Physics 151: Lecture 25 Today

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Physics 151 Lecture 25Todays Agenda

• Todays Topics
• Finish Chapter 11
• Statics (Chapter 12)

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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

3
Example Throwing ball from stool
See text Ex. 11.11

• 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
d
?F
I
I
top view before after
See example 13-9
4
Gyroscopic Motion...
See text 11.6

• 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
5
Gyroscopic Motion...
See text 11.6

• ... 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.

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Gyroscopic Motion...
See text 11.6

• 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
L
pivot
?
mg
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Gyroscopic Motion...
See text 11.6

• Consider a view looking down on the gyroscope.
• The magnitude of the change in angular momentum
  in a time dt is dL Ld?.
• So
• where?? is the precession frequency

top view
L(t)
dL
d?????????
pivot
L(tdt)
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Gyroscopic Motion...
See text 11.6

• 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
pivot
?
mg
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Lecture 24, Act 1Statics

• 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
wont precess
?
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Lecture 24, Act 1Statics

• Remember that W t/L.
• So what is t?
• t r x F
• r in this case is zero. Why?
• Thus W 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 wont precess
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Summary Comparison between Rotation and Linear
Motion
See text 10.3

• Angular Linear

x
? x / R
? v / R
v
a
? a / R
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Comparison Kinematics

• Angular Linear

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Comparison Dynamics

• Angular Linear

m
I Si mi ri2
F a m
t r x F a I
L r x p I w
p mv
W F ?x
W ? D?
?K WNET
?K WNET
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Statics (Chapter 12)
See text 12.1-3

• As the name implies, statics is the study of
  systems that dont move.
• Ladders, sign-posts, balanced beams, buildings,
  bridges, etc...
• Example What are all ofthe forces acting on a
  carparked on a hill ?

N
f
mg
?
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Car on Hill
See text 12.1-3

• Use Newtons 2nd Law FTOT MACM 0
• Resolve this into x and y components

x f - mg sin ? 0 f mg sin ?
N
y N - mg cos ? 0 N mg cos ?
f
mg
?
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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. A
• b. B
• c. C
• d. D
• e. A and

F
F
F
2F
F
F
F
F
F
F
F
F
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Statics Using Torque
See text 12.1-3

• Now consider a plank of mass M suspended by two
  strings as shown. We want to find the tension in
  each string

T1
T2
M
x cm

• This is no longer enough tosolve the problem !
• 1 equation, 2 unknowns.
• We need more information !!

L/2
L/4
Mg
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Using Torque...
See text 12.1-3

• We do have more information
• We know the plank is not rotating.
• ?TOT I? 0

T1
T2
M
x cm
L/2
L/4

• The sum of all torques is zero.
• This is true about any axiswe choose !

Mg
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Using Torque...
See text 12.1-3

• Choose the rotation axis to be along the z
  direction (out of the page) through the CM

T1
T2
M
x cm
L/2
L/4
Mg
Gravity exerts notorque about CM
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Using Torque...

• Since the sum of all torques must be 0

T1
T2
M
x cm
L/2
L/4
Mg
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Approach to Statics
See text 12.1-3

• In general, we can use the two equations
• to solve any statics problems.
• When choosing axes about which to calculate
  torque, we can be clever and make the problem
  easy....

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Lecture 25, Act 2Statics

• 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
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Lecture 25, Act 2Solution A

• The total torque about the pivot must be zero.

1kg
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Lecture 25, Act 2Solution B

• Since the system is not rotating, the
  x-coordinate of the CM must be the same as the
  pivot.

1kg
x
25
Example Problem Hanging Lamp
See text Ex.12.3

• 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 ?

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Example Hanging Lamp
See text Ex.12.3

• 1. You need to solve for the forces on the string
  and the hinge
• Use statics equations.

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Example Hanging Lamp
See text Ex.12.3

• 1. You need to solve for T and components of FH.
• Use SF 0 in x and y.
• Use St 0 in z.

T
FHy
FHx
??
m
L/2
L/2
mg
Mg
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Hanging Lamp...
See text Ex.12.3

• 3. First use the fact that in both
  x and y directions

x T cos ? - Fx 0 y T sin ? Fy - Mg - mg
0
T
Fy
??
m
Fx
L/2
L/2
M
mg
Mg
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Hanging Lamp...
See text Ex.12.3

• 3 (Cont.) So we have three equations and three
  unknowns
• T cos ? Fx 0
• T sin ? Fy - Mg - mg 0
• LMg (L/2)mg LTsinf 0
• Which we can solve to find,

T
Fy
??
m
Fx
L/2
L/2
M
mg
Mg
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Hanging Lamp...
See text Ex.12.3

• 4. Put in numbers

T
Fy
??
m
Fx
L/2
L/2
M
mg
Mg
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Hanging Lamp...
See text Ex.12.3

• 4. Have we answered the question?
• Well the string must be strong enough to exert a
  force of 50 N without breaking.
• We dont yet have the total force the hinge must
  withstand.

T
Fy
??
m
Fx
Buy a hinge that can take more than 43 N
L/2
L/2
M
mg
Mg
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Lecture 26, Act 1Statics

• 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 (b) 2 3 (c) 3 only

3
1
2
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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.

• If the box can rotate in such a way that the
  center of mass islowered, it will !

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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 islowered, it will !

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Lecture 26, Act 1 Addendum

• What are the torques ??
• (where do the forces act ?)

t goes to zero at critical point
t switches sign at critical point
t always zero
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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

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Recap of todays lecture

• Chapter 12 - Statics