Download Lecture 19: Rotational Motion in Phys141 - Topics: Kinematics, Energy, Forces - Prof. Wolf and more Study notes Physics in PDF only on Docsity!
Phys141 – Mon 10/16 – Lecture 19
- Today: Chapter 10: rotational Motion
- Administrative:
Reading for Wed: Rest of Chapter 10
Reading for Fri: Chapter 11.1-
Lab: Make-up week – you only need to come if you
missed one lab. PLEASE CHECK WITH YOUR
TA if you are not sure whether you missed a lab!!
Discussions will still take place to go over HW and
exam related problems
Chapter 10: Rigid Object rotation
In chapter 9 -> describe the center of mass motion of
large objects, or of systems of several particles
as if they were “point” particles.
BUT: Large objects, or systems of many particles
are not points, they can also ROTATE
(1) Kinematics
(2) Energy for rigid, rotating objects
(3) Forces
Reminder: Angular Position
- Fixed reference line
- P is located at ( r , θ)
- arc length s
- radius r
s = θ r
θ θ θ
θ
f i
s r
A ngular displacement ΔΘ
Δs
Angular Speed, angular acceleration
angular speed, ω
Units of ω: radians/sec or s - (NOTE: radians have no dimension)
lim t 0
d
t dt
θ θ ω (^) Δ →
lim t 0
d
t dt
angular acceleration
Units of a: rad/s² or s -
Δθ
Rotational Kinematic Equations
ω (^) f = ω i (^) +α t
θ (^) f = θ i (^) + ω i (^) t + α t
Acceleration
In terms of “real” tangential acceleration:
In addition: Centripetal acceleration
-> Total acceleration (the directions of tangential acceleration and centripetal acceleration are perpendicular)
( ) 2 2
C =^ =^ =ω
v r
a r
r r
2 2 2 2 2 4 2 4
a = at + ar = r α + r ω = r α +ω
at = r α
Moment of Inertia
of inertia:
- Dimensions: ML 2
- SI units: kg.m 2
- Calculation:
2 i i i
I = ∑ r m
lim 2 2 mi 0 i i i
I = Δ →∑ r Δ m = r dm
2 2
2 2 2
1 2 1 1 2 2
R i i i i i
R i i i
K K m r
K m r I
ω
ω ω
= =
= ⎛^ ⎞ = ⎜ ⎟ ⎝ ⎠
Rotational energy
Moment of Inertia of a Uniform Thin Hoop
hoop, all mass
elements are the
same distance from
the center
2 2
2
I r dm R dm
I MR
∫ ∫
Moment of Inertia of a Uniform Solid Cylinder
- Divide the cylinder into concentric shells with radius r , thickness dr and length L
- Then for I
DEMO
2 2
2
2 1 z 2
I r dm r Lr dr
I MR
= = πρ
=
Moment of inertia in terms of densities
Linear Mass Density
mass per unit length of a rod of uniform cross- sectional area A:
Area mass density:
Mass per unit area of a sheet of thickness L
Volumetric Mass Density mass per unit volume:
2 I = (^) ∫ ρ r dV
m
V
ρ =
m A L
m
L
A
Calculate inertia by integrating over length, area, or volume instead of mass:
2 I = (^) ∫ r dm
2 I = (^) ∫ λ r dL
2
∫
I r dA
Moment of Inertia of a Uniform Rigid Rod
- The shaded area has a mass dm = Μ/ Ldx
- Then the moment of inertia is
Note: Careful about the choice of origin. That should be the point of rotation
2 / 2 2 / 2 3 / 2 2 / 2
−
−
∫ ∫
L L L L
M
I r dm x dx L M I x ML L
R
Parallel-Axis Theorem
- Previous examples, axis of rotation ~ axis of symmetry of the object and therefore the axis of rotation went through the center of mass.
- Arbitrary axis -> integral difficult -> the parallel-axis theorem helps:
I = I CM + MD^2
- I is about any axis parallel to the axis through the center of mass of the object
- I CM is about the axis through the center of mass
- D is the distance from the center of mass axis to the arbitrary axis
