Transcript File

Chapter 8
Rotational Motion
Rotational Motion
Rigid body = a body with a definite
shape that doesn’t change, so that the
particles composing it stays in fixed
positions relative to one another.
Angular Quantities
In purely rotational motion, all
points on the object move in
circles around the axis of
rotation (“O”). The radius of
the circle is r. All points on a
straight line drawn through the
axis move through the same
angle in the same time. The
angle θ in radians is defined:
where l is the arc length. (some
times given as s also)
We see that when the disk rotates so that the arc length (s)
equals the length of the radius of the disk (r), angle (θ) will
equal 1 radian. The resulting equation is
s = rθ
θ = s/r
If the disk rotates through one
complete revolution, then s equals
the entire circumference and θ
equals 2π radians
s = rθ
2πr = rθ
θ = 2π radians
where the unit of a radian
represents the
dimensionless measure of
the ratio of the circle's arc
length to its radius
Since one complete revolution
equals 360º, we now have the
conversion that
360º = 2π radians
1 radian = 180/π or
approximately 57.3º
Angular Quantities
Angular displacement:
The average angular velocity is
defined as the total angular
displacement divided by time:
Angular Quantities
The angular acceleration is the rate at which the
angular velocity changes with time:
Angular Quantities
Every point on a rotating body has an angular
velocity ω and a linear velocity v.
They are related:
Angular Quantities
Therefore, objects
farther from the axis
of rotation will move
faster.
Buzz Lightyear!
Ladybugs
Angular Quantities
If the angular velocity of a
rotating object changes, it
has a tangential
acceleration:
Even if the angular
velocity is constant, each
point on the object has a
centripetal acceleration:
Angular Quantities
Here is the correspondence between linear
and rotational quantities:
Constant Angular Acceleration
The equations of motion for constant angular
acceleration are the same as those for linear
motion, with the substitution of the angular
quantities for the linear ones.
Problem 1
An old phonograph record revolves at 45
rpm.
What is its angular
Once the motor is turned off,
velocity in rad/sec?
it takes 0.75 seconds to come
ω = 45 rev/min = 4.71 rad/sec to a stop. What is its average
angular acceleration?
givens:
ωf = 0, ωo = 4.71 rad/sec, t = 0.75
seconds.
using the equation
ωf = ωo + αt
we can determine that
α = (0 - 4.71)/0.75 = -6.28 rad/sec2
How many revolutions did it make while
coming to a stop?
Combining α = ω/t and θ = ωot +½αt2
we get:
θ = ½(ωf + ωo)t
we can determine that
θ = ½ (0 + 4.71) 0.75
θ = 1.77 radians
since there are 2π radians in every
revolution,
θ = 0.281 rev.
A fan that is turning at 10
rev/min speeds up to 25
rev/min in 10 seconds.
b. If the tip of one blade is 30
cm from the center, what is
the final tangential velocity of
the tip?
a. How many revolutions
does the blade require to alter
using the equation
its speed?
v = rω
allows us to determine that
using the equation
v = (0.30)(2.63)
θ = ½(ωf + ωo)t
v = 0.789 m/sec
we can determine that
θ = ½(1.05 + 2.63)10
θ = 18.4 radians
since there are 2π radians in every
revolution, θ = 2.92 rev
Consider a standard analog wall clock with a second
hand, minute hand, and hour hand
a. Calculate the angular velocity of the second
hand of a clock
ω = 1 rev/min = 0.105 rad/sec
b. If the second hand is 8" long (there are 2.54 cm in
every inch), what is the linear velocity of the tip of the
second hand?
using the equation
v = rω
allows us to determine that
v = (0.203)(0.105)
v = 0.0213 m/sec
Two wheels are connected by a common cord.
One wheel has a radius of 30 cm, the other has a
radius of 10 cm
When the small wheel is revolving at
10 rev/min, how fast is the larger
wheel rotating?
Since the two wheels share the
same tangential velocity, their
angular velocities will be
inversely proportional to their
radii.
vlarge=vsmall
rlargeωlarge= rsmallωsmall
ωlarge = (rsmall/rlarge) ωsmall
ωlarge = (0.10/0.30)(10) = 3.33
rev/min
A rotor turning at 1200 rev/min has a diameter
of 5 cm. As it turns, a string is to be wound
onto its rim
How long of a piece of string
will be wrapped in 10
seconds?
since the wheel is turning at a
constant angular velocity, we can
use the equations
θ = ωt
s = rθ
substituting gives us the equation
s = r(ωt)
and we can calculate the amount of
string wrapped around the exterior
of the rotor
s = (0.025)(1200)(0.105)(10)
s = 31.5 meters
Rolling Motion (Without Slipping)
In (a), a wheel is rolling without
slipping. The point P, touching
the ground, is instantaneously
at rest, and the center moves
with velocity v.
In (b) the same wheel is seen
from a reference frame where C
is at rest. Now point P is
moving with velocity –v.
The linear speed of the wheel is
related to its angular speed:
Example
A Bicycle slows down from 8.4m/s to rest over a distance of 115m.
a) Find the angular velocity of the wheels when the bike tis traveling 8.4m/s.
b) The total number of revolutions each wheel makes before coming to a stop.
c) The angular acceleration of the wheel?
d) The time it took to stop?
Torque
To make an object start rotating, a force is needed;
the position and direction of the force matter as well.
The perpendicular distance from the axis of rotation
to the line along which the force acts is called the
lever arm.
Units for Torque
Torque is proportional to the
magnitude of F and to the distance r
from the axis. Thus, a tentative
formula might be:
t = Fr
Units: Nm or
lbft
t = (40 N)(0.60 m)
= 24.0 Nm, cw
t = 24.0 Nm, cw
6 cm
40 N
Torque
A longer lever
arm is very
helpful in
rotating objects.
Torque
Here, the lever arm for FA is the distance from the
knob to the hinge; the lever arm for FD is zero;
and the lever arm for FC is as shown.
8-4 Torque
The torque is defined
as:
Example 1: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
• Extend line of action, draw, calculate
r.
r = 12 cm sin 600
= 10.4 cm
t = (80 N)(0.104 m) =
8.31 N m
Alternate: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
positive
12 cm
Resolve 80-N force into components as
shown.
Note from figure: rx = 0 and ry = 12
cm
t = (69.3 N)(0.12 m)
t = 8.31 N m as before
Example 2: Find resultant torque about
axis A for the arrangement shown below:
Find t due to
each force.
Consider 20-N
force first:
r = (4 m) sin
= 2.00 m
negative
30 N
r
300
6m
40 N
300
t = Fr = (20 N)(2 m)
= 40 N m, cw
2m
20 N
300
A
4m
The torque about A
is clockwise and
negative.
t20 = -40 N m
Example 2 (Cont.): Next we find torque
due to 30-N force about same axis A.
Find t due to
each force.
Consider 30-N
force next.
r = (8 m) sin
= 4.00 m
300
t = Fr = (30 N)(4 m)
= 120 N m, cw
r
negative
30 N
300
20 N
300
6m
40 N
2m
A
4m
The torque about A
is clockwise and
negative.
t30 = -120 N m
Example 2 (Cont.): Finally, we consider
the torque due to the 40-N force.
Find t due to
each force.
Consider 40-N
force next:
r = (2 m) sin 900
= 2.00 m
t = Fr = (40 N)(2 m)
= 80 N m, ccw
positive
30 N
r
300
2m
6m
40 N
20 N
300
A
4m
The torque about A
is CCW and positive.
t40 = +80 N m
Example 2 (Conclusion): Find resultant
torque about axis A for the arrangement
shown below:
Resultant torque is
the sum of individual
torques.
20 N
30 N
300
300
2m
6m
40 N
A
4m
tR = t20 + t20 + t20 = -40 N m -120 N m + 80 N m
tR = - 80 N m
Clockwise
Rank the wrenches from the one that is experiencing the
most torque to the one experiencing the least
Where should the 100 g mass be
placed to balance the system?
Challenge Problem: What is the mass of the bolt in the
picture below
Challenge Problem: Where do you
need to put the support to balance the
meters tick shown below? Ignore the
mass of the meter-stick.
Rotational Dynamics; Torque and
Rotational Inertia
Knowing that
, we see that
This is for a single point
mass; what about an
extended object?
As the angular
acceleration is the same
for the whole object, we
can write:
Rotational Dynamics; Torque and
Rotational Inertia
The quantity
is called the
rotational inertia of an object.
The distribution of mass matters here – these
two objects have the same mass, but the one on
the left has a greater rotational inertia, as so
much of its mass is far from the axis of rotation.
Rotational
Dynamics; Torque
and Rotational
Inertia
The rotational inertia of
an object depends not
only on its mass
distribution but also the
location of the axis of
rotation – compare (f)
and (g), for example.
A merry go round accelerates from rest to 0.68 rad/s
in 34s. Assuming the merry go round is a uniform
disk of radius 7.0m and mass 31,000kg, calculate the
net torque required to accelerate it. I = ½ MR2.
Rotational Kinetic Energy
The kinetic energy of a rotating object is given
by
By substituting the rotational quantities, we find
that the rotational kinetic energy can be written:
A object that has both translational and
rotational motion also has both translational and
rotational kinetic energy:
Rotational Kinetic Energy
When using conservation of energy, both
rotational and translational kinetic energy must
be taken into account.
All these objects have the same potential energy
at the top, but the time it takes them to get down
the incline depends on how much rotational
inertia they have.
A bowling ball of mass 7.25kg and radius
10.8cm rolls without slipping down a lane at
3.1m/s. Calculate the total Kinetic Energy of
the ball. The lane is 18m long. Hope many
rotations does the ball make on the way down
the lane?
Rotational Kinetic Energy
The torque does work as it moves the wheel
through an angle θ:
Angular Momentum and Its Conservation
In analogy with linear momentum, we can define
angular momentum L:
We can then write the total torque as being the
rate of change of angular momentum.
If the net torque on an object is zero, the total
angular momentum is constant.
Angular Momentum and Its Conservation
Therefore, systems that can change their
rotational inertia through internal forces will also
change their rate of rotation:
video
What is the angular momentum of a 2.8kg uniform
cylindrical grinding wheel of radius 28cm rotating
at 1300rpm? How much torque is required to stop
it in 6.0s?