Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 21
Today’s Agenda

Topics
Moments of Inertia
Torque
Ch. 10.5
Ch. 10.6, 10.7
Physics 151: Lecture 21, Pg 1
Lecture 22, ACT 1
Rotational Definitions

Your goofy friend likes to talk in physics speak. She
sees a disk spinning and says “ooh, look! There’s a
wheel with a negative w and with antiparallel w and a !!”
Which of the following is a true statement ?
(a) The wheel is spinning counter-clockwise
and slowing down.
(b) The wheel is spinning counter-clockwise
and speeding up.
?
(c) The wheel is spinning clockwise and
slowing down.
(d) The wheel is spinning clockwise and
speeding up
Physics 151: Lecture 21, Pg 2
See text: 10.1
Example:

A wheel rotates about a fixed axis with a constant
angular acceleration of 4.0 rad/s2. The diameter
of the wheel is 40 cm. What is the linear speed of
a point on the rim of this wheel at an instant when
that point has a total linear acceleration with a
magnitude of 1.2 m/s2?
a.
b.
c.
d.
e.
39 cm/s
42 cm/s
45 cm/s
35 cm/s
53 cm/s
Physics 151: Lecture 21, Pg 3
See text: 10.4
Moment of Inertia

So K 
1
I w2
2
2
where I   mi ri
i

Notice that the moment of inertia I depends on the
distribution of mass in the system.
The further the mass is from the rotation axis, the bigger
the moment of inertia.

For a given object, the moment of inertia will depend on
where we choose the rotation axis (unlike the center of
mass).

We will see that in rotational dynamics, the moment of
inertia I appears in the same way that mass m does when
we study linear dynamics !
Physics 151: Lecture 21, Pg 4
See text: 10.5
Parallel Axis Theorem


Suppose the moment of inertia of a solid object of mass M
about an axis through the center of mass is known, = ICM
The moment of inertia about an axis parallel to this axis but a
distance R away is given by:
IPARALLEL = ICM + MR2

So if we know ICM , it is easy to calculate the moment of inertia
about a parallel axis.
Physics 151: Lecture 21, Pg 5
See text: 10.5
Parallel Axis Theorem: Example

Consider a thin uniform rod of mass M and length D. Figure
out the moment of inertia about an axis through the end of
the rod.
D=L/2
M
CM
2
IPARALLEL = ICM + MD
x
L
We know ICM 
So
1
ML2
12
IEND
IEND
ICM
1
L2 1

2
 ML  M    ML2
2
12
3
which agrees with the result from the board.
Physics 151: Lecture 21, Pg 6
Direction of Rotation:


In general, the rotation variables are vectors (have a direction)
If the plane of rotation is in the x-y plane, then the convention
is
y
CCW rotation is in
the + z direction
x
z
y
CW rotation is in
the - z direction
x
z
Physics 151: Lecture 21, Pg 7
See text: 10.1
Direction of Rotation:
The Right Hand Rule
y

To figure out in which direction the rotation
vector points, curl the fingers of your right
hand the same way the object turns, and
your thumb will point in the direction of the
rotation vector !
x
z


We normally pick the z-axis to be the
rotation axis as shown.
 = z
w = wz
a = az
y
x
z
For simplicity we omit the subscripts
unless explicitly needed.
Physics 151: Lecture 21, Pg 8
See text: 10.6 and 10.7
Rotational Dynamics:
What makes it spin?



video
Suppose a force acts on a mass constrained to move in a
circle. Consider its acceleration in the ^
 direction at some
instant:
^
r
^
a = ar

F
Now use Newton’s 2nd Law in the ^

direction:
F = ma = mar
F
a
r
Multiply by r :
rF =
mr2a
m
a
Physics 151: Lecture 21, Pg 9
See text: 10.6 and 10.7
Rotational Dynamics:
What makes it spin?
rF = mr2a use
= Ia

I
= mr 2
^
r
^

Define torque:  = rF.
 is the tangential force F
times the lever arm r.
F
F
a
 = Ia
m
r

Torque has a direction:
+ z if it tries to make the system
spin CCW.
- z if it tries to make the system
spin CW.
a
Physics 151: Lecture 21, Pg 10
See text: 10.6 and 10.7
Rotational Dynamics:
What makes it spin?

So for a collection of many particles
arranged in a rigid configuration:
2
 ri Fi ,   mi ri a i
i
i
i

I
Since the particles are connected rigidly,
they all have the same a.
 i  I a
i
tot = Ia
m4
F4
F1
m3
F3
r1 m1
w
r4
r3
r2
m2
F2
Physics 151: Lecture 21, Pg 11
See text: 10.6 and 10.7
Rotational Dynamics:
What makes it spin?
TOT = Ia




This is the rotational version
of FTOT = ma
Torque is the rotational cousin of force:
 The amount of “twist” provided by a force.
Moment of inertia I is the rotational cousin of mass.
 If I is big, more torque is required to achieve a given
angular acceleration.
Torque has units of kg m2/s2 = (kg m/s2) m = Nm.
Physics 151: Lecture 21, Pg 12
See text: 10.6, 10.7, 11.2
Torque

Recall the definition of torque:

= rF

= r Fsin 

 r sin F
Fr  F

F
  r F
r = “distance of closest approach”

r
r
See Figure 10.13
Physics 151: Lecture 21, Pg 13
See text: 10.1
Example:

You throw a Frisbee of mass m and radius r so
that it is spinning about a horizontal axis
perpendicular to the plane of the Frisbee. Ignoring
air resistance, the torque exerted about its center
of mass by gravity is :
a.
b.
c.
d.
e.
0.
mgr.
2mgr.
a function of the angular velocity.
small at first, then increasing as the Frisbee
loses the torque given it by your hand.
Physics 151: Lecture 21, Pg 14
See text: 10.6 and 10.7
Torque
 = r Fsin 

So if  = 0o, then  = 0

And if  = 90o, then  = maximum
F
r
F
r
See Figure 10.13
Physics 151: Lecture 21, Pg 15
Lecture 21, Act 1
Torque

In which of the cases shown below is the torque
provided by the applied force about the rotation axis
biggest? In both cases the magnitude and direction
of the applied force is the same.
(a) case 1
L
(b) case 2
(c) same
F
F
L
axis
case 1
case 2
Physics 151: Lecture 21, Pg 16
Lecture 21, ACT 2

A uniform rod of mass M = 1.2kg and length L = 0.80 m,
lying on a frictionless horizontal plane, is free to pivot
about a vertical axis through one end, as shown. If a
force (F = 5.0 N,  = 40°) acts as shown, what is the
resulting angular acceleration about the pivot point ?
a.
b.
c.
d.
e.
16 rad/s2
12 rad/s2
14 rad/s2
10 rad/s2
33 rad/s2
Physics 151: Lecture 21, Pg 17
See text: 11.2
Torque and the
Right Hand Rule:

The right hand rule can tell you the direction of torque:
Point your hand along the direction from the axis to the
point where the force is applied.
Curl your fingers in the direction of the force.
Your thumb will point in the direction
of the torque.
F
y
=rXF
MAGNITUDE:  = r F sin 
DIRECTION:
r
x

z
Physics 151: Lecture 21, Pg 18
See text: 11.2
Torque & the Cross Product:

So we can define torque as:
 =r xF
F
= r F sin 
X = y FZ - z FY
Y = z FX - x FZ
Z = x FY - y FX


r
y
z
x
Physics 151: Lecture 21, Pg 19
See text: 10.8
How Much WORK is Done ?

Consider the work done by a force F acting on an object
constrained to move around a fixed axis. For an
infinitesimal angular displacement d:
.
dW = F dr = FRdcos()
= FRdcos(90-)
= FRdsin()
= FRsin() d
dW = d




F

R
d
dr=Rd
axis
We can integrate this to find: W = 
Analogue of W = F •r
W will be negative if  and  have opposite sign !
Physics 151: Lecture 21, Pg 20
See text: 10.8
Work & Kinetic Energy:
K = WNET

Recall the Work Kinetic-Energy Theorem:

This is true in general, and hence applies to rotational
motion as well as linear motion.

So for an object that rotates about a fixed axis:
K 


1
I w 2f  w 2i  Wnet
2
Physics 151: Lecture 21, Pg 21
See text: 10.8
Example: Disk & String

A massless string is wrapped 10 times around a
disk of mass M=40 g and radius R=10cm. The
disk is constrained to rotate without friction about
a fixed axis though its center. The string is pulled
with a force F=10N until it has unwound.
(Assume the string does not slip, and that the
disk is initially not spinning).
How fast is the disk spinning after the string
has unwound?
R
M
w = 792.5 rad/s
F
See example 10.15
Physics 151: Lecture 21, Pg 22
Lecture 21, ACT 2

Strings are wrapped around the circumference of two
solid disks and pulled with identical forces for the
same distance. Disk 1 has a bigger radius, but both
are made of identical material (i.e. their density r =
M/V is the same). Both disks rotate freely around
axes though their centers, and start at rest.
Which disk has the
biggest angular velocity
after the pull ?
(a) disk 1
w2
w1
F
F
(b) disk 2
(c) same
Physics 151: Lecture 21, Pg 23
Example 2

A rope is wrapped around the circumference of a solid
disk (R=0.2m) of mass M=10kg and an object of mass
m=10 kg is attached to the end of the rope 10m above
the ground, as shown in the figure.
a) How long will it take
for the object to hit
the ground ?
1.7 s
a) What will be the
velocity of the object
when it hits the
ground ?
11m/s
w
M
T
m
h =10 m
a) What is the tension
on the cord ?
32 N
Physics 151: Lecture 21, Pg 24
See text: 10.8
Example: Rotating Road

A uniform rod of length L=0.5m and mass m=1 kg is
free to rotate on a frictionless pin passing through one
end as in the Figure. The rod is released from rest in
the horizontal position. What is
a) angular speed when it reaches the lowest point ?
b) initial angular acceleration ?
c) initial linear acceleration of its free end ?
L
m
See example 10.14
a)
w = 7.67 rad/s
b)
a=
c)
a = 15 m/s2
30 rad/s2
Physics 151: Lecture 21, Pg 25
Lecture 22, ACT 2
A campus bird spots a member of an opposing football
team in an amusement park. The football player is on a
ride where he goes around at angular velocity w at
distance R from the center. The bird flies in a
horizontal circle above him. Will a dropping the bird
releases while flying directly above the person’s head
hit him?
a. Yes, because it falls straight down.
b. Yes, because it maintains the acceleration of the bird
as it falls.
c. No, because it falls straight down and will land behind
the person.
d. Yes, because it mainatins the angular velocity of the
bird as it falls.
e. No, because it maintains the tangential velocity the
bird had at the instant it started falling.

Physics 151: Lecture 21, Pg 26
See text: 10.1
Example:

A mass m = 4.0 kg is connected, as shown, by a
light cord to a mass M = 6.0 kg, which slides on a
smooth horizontal surface. The pulley rotates
about a frictionless axle and has a radius R =
0.12 m and a moment of inertia I = 0.090 kg m2.
The cord does not slip on the pulley. What is the
magnitude of the acceleration of m?
a.
b.
c.
d.
e.
2.4 m/s2
2.8 m/s2
3.2 m/s2
4.2 m/s2
1.7 m/s2
Physics 151: Lecture 21, Pg 27
Recap of today’s lecture

Chapter 10,
Calculating moments of inertia
Tourque
Right Hand Rule
Physics 151: Lecture 21, Pg 28