review2

```SPECIAL OFFICE HOURS:
Tuesday: 11AM-1PM
Uniform circular motion is due to a centripetal acceleration
This acceleration is always pointing to the center
This acceleration is due to a net force
A diagram of gravitational force
We want to
place a
satellite into
circular orbit
300km
above the
earth
surface.
What speed,
period and
acceleration
it must
have?
What is “work” as defined in Physics?
•Formally, work is the product of a constant force F
through a parallel displacement s.
W is N.m
W = F.s
1 Joule (J) = (1N.m)
W&gt;0
W&lt;0
W=0
Work and Kinetic Energy
work-energy theorem
W = Kf - Ki
Gravitational Potential Energy (Near Earth’s
surface)
U = mgy
Uel = (1/2) kx2
• Work-Energy Theorem
Wtotal = Kf – Ki
• Conservatives force
Kf + Uf = Ki + Ui
• Non-conservative forces
Kf + Uf = Ki + Ui + Wother
Off center collisions
Suppose we have several particles A, B, etc., with masses
mA, mB, …. Let the coordinates of A be (xA, yA), let those
of B be (xB, yB), and so on. We define the center of
mass of the system as the point having coordinates
(xcm,ycm) given by
xcm = (mAxA + mBxB + ……….)/(mA + mB + ………),
Ycm = (mAyA + mByB +……….)/(mA + mB + ………).
v = rω
atan = rα
2
2
a  atan
Kinetic Energy of Rotating Rigid Body
Moment of Inertia
KA = (1/2)mAvA2
vA = rA ω
vA2 = rA2 ω2
KA = (1/2)(mArA2)ω2
KB = (1/2)(mBrB2)ω2
KC = (1/2)(mCrC2)ω2
..
K = KA + KB + KC + KD ….
K = (1/2)(mArA2)ω2 + (1/2)(mBrB2)ω2 …..
K = (1/2)[(mArA2) + (mBrB2)+ …] ω2
K = (1/2) I ω2
I = mArA2 + mBrB2 + mCrC2) + mDrD2 + …
Unit: kg.m2
• Every motion of of a rigid body can be represented
as a combination of motion of the center of mass
(translation) and rotation about an axis through the
center of mass
• The total kinetic energy can always be represented as
the sum of a part associated with motion of the center of
mass (treated as a point) plus a part asociated with
rotation about an axis through the center of mass
Total Kinetic Energy
Ktotal = (1/2)Mvcm2 + (1/2)Icmω2