Transcript Physics Ch. 7 Rotational Motion

Physics Ch. 7 Rotational Motion
Rotational motion-involves objects that spin around a
fixed point (axis)
 Radius-distance from the center to the edge of the circle or
object
 Arc length (s) is the distance the object moves, measured
along the circumference of the circle
 Angle (θ) is the angle through which the object moves
 Arc length and angle are related- if the angle doubles, then
the arc length or distance also doubles

 Angles
can be measured in radians as well as
degrees. All angles for this chapter will need to be
in radians.
 When the arc length is equal to the length of the
radius, then the angle swept by r is equal to 1 rad,
which is equal to about 57.3°.

θ = s/r = 2π rad = 2 (3.14) = 6.28 rad = 360°
Converting degrees to radians
 Multiply
the angle in degrees by 2π/360°
OR
θ (rad) = π/180° θ (deg)
Angular Displacement

The change in the arc length, Δs, divided by the distance
of the object from the axis.
Δθ = Δs/r
 Angular displacement (radians) = change in arc
length/distance from axis
 S is considered positive when the point rotates
counterclockwise and negative when it rotates clockwise

Angular Displacement
 Sample
Problem 7A pg. 246
You Try It
 Earth
has an equatorial radius of approximately
6380 km and rotates 360° every 24 h.
 A.
What is the angular displacement (in degrees)
of a person standing at the equator for 1.0 h?
 B. Convert this angular displacement to radians.
 C. What is the arc length traveled by this person?
Angular Speed
angular speed ωavg is the ratio of the
angular displacement, Δt, that the object takes to
undergo that displacement. Angular speed
describes how quickly the rotation occurs.
 Average
ωavg = Δ θ/ Δ t
Average angular speed = angular displacement/time
 Angular
speed is given in units of radians per
seconds. Sometimes they are given in revolutions
per unit time. 1 rev = 2 π rad
Angular Speed
 Sample
Problem 7B on pg. 248
You Try It-Angular Speed
 An
Indy car can complete 120 laps in 1.5 h. Even
though the track is an oval rather than a circle, you
can still find the average angular speed. Calculate
the average angular speed of the Indy car.
Angular Acceleration
 Angular
acceleration has the units radians per
second per second
 Average angular acceleration = change in angular
speed/ time interval
 α avg =
ω2 – ω1
t2 – t1
Angular Acceleration
 Sample
Problem 7C pg. 249
You Try it-Angular Acceleration
 A top
that is spinning at 15 rev/s spins for 55 s
before coming to a stop. What is the average
angular acceleration of the top while it is slowing?
 Do
two points on a wheel with two different radi
have the same angular speed and angular
acceleration?
 Demo
1
 Compare
the variables of rotational motion to
linear motion from ch. 2
 Linear
x
v
a
Angular
θ
ω
α
Kinematics Formulas
 Formulas
in Table 7-2 on pg. 251
ω in these equations represents the instantaneous
angular speed of the rotating object rather than the
average angular speed.
Angular Kinematics
 Sample
Problem 7D on pg. 251
You Try it-Angular Kinematics
 A barrel
is given a downhill rolling start of 1.5
rad/s at the top of a hill. Assume a constant
angular acceleration of 2.9 rad/s2
 A. If the barrel takes 11.5 s to get to the bottom of
the hill, what is the final angular speed of the
barrel?
 B. What angular displacement does the barrel
experience during the 11.5 s ride?
Ch. 7-2 Tangential and Centripetal
Acceleration
 How
does angular speed and acceleration relate to
golfing?
 Angular speed and acceleration are related to the
linear speed and acceleration.
 For example, the golfer wants the club to travel an
approximate circle around the body. A large
angular acceleration will also produce a large
linear acceleration of the club on the ball.
Tangential Speed
 Two
objects with different distances from the axis
of rotation will have the same angular speed and
acceleration.
 However, two objects with different distances from
the axis of rotation will have different tangential
speeds.
 Tangential speed=instantaneous linear speed
The object on the outside of a circle must travel the same
angular displacement during the same amount of time as
the object on the inside.
 So, the object on the outside must travel a greater distance,
Δs, than the object on the inside.
 An object farther from the axis must travel at a higher
tangential speed to achieve the same angular displacement
as an object closer to the axis.

Tangential Speed Formula

Tangential speed = distance from axis x angular speed
 Vt =
ω
rω
is equal to instantaneous angular speed
and is measured in radians per unit of time
only
Tangential Speed
 Sample
problem 7E on pg. 254
You try it!
A golfer has a maximum angular speed of 6.3 rad/s for her
swing. She can choose between two drivers, one placing
the club head 1.9 m from her axis of rotation and the other
placing it 1.7 m from the axis.
 A. Find the tangential speed of the club head for each
driver.
 All other factors being equal, which driver is likely to hit
the ball farther?

Tangential Acceleration
 The
instantaneous linear acceleration of an object
directed along the tangent to the object’s circular
path.
 If
an object traveling along a circular path speeds
up, it experiences an angular acceleration. The
linear acceleration related to this angular
acceleration is tangent to the circular path.
Tangential Acceleration

Tangential acceleration = distance from axis x angular acceleration
 at
= rα
Tangential Acceleration
 Sample
problem 7F on pg. 256
You Try It!
has a tangential acceleration of 0.98 m/s2
when it is released. The string is wound around a
central shaft of radius 0.35 cm. What is the
angular acceleration of the yo-yo?
 A yo-yo
Centripetal Acceleration
 As
an object moves in a circular path at constant
speed, it still has an acceleration.
 Remember that acceleration can be produced by a
change in magnitude or a change in direction.
 For an object traveling at a constant speed in a
circular path, the acceleration is due to a change in
direction, centripetal acceleration (center seeking).
 Fig.
7-8
 Δv = vf – vi
 Notice that V initial is negative so the vector is
reversed. Also, if the change in time is very small,
these two vectors will almost be perpendicular to
each other, making the resultant pointing toward
the center of the circle.
Centripetal Acceleration
 Centripetal
acceleration = (tangential speed)2
distance from axis
ac = vt2
r
Centripetal acceleration = distance from axis x
(angular speed)2
ac = rω2
Centripetal Acceleration
 Sample
problem 7C on pg. 258
You Try It!
 A cylindrical
space station with a 115 m radius
rotates around its longitudinal axis at an angular
speed of 0.292 rad/s. Calculate the centripetal
acceleration:
 A. Halfway to the rim of the station
 B. At the rim of the station
Tangential vs. Centripetal
Acceleration
 Tangential
acceleration is due to changing speed.
 Centripetal acceleration is due to changing
direction.
 Fig. 7-9
 When both types exist simultaneously, the total
acceleration can be found using the Pythagorean
theorem.
 The
direction of the total acceleration depends on
the magnitude of each component of acceleration
and can be found using the inverse of the tangent
function.
θ
= tan-1 a
c
at
Causes of Circular Motion
 A ball
connected to a string that is whirled in a
circular path has a centripetal acceleration because
the direction of the ball is constantly changing.
 The
inertia of the ball tends to maintain the ball’s
motion in a straight path (tangent).
 However, the string is forcing the ball to follow a
circular path, centripetal force.
 This is the force that maintains the circular motion
of the object.
Force that Maintains the Circular Motion
 Fc = mvt 2
r

 Force that maintains circular motion = mass X (tangential speed)2
distance to radius
Fc = mrω2
Force that maintains circular motion = mass X distance to axis X
(angular speed)2
Demo 3
 Any
object with a constant speed will still have a
centripetal acceleration due to the change in
direction.
 The force that keeps the object moving circular is
the centripetal force.
 Try this with a ball attached to string.
 Try this with an object that has a constant velocity.
Sample Problem 7H
 Force
that maintains the circular motion on pg. 261
Challenge Problem
moon (mass = 7.36 X 1022 kg) orbits Earth at a
range of 3.84 X 105 km with a period of
approximately 28 days. Determine the force that
maintains the circular motion of the moon.
 The
 The
force that maintains circular motion is
necessary for circular motion.
 If the force vanishes, then the object will travel in a
straight line path tangent to the circle.
 Notice that the force is always at a right angle to
the direction of motion.
 Fig. 7-11 shows the direction the ball follows if the
force vanishes.
 Consider
a car making a sharp left turn with a
passenger inside the car. The passenger slides
right and hits the door. Is there a force pushing an
object outward?
 Centrifugal force is often used to explain this but
not used in this book.
 Could it be inertia?
Inertia can explain
 The
passenger is traveling straight. When the car
turns left, the passenger wants to continue straight
because of inertia (Newton’s first Law).
 However, if a larger force, that maintains the
circular motion, acted on the passenger, it would
force the person to move in a circular motion with
the car. Friction is the source of the force (between
the person and the seat).
 If
the force is not sufficient, the person slides
across the seat.
 Eventually,
the door will provide a large enough
force to allow the passenger to follow the circular
motion with the car.
Newton’s Law of Universal Gravitation
 Planets
move in a circular orbit around the sun.
 The force keeping these planets from moving off in
a straight path is called the gravitational force.
 The gravitational force is the force between any
two objects.
 This force is a force of attraction and equal but
opposite in magnitude between two objects.
Newton’s Law of Universal Gravitation
 Fg =
G m1 m2
r2

 Graviational

Force = constant X mass1 X mass 2
(distance b/t center of
masses)2
G
is the constant of universal gravitaton and equal
to 6.673 X 10-11
 The
force between two masses decreases as the
masses move farther apart.
 By
substituting the actual values of the mass of
Earth and radius of Earth, we can find the free-fall
acceleration of 9.83 m/s2
Gravitational Force
 Sample
Problem 7I on pg. 264
You Try It!
 Find
the gravitational force exerted on the moon
(mass = 7.36 X 1022 kg) by Earth (mass = 5.98 X
1024 kg) when the distance between them is 3.84 X
108 m).