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
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
F
Rotational Equilibrium
Fy  0 Fx  0
  0
Example: A hungry 700N bear walks
out on a uniform beam in an attempt
to retrieve some goodies hanging at
the end. The beam weighs 200N and
is 6.0m long; the goodies weigh 80N.
a) Draw a force diagram.
b) When the bear is at 1.00 m,
determine the tension in the wire.
c) Determine angle of the reaction
force from wall on beam.
A rod is held in place by
a light wire attached to a
wall as shown. The
weight of rod is 1000N.
Hanging from rod is
2000N crate.
a) Find the tension in the wire. Diagram forces.
b) Determine reaction force (mag & dir)
Example 3: A ladder having a uniform density
and a mass, m, rests against a frictionless vertical
wall at an angle of 60◦ . The lower end rests on a
flat surface where the coefficient of static friction
is µs = 0.40. A student with a mass M = 2m
attempts to climb the ladder. What fraction of
the length L of the ladder will the student have
reached when the ladder begins to slip?
Draw a force diagram.
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
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
When a rigid object rolls across a perfectly level surface,
the object’s contact point with the surface is
instantaneously at rest. If this were untrue, the object
would be slipping or skidding. Because the contact point
is at rest, you can think of this as an instantaneous axis
of rotation.
Relative to the contact point, all points on the object
have the same angular velocity even though they have
different linear velocities.
You can think of rolling as a combination of a
pure rotation about the CM and a pure translation
of the CM.
The velocity of any point on the disk as seen by an
observer on the ground is the vector sum of the velocity
with respect to the center of mass and the velocity of the
center of mass with respect to the ground:
v pt on disk rel ground = v pt on disk rel cm + v cm rel to ground
FOR EXAMPLE…
vtop rel to cm  R
vCM rel to grnd  R
vtop rel to grnd  2vCM
The point on the top of the wheel has a speed (rel to
ground) that is twice the velocity of the center of mass.
Consider the point in contact with the ground:
vbot rel to cm  R
vCM rel to grnd  R
vbot rel to grnd  R  R  0
The point in contact with the ground has a speed of zero,
momentarily at rest. If your car is traveling down the highway at 70
mph, the tops of your wheels are going 140 mph while the
bottoms of the wheels are going 0 mph.
Rolling Motion is considered a combination of
both translational and rotational
Consider a solid sphere rolling from rest
down an incline.
Question: What forces would act on the rolling
sphere? Would friction have to act for it to roll?
Example: A solid sphere rolls down an incline.
Determine aCM and the static frictional force on the
sphere of mass, m, & radius R.
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 both sides with
respect to time (ds/dt = Rdθ/dt,
where R=constant) yields
vcm = Rω
(R = distance from pt of contact)
acm = Rα
**These are the conditions
for rolling without slipping or
smooth rolling.
Violating pure rolling condition:
If a force were applied to CM of a sphere (no
friction), the sphere would start to move,
changing vCM, but without changing ω.
F
Static friction needs to be present to initiate rolling.
In pure translation like in sliding, the sole purpose of
friction is to oppose relative motion between
surfaces.
In the case of rolling, friction converts a part of one
type of acceleration to another (from linear to
angular as in this case). Probably more apropos
would be saying that friction changes a part of
translational kinetic energy into rotational kinetic
energy.
Without friction, the force passing through the CM
would have only caused linear acceleration.
Example 1: A solid cylinder is at rest on a flat surface.
When a horizontal force, F, is exerted on the cylinder’s
axle, what is the minimum coefficient of static friction to
keep the cylinder from slipping? Cylinder has mass, m.
F
fs
Why would friction be to the left?
Example 2: A cylinder of mass M and radius R has a
string wrapped around it, with the string coming off the
cylinder above the cylinder. If the string is pulled to the
right with a force F, what is the acceleration of the
cylinder if the cylinder rolls without slipping? What is the
frictional force acting on the cylinder?
F
QUESTION: Must there be friction? If so, what kind?
What direction?
Rolling with Slipping
Example 2: 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?