Download PHYSICS 231 INTRODUCTORY PHYSICS I Lecture 12

PHYSICS 231
INTRODUCTORY PHYSICS I
Lecture 12
Last Lecture
GMm
Fgrav =
R2
G = 6.67 "10#11 Nm 2 /kg 2
•
Newton’s Law of gravitation
•
Kepler’s Laws of Planetary motion
!
1. Ellipses with sun at focus
2. Sweep out equal areas in equal times
3
R
GM
3.
=
= Constant
2
2
T
4"
!
Gravitational Potential Energy
• PE = mgh valid only near
Earth’s surface
• For arbitrary altitude
Mm
PE = !G
r
• Zero reference level is
at r=!
Example 7.18
You wish to hurl a projectile from the surface of the
Earth (Re= 6.38x106 m) to an altitude of 20x106 m
above the surface of the Earth. Ignore rotation of the
Earth and air resistance.
a) What initial velocity is required?
a) 9,736 m/s
b) What velocity would be required in order for the
projectile to reach infinitely high? I.e., what is the
escape velocity?
b) 11,181 m/s
c) (skip) How does the escape velocity compare to the
velocity required for a low earth orbit?
c) 7,906 m/s
Chapter 8
Rotational Equilibrium
and
Rotational Dynamics
Wrench Demo
Torque
• Torque, " , is tendency of a force to
rotate object about some axis
! = Fd
• F is the force
• d is the lever arm (or moment arm)
• Units are Newton-meters
Door Demo
Torque is vector quantity
• Direction determined by axis of twist
• Perpendicular to both r and F
• Clockwise torques point into paper.
Defined as negative
• Counter-clockwise torques point out of paper.
Defined as positive
r
-
r
F
+
F
Non-perpendicular forces
! = Fr sin "
! is the angle between F and r
Torque and Equilibrium
• Forces sum to zero (no linear motion)
!Fx = 0 and !Fy = 0
• Torques sum to zero
(no rotation)
!" = 0
Axis of Rotation
• Torques require point of reference
• Point can be anywhere
• Use same point for all torques
• Pick the point to make problem easiest
(eliminate unwanted Forces from equation)
Example 8.1
Given M = 120 kg.
Neglect the mass of the beam.
a) Find the tension
in the cable
b) What is the force
between the beam and
the wall
a) T=824 N
b) f=353 N
Another Example
Given: W=50 N, L=0.35 m,
x=0.03 m
Find the tension in the muscle
W
x
L
F = 583 N
Center of Gravity
• Gravitational force acts on all points of an
extended object
• However, one can treat gravity as if it acts at
one point: the center-of-gravity.
$#m x '
i i
&
) = M tot gX
" = # (mi g)x i = M tot g&
)
M
tot
%
(
• Center of gravity:
!
!
mx
"
X=
i
M tot
i
Example 8.2
Given: x = 1.5 m, L = 5.0 m,
wbeam = 300 N,
wman = 600 N
Find: T
Fig 8.12, p.228
Slide 17
T = 413 N
x
L
Example 8.3
Consider the 400-kg
beam shown below.
Find TR
TR = 1 121 N
Example 8.4a
Tleft
Wbeam
B
A
Given:
Wbeam=300
Wbox=200
Find:
Tleft
Tright
D
C
8m
2m
Wbox
What point should I use for torque origin?
A
B
C
D
Example 8.4b
Tleft
Wbeam
B
A
Given:
Tleft=300
Tright=500
Find:
Wbeam
Tright
D
C
8m
2m
Wbox
What point should I use for torque origin?
A
B
C
D
Example 8.4c
Tleft
Wbeam
B
A
Given:
Tleft=250
Tright=400
Find:
Wbox
Tright
D
C
8m
2m
Wbox
What point should I use for torque origin?
A
B
C
D
Example 8.4d
Tleft
Wbeam
B
A
Given:
Wbeam=300
Wbox=200
Find:
Tright
Tright
D
C
8m
2m
Wbox
What point should I use for torque origin?
A
B
C
D
Example 8.4e
Tleft
Wbeam
B
A
Given:
Tleft=250
Wbeam=250
Find:
Wbox
Tright
D
C
8m
2m
Wbox
What point should I use for torque origin?
A
B
C
D
Torque and Angular Acceleration
Analogous to relation between F and a
F = ma,
! = I"
Moment of Inertia
F
m
R
" = FR = (mat )R = m(#R)R
" = mR 2#
!
Moment of Inertia
• Moment of inertia, I: rotational analog to mass
I = ! mi ri2
i
• r defined relative to rotation axis
• SI units are kg m2
Baton Demo
Moment-of-Inertia Demo
More About Moment of Inertia
• Depends on mass and its distribution.
• If mass is distributed further from axis of
rotation, moment of inertia will be larger.
Moment of Inertia of a Uniform Ring
• Divide ring into segments
• The radius of each
segment is R
I = !mi ri2 = MR 2
Example 8.6
What is the moment of inertia of the following point
masses arranged in a square?
a) about the x-axis?
b) about the y-axis?
c) about the z-axis?
a) 0.72 kg#m2
b) 1.08 kg#m2
c) 1.8 kg#m2
Other Moments of Inertia
Other Moments of Inertia
cylindrical shell : I = MR 2
1
solid cylinder : I = MR 2
2
1
rod about center : I = ML2
12
1
rod about end : I = ML2
3
2
2
spherical shell : I = MR
3
2
solidsphere : I = MR 2
5
bicycle rim
filled can of coke
baton
baseball bat
basketball
boulder
Example 8.7
Treat the spindle as a solid cylinder.
a) What is the moment of Inertia of the
spindle? (M=5.0 kg, R=0.6 m)
b) If the tension in the rope is 10 N,
what is the angular acceleration of the
wheel?
c) What is the acceleration of the
bucket?
M
d) What is the mass of the bucket?
a) 0.9 kg#m2
b) 6.67 rad/s2
c) 4 m/s2
d) 1.72 kg
Example 8.9
A 600-kg solid cylinder of radius 0.6 m which can
rotate freely about its axis is accelerated
by hanging a 240 kg mass from the end by a string
which is wrapped about the cylinder.
a) Find the linear acceleration of the mass.
4.36 m/s2
b) What is the speed of the mass after it has
dropped 2.5 m?
4.67 m/s