of rotation.
Torque
• For a force applied perpendicularly
  rF
τ = magnitude of torque (Greek letter tau)
r = distance from axis to force
F = force
•Direction:
•Torque resulting from a force is + if the rotation is
counterclockwise and – if the rotation is clockwise
•Counterclockwise +, clockwise −
•S.I. Units: newton-meters (N∙m)
Torque
• Imagine a dog-flap door
• Rotates on a hinge
• Allowing pets to enter and leave a house at will
Axis of Rotation
Lever Arm:
Perpendicular
distance from the
axis of rotation to
a line draw along
the direction of
the force
Force:
* If the dog pushed on the door with the same
force but at a point closer to the hinge, the door
would be more difficult to rotate
Torque, angle & lever
arm
• * Preview Kinetic Books 11.2
•Forces do not have to be perpendicular to an
object to cause the object to rotate
τ=r×F
τ = rF sin θ
τ = Torque (N x m)  Vector quantity
r = Position vector (m) (AKA: Displacement)
F = Force (N)  Vector quantity
θ = angle between r and F
Calculating Torque using
the Lever Arm
• Torque can also be calculated using the
concept of a lever arm
• Lever arm is the perpendicular distance from
the axis of rotation to the line containing the
force vector
• “Line of action”
τ = F (r sin θ)
* r sin θ = lever arm
Mathematically:
τ = F (r sin θ) is the same as τ = rF sin θ
Torque
Problem
• A basketball is being pushed by two players
during a tip-off. One player exerts an upward
force of 15 N at a perpendicular distance of 14
cm from the axis of rotation. The second
player applies a downward force of 11 N at a
perpendicular distance of 7 cm from the axis of
rotation. Find the net torque acting on the ball
about its center of mass.
Answer
• Given:
• F1 = 15N
• F2 = 11N
• τnet = ?
r1 = 0.14m
r2 = 0.07m
τ = F1r1 + F2r2
(** The factor sin Θ is not included because each distance is the
perpendicular distance from the axis of rotation to a line drawn
along the direction of the force)
Answer Continued
• τ = F1r1 + F2r2
• * Use the standard convention of signs…
• τ = (15N)(-0.14m) + (-11N)(0.070m)
• τ = -2.9 Nxm
• Net torque is negative, so the ball rotates in a
clockwise direction
The Achilles Tendon
The tendon exerts a force of
magnitude 790 N. Determine
the torque (magnitude and
direction) of this force about
the ankle joint.
Answer
τ = rF sin θ
790 N
   3.6 102 m   790 N  sin145
 16.3 N  m