Transcript Physics 131: Lecture 14 Notes

Physics 151: Lecture 25
Today’s Agenda

Today’s Topics:
Finish Chapter 11
Statics (Chapter 12)
Physics 151: Lecture 25, Pg 1
Lecture 25, Act 3

A particle whose mass is 2 kg moves in the xy
plane with a constant speed of 3 m/s along the
direction r = i + j. What is its angular momentum
(in kg · m2/s) relative to the origin?
a. 0 k
b. 6 (2)1/2 k
c. –6 (2)1/2 k
d. 6 k
e. –6 k
Physics 151: Lecture 25, Pg 2
See text: Ex. 11.11
Example: Throwing ball from stool

A student sits on a stool which is free to rotate. The
moment of inertia of the student plus the stool is I. She
throws a heavy ball of mass M with speed v such that its
velocity vector passes a distance d from the axis of
rotation.
What is the angular speed F of the student-stool
system after she throws the ball ?
M
v
F
I
top view: before
See example 13-9
d
I
after
Physics 151: Lecture 25, Pg 3
See text: 11.6
Gyroscopic Motion...


Suppose you have a spinning gyroscope in the
configuration shown below:
If the left support is removed, what will happen ?
The gyroscope does not fall down !

pivot
g
Physics 151: Lecture 25, Pg 4
See text: 11.6
Gyroscopic Motion...

... instead it precesses around its pivot axis !

This rather odd phenomenon can be easily understood
using the simple relation between torque and angular
momentum we derived in a previous lecture.

pivot
Physics 151: Lecture 25, Pg 5
See text: 11.6
Gyroscopic Motion...


The magnitude of the torque about the pivot is  = mgd.
The direction of this torque at the instant shown is out of the page
(using the right hand rule).
The change in angular momentum at the instant shown must
also be out of the page!
d
dL

dt
L

pivot
mg
Physics 151: Lecture 25, Pg 6
See text: 11.6
Gyroscopic Motion...

Consider a view looking down on the gyroscope.
The magnitude of the change in angular momentum in a time
dt is dL = Ld.
So
dL
d
L
 L
dt
dt
where  is the “precession frequency”
L(t)
dL
top view
d
pivot
L(t+dt)
Physics 151: Lecture 25, Pg 7
See text: 11.6
Gyroscopic Motion...
dL

 L
dt


L

So

In this example  = mgd and L = I:

The direction of precession is given by applying the right
hand rule to find the direction of  and hence of dL/dt.
d

L


mgd
I
pivot
mg
Physics 151: Lecture 25, Pg 8
Lecture 24, Act 1
Statics
Suppose you have a gyroscope that is supported on
a gymbals such that it is free to move in all angles,
but the rotation axes all go through the center of
mass. As pictured, looking down from the top, which
way will the gyroscope precess?
(a) clockwise (b) counterclockwise (c) it won’t precess


Physics 151: Lecture 25, Pg 9
Lecture 24, Act 1
Statics
Remember that /L.
So what is ?
=rxF
r in this case is zero. Why?
Thus  is zero.
It will not precess. At All. Even if you move the base.
This is how you make a direction finder for an airplane.

Answer (c) it won’t precess

Physics 151: Lecture 25, Pg 10
See text: 10.3
Summary:
Comparison between Rotation and Linear Motion
Angular
Linear
=x/R
x
=v/R
v
 =a/R
a
Physics 151: Lecture 25, Pg 11
Comparison
Kinematics
Angular

Linear
  constant
a  constant
  0  t
v  v 0  at
x  x 0  v 0t 
 2  02  2
v 2  v 0  2ax
1
2
1
v AVE  (v  v 0 )
2
2
 AVE  (   0 )


1 2
at
2
1
  0   0 t  t 2
2

Physics 151: Lecture 25, Pg 12
Comparison:
Dynamics
Angular
Linear
I = Si mi ri2
m
r x F =  I
F=am
L = r x p = I 
 EXT 
p = mv
dL
dt
FEXT 
W = F •Dx
W =  D

1
K  I 2
2

DK = WNET

dp
dt
K
1 2
mv
2
DK = WNET

Physics 151: Lecture 25, Pg 13
See text: 12.1-3
Statics
(Chapter 12)

As the name implies, “statics” is the study of systems that
don’t move.
Ladders, sign-posts, balanced beams, buildings,
bridges, etc...

Example: What are all of
the forces acting on a car
parked on a hill ?
y
x
N
f

mg
Physics 151: Lecture 25, Pg 14
See text: 12.1-3
Car on Hill


Use Newton’s 2nd Law: FTOT = MACM = 0
Resolve this into x and y components:
x:
F  0
f - mg sin  = 0
f = mg sin 
y:
N - mg cos  = 0
y
x
N
f
N= mg cos 

mg
Physics 151: Lecture 25, Pg 15
Example 1

The diagrams below show forces applied to an equilateral
triangular block of uniform thickness. In which diagram(s) is
the block in equilibrium?

a.
b.
c.
d.
e.




A
B
C
D
A and
F
F
F
2F
F
F
F
F
F
F
F
F
Physics 151: Lecture 25, Pg 16
See text: 12.1-3
Statics: Using Torque


Now consider a plank of mass M suspended by two strings
as shown. We want to find the tension in each string:
First use  F  0
T1
T2
T1 + T2 = Mg

This is no longer enough to
solve the problem !
1 equation, 2 unknowns.
M
x cm
L/2
L/4
y
Mg

We need more information !!
x
Physics 151: Lecture 25, Pg 17
See text: 12.1-3
Using Torque...

We do have more information:
We know the plank is not rotating.
TOT = I = 0
  0
T1
T2
M
x cm
L/2

L/4
The sum of all torques is zero.
y

This is true about any axis
we choose !
Mg
x
Physics 151: Lecture 25, Pg 18
See text: 12.1-3
Using Torque...

Choose the rotation axis to
be along the z direction (out
of the page) through the CM:

The torque due to the string T1
on the right about this axis is:
 2  T2

L
4
The torque due to the string on
the left about this axis is:
L
1  T1
2
T2
M
x cm
L/2
L/4
y
Mg
Gravity exerts no
torque about CM
x
Physics 151: Lecture 25, Pg 19
Using Torque...

Since the sum of all
torques must be 0:
L
L
T 2  T1  0
4
2
T 2  2T1
T1
T2
M
x cm
We already found that
T1 + T2 = Mg
L/2
T1 
T2 
1
Mg
3
2
Mg
3
L/4
y
Mg
x
Physics 151: Lecture 25, Pg 20
See text: 12.1-3
Approach to Statics:

In general, we can use the two equations
F  0
  0
to solve any statics problems.

When choosing axes about which to calculate torque, we
can be clever and make the problem easy....
Physics 151: Lecture 25, Pg 21
Lecture 25, Act 2
Statics

A 1kg ball is hung at the end of a rod 1m long. The system balances
at a point on the rod 0.25m from the end holding the mass.
What is the mass of the rod ?
(a) 0.5 kg
(b) 1 kg
(c) 2 kg
1m
1kg
Physics 151: Lecture 25, Pg 22
Lecture 25, Act 2
Solution A

The total torque about the pivot must be zero.

The center of mass of the rod is at its center, 0.25m to the
right of the pivot.

Since this must balance the ball, which is the same distance
to the left of the pivot, the masses must be the same !
same distance
mROD = 1kg
X
1kg
CM of rod
Physics 151: Lecture 25, Pg 23
Lecture 25, Act 2
Solution B

Since the system is not rotating, the x-coordinate of the CM must be
the same as the pivot.

The center of mass of the rod is at its center, 0.25m to the
right of the pivot.

Since the CM of the ball is 0.25m to the left of the pivot, the
mass of the rod must be 1kg to make xCM = 0.
-.25m
1kg
mROD = 1kg
.25m
X
CM of rod
x
Physics 151: Lecture 25, Pg 24
See text: Ex.12.3
Example Problem: Hanging Lamp

Your folks are making you help out on fixing up your house.
They have always been worried that the walk around back
is just too dark, so they want to hang a lamp. You go to the
hardware store and try to put together a decorative light
fixture. At the store you find a bunch of massless string
(kind of a surprising find?), a lamp of mass 2 kg, a plank of
mass 1 kg and length 2 m, and a hinge to hold the plank to
the wall. Your design is for the lamp to hang off one end of
the plank and the other to be held to a wall by a hinge. The
lamp end is supported by a massless string that makes an
angle of 30o with the plank. (The hinge supplies a force to
hold the end of the plank in place.) How strong must the
string and the hinge be for this design to work ?
Physics 151: Lecture 25, Pg 25
See text: Ex.12.3
Example: Hanging Lamp
1. You need to solve for the forces on the string and the hinge
Use statics equations.

m
L
hinge
M
Physics 151: Lecture 25, Pg 26
See text: Ex.12.3
Example: Hanging Lamp
1. You need to solve for T and components of FH.
Use SF = 0 in x and y.
Use S = 0 in z.
T
FH

FHx
y
m
L/2
Mg
L/2
mg
Physics 151: Lecture 25, Pg 27
See text: Ex.12.3
Hanging Lamp...
3. First use the fact that
x:
y:

F  0
in both x and y directions:
y
T cos  - Fx = 0
T sin  + Fy - Mg - mg = 0
x
Now use    0 in the z direction.
If we choose the rotation axis to
be through the hinge then the
hinge forces Fx and Fy will not
enter into the torque equation:
T
Fy

m
Fx
L
LMg  mg - LTsin  0
2
L/2
M
L/2
mg
Mg
Physics 151: Lecture 25, Pg 28
See text: Ex.12.3
Hanging Lamp...
3 (Cont.) So we have three equations and three unknowns:
T cos  + Fx = 0
y
T sin  + Fy - Mg - mg = 0
LMg + (L/2)mg – LTsin = 0
x
Which we can solve to find,
T =
Fx =
T
(M + m 2 ) g
sin 
M +m
(
m
)
2 g
Fx
tan 
1
F y  mg
2
Fy

L/2
M
L/2
mg
Mg
Physics 151: Lecture 25, Pg 29
See text: Ex.12.3
Hanging Lamp...
4. Put in numbers

M  m  g 2.5kg ~ 10m / s 
2 
T
 50 N
y
2
Fx 
sin 
M m

sin( 30)
x
g
2  2.5kg ~ 10m / s   43N
tan 
2
T
tan(30)
Fy

1
Fy  mg  0.5kg (~ 10m / s 2 )  5 N
2
m
Fx
L/2
M
L/2
mg
Mg
Physics 151: Lecture 25, Pg 30
See text: Ex.12.3
Hanging Lamp...
4. Have we answered the question?
Well the string must be strong enough to exert a force of 50 N
y
without breaking.
x
We don’t yet have the total force the hinge must withstand.
F  Fx2  Fy2  52  432 N  43N
T
Fy

m
Fx
Buy a hinge that can take more than 43 N
L/2
M
L/2
mg
Mg
Physics 151: Lecture 25, Pg 31
Lecture 26, Act 1
Statics

A box is placed on a ramp in the configurations
shown below. Friction prevents it from sliding. The
center of mass of the box is indicated by a white dot
in each case.
In which cases does the box tip over ?
(a) all
1
(b) 2 & 3
(c) 3 only
2
3
Physics 151: Lecture 25, Pg 32
Lecture 26, Act 1
Solution

We have seen that the torque due to gravity acts as though all
the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass is
lowered, it will !
Physics 151: Lecture 25, Pg 33
Lecture 26, ACT 1
Solution

We have seen that the torque due to gravity acts as though all
the mass of an object is concentrated at the center of mass.

Consider the bottom right corner of the box to be a pivot point.

If the box can rotate in such a way that the center of mass is
lowered, it will !
Physics 151: Lecture 25, Pg 34
Lecture 26, Act 1
Addendum

What are the torques ??
(where do the forces act ?)
 switches sign
at critical point
 goes to zero at
critical point
 always zero
N
f
rG
mg
g 
rf
f 0
rN
N 
Physics 151: Lecture 25, Pg 35
Example 3

A square of side L/2 is removed from one corner of a
square sandwich that has sides of length L. The center of
mass of the remainder of the sandwich moves from C to C’.
The distance from C to C’ is :
Physics 151: Lecture 25, Pg 36
Recap of today’s lecture

Chapter 12 - Statics
Physics 151: Lecture 25, Pg 37