Transcript AP C UNIT 4 - student handout

Translational motion is movement in a straight line
Rotational motion is about an axis
>Rotation is about an internal axis (earth spins)
>Revolution is about an external axis (earth orbits)
Radian (θ) measure…ratio of arc length (s) to radius
r. When s = r, we have defined 1 radian.
r
θ
s
Δs = r Δθ
ds
d
r
dt
dt
dv
d
r
dt
dt
at
ac
Total acceleration,
Direction for ω and α
RHR: direction is along axis of rotation - curl
fingers along direction of spin and direction
of thumb is direction of ω and α. Explanation
is in your book as to why…it stems from
mathematical definition of vector in rotation.
If object is speeding up, α is in direction of ω
If object is slowing down, α is opposite ω
Constant acceleration equations:
Linear:
v = vo +at
v2 = vo2 + 2aΔx
Δx = vot + ½at2
Rotation for fixed axis:
ω = ωo + αt
ω2 = ωo2 + 2αΔθ
Δθ = ωot + ½αt2
A disc rotates about an axis thru its center according to
the equation:
1 3
 (t )  t  6t
3
A) Find the angular velocity and acceleration for general time, t.
B) Find the mag. of total linear acc. of a point 0.5m from center at t = 1s.
C)Find the linear speed of a point on disc 20cm from center at t = 2s
Rotational Inertia and Kinetic energy
The kinetic energy of a rotating object will be the sum of
the kinetic energy of every point on the object:
K  12 m1v12  12 m2 v 22  12 m3 v32  ...
n
n
K   m v   mi r 
i 1
1
2
2
i i
i 1
1
2
2
i
n

2 2
1
K  2   mi ri  i ( is constant so it can be pulled out)
 i 1

we define the rotational inertia
(or moment of inertia) as
where I is measured in kg*m2
n
I   mi ri
i 1
2
Rotational Inertia for a SYSTEM:
A uniform rod of length L and mass M that can pivot
about its center of mass has 2 masses m1 and m2 placed
at each end. The moment of inertia of a uniform rod
about its center is (1/12) ML2. Find moment of inertia for
the system.
L
m1
m2
Calculus to find I
To sum up the infinite points on a solid object you must
integrate the equation for rotational inertia (I)
What is the rotational inertia of a rod of length L and
linear mass density, λ, spinning around an axis through
its center of mass?
Parallel Axis Theorem
It is used when you already know I of body about an axis
that is parallel to another axis you are trying to find.
What is the moment of inertia for a uniform rod of
length L and mass M spinning halfway between the
center of mass & its end?
Rotational kinetic energy
If object has only translational motion then its kinetic energy is just
K trans  mv
1
2
2
If an object has only rotational motion then its
kinetic energy is just
If it has both translational & rotational motion
then its kinetic energy is
The hollow ball has more R.I., therefore, more rotational KE
for the same rotational speed. Both balls have the same
translational speeds and same translational KE. Hollow ball
will go higher due to higher rotational KE.
Mass doesn’t matter. Just as with freefalling masses of
different value. With equal speeds, a tennis ball will out roll
a bowling ball or marble.
A uniform rod of mass m and length L can
rotate about a frictionless hinge that is fixed.
If the rod is released from rest where it rotates
downward about the hinge, find tangential & angular
speed of the edge of the rod at the bottom of the swing.
Cross Product
The cross product is a vector product (recall dot product was a
scalar product). The cross product of two vectors produces a
third vector which is perpendicular to the plane in which the
first two lie.
That is, for the cross of two vectors, A and B,
we place A and B so that their tails are at
a common point (tail to tail). Their cross
product, A x B, gives a third vector, C, whose
tail is also at the same point as those of A
and B. The vector C points in a direction
perpendicular (or normal) to both A and B.
The cross product is defined by the
formula A x B = |AB|sinθ
î × î = ĵ × ĵ = k×k = (1)(1)(sin 0°) = 0
Newton’s 2nd Law for Rotation
Torque can cause a change in rotational motion or can
cause a rotational acceleration. The distance from the
pivot that the force acts is called the leverarm or
moment arm, r.
r
O
The moment arm or lever arm
is the perpendicular distance
from the axis of rotation to the
line of force
The line of force or line of
action is a line drawn at the
point of application in the
direction of the application
of the force
F
Two masses hang over a fixed pulley. The pulley
has mass 1.5 kg & radius 15cm where m1=15 kg
and m2 = 10kg. The rope moves through the
pulley without slipping.
A. What is the acceleration of the boxes?
m1
m2
B. Determine the two tensions.
Yo-Yo
A string is wound around a
Yo-Yo of mass M and radius
R. The Yo-Yo is released
and allowed to fall from rest.
Find acceleration and
tension in string as it falls.
Make rotation equation and
force equation for Yo-Yo.
Work & Power
Work done on an object can change either its translational
kinetic energy or rotational kinetic energy or both
More calculus…just in time
for the holiday 
d
sin x  cos x
dx
sin
xdx


cos
x

C

d
cos x   sin x
dx
 cos xdx  sin x  C
ANGULAR
MOMENTUM
Consider a particle of mass, m, instantaneous
velocity, v, and position vector, r, where particle
moves in the xy plane about origin O.
y
v
r
O
m
x
The particle therefore has momentum, p=mv.
We extend the position vector, r, to see the
angle between r and p.
Angular Momentum
of a particle moving
about point O, is
defined as:
y
p
θ
r
θ
O
rsinθ
x
Direction of L is out of page using
RHR.
Continuing with the formula for
angular momentum…
Assuming the angle between r and p is 90o then
L = rmv
L = rm(rω)
(using our linear-angular conversion)
L = mr2ω
where I=mr2, so we now get

L
L=Iω
NOTE: An object can possess angular
momentum about any point, regardless if it’s
moving in a circle, orbit, or line about some
point.
In the figure to left, a
dropped object can
have angular
momentum about the
origin where r is
increasing along with
the velocity and
therefore the angular
momentum.
Torque and Angular Momentum
 net  I
Conservation of Angular Momentum
Assuming no net
external torques
Relationship between force (F),
torque (τ), and momentum vectors
(p and L) in a rotating system
Consider the next example in regards to
torque and angular momentum vectors
EXAMPLE 1
Consider a thin rod of mass, M, and length, L,
lying on a frictionless table. There is a frictionless
pivot at the top end of the rod
A mass, m, slides in a
speed, vo, and collides
with the rod a distance
2/3L from pivot.
The mass rebounds
with speed, ¼vo, where
moment of inertia of rod
is 1/12ML2 about CM.
Top View
Find the
angular
velocity after
the collision
EXAMPLE 2
A dart of mass, m, is shot with speed, vo, at a
hoop of mass, M, and radius R. The hoop
can be considered a ring. The dart strikes
and sticks into top of hoop.
a) Find speed that wheel and dart rotate with
after collision.
b) Find KE lost due to collision.
ROLLING
Consider a wheel that only has pure
translational movement. All points on
wheel have the same velocity.
Consider a wheel that has pure rotation. All points have the
same angular velocity, but the linear velocity depends on the
distance from the center.
If a wheel has both translational and rotational
movement across a level surface, this is what is referred
to as rolling motion.
Relative to the point of contact with surface (bottom of
the wheel), the top of wheel is moving twice as fast as
center of wheel (vedge > vCM ). The bottom of wheel is
considered to be instantaneously at rest with no
relative motion with ground.
Rolling without Slipping
If object rolls without slipping, the arc-length of a path along the
surface of the object as it rotates matches the translational
distance traveled by the center of the object.
Since s = Rθ
Differentiating with respect to time
(ds/dt = Rdθ/dt) yields
vcm = Rω (R = dist from pt of
contact)
acm = Rα
**These are the conditions
for rolling without slipping
or smooth rolling.
If a circular object rolls without slipping where vcm=
constant, then acm = 0. Consequently, angular
acceleration is zero and the net torque is zero.
Thus, no net torque is needed to maintain rolling
without slipping at constant speed. (angular
momentum is enough to keep object turning)
An object that rolls without slipping over a
frictionless surface continues to do so, and an
object that rolls without slipping at constant
speed with friction experiences no frictional
force…meaning, you don’t need friction to
continue to roll.
Violating pure rolling condition:
If a force were applied to COM of a sphere
in the absence of friction, the sphere would
start to move, changing vCM, but without
changing ω.
This violates our condition, vCM =Rω,
therefore the sphere starts to slide.
vcm = Rω & acm = Rα are invalid.
F
If a net force acts on a rolling wheel to speed it up or
slow it down, than that net force causes an aCM
along the direction of travel. In conjunction, this
causes an angular acceleration.
In order to counteract any tendency to slip if α is too
large, static friction acts to maintain pure rolling.
Below is an example if the wheel was made (by axle
turning faster) to move faster to the right while
already in motion. Static prevents slipping to left
If the wheel was made to
slow down, both vectors
would be reversed.
aC
M
fs
Example
A solid cylinder is at rest on a flat surface. When a horizontal
force is exerted on the cylinder’s axle, what is the minimum
coefficient of static friction to keep the cylinder from slipping?
Fapp
fs
Consider a solid sphere rolling from rest
down an incline. What forces would act on
rolling sphere? Would friction have to act?
Determine aCM and the static frictional
force on sphere of mass, m, & radius R.
Rolling with Slipping
A bowling ball of mass, m, and radius, R, is
initially thrown so that it only slides with speed vo
but doesn’t rotate. As it slides, it begins to spin,
and eventually rolls without slipping. How long
will it take to stop sliding with only pure rolling
without slipping?