PES 1110 Fall 2013, Spendier
Lecture 34/Page 1
Today:
- Conservation of Angular Momentum (11.11)
- Equilibrium (12.1-12.4)
- HW 8 due, Wednesday Nov 20th
- Quiz 5, Friday Nov 22nd (covers lectures 29-33, HW 8)
Conservation of Angular Momentum:
n 
 Li  constant
i 1
In the absence of external torques, the total angular momentum is conserved!

L  I




I Ai Ai  I BiBi  I Af  Af  I Bf Bf
Example 1:
An amusement park ride has a car, initial speed
6 m/s
and distance R = 4 m, traveling in a horizontal circular path.
If the distance R is slowly decreased (so the car always goes
around in a circle) to 1.5 m, how fast will the car be going?
Ignore friction.
PES 1110 Fall 2013, Spendier
Lecture 34/Page 2
Equilibrium
We have devoted a lot of time in understanding why and how bodies accelerate in
response to the forces that act on them. But very often we are interested in making sure
that bodies don’t accelerate. Any building, from a multistory skyscraper to a small shed,
must be designed so that it won’t topple over. Similar concerns arise with s suspension
bridge, a ladder leaning against a wall, a hanging sign, or a crane hoisting a bucket full of
concrete.
A body that can be modeled as a particle is in equilibrium whenever the vector sum of the
forces acting on it is zero. But for situations we just discussed that condition is not
enough. If forces act at different points on an extended body, an additional requirement
must be satisfied to ensure that the body has no tendency to rotate: The sum of the
torques about any point must be zero.
The Requirements for Equilibrium (12.3)
An object is said to be in equilibrium if:
1) P = constant (the linear momentum of its COM)

2) L = constant (its angular momentum about any axis)
A special case we will focus on is “static equilibrium”. Objects do not move in any way –
either
 in translation or in rotation. In this case:
1) P = 0
2) L = 0
(If you study mechanical or civil engineering you will study “statics” a lot more than we
will here.)
We discussed earlier that in equilibrium the vector sum of all forces/torques acting on the
object must be zero.
1) Linear forces:


dP
Fnet 
(Newton’s 2nd law for linear motion)
dt
since P = constant

Fnet  0 (Newton’s 2nd law equivalent for linear acceleration a = 0)
The linear forces all balance! If we are dealing with 2D motion, then we can apply this to
the x and y components of the force acting on the object

Fnet , x   Fi , x  0
i

Fnet , y   Fi , y  0
i
We did some similar problems earlier in the course but we did not use the word
“equilibrium”. Should the vector sum of all forces acting on an object be equal to zero,



dp
then
 constant and the object is said to be in static ( p  0 ) or dynamic ( p  0 )
dt
equilibrium.
PES 1110 Fall 2013, Spendier
Lecture 34/Page 3
2) Rotational Forces:


dL
t net 
(Newton’s 2nd law for rotation)
dt
since L = constant

t net  0 (Newton’s 2nd law equivalent for rotation acceleration α = 0)
The rotational torques all balance. We will only consider cases where there is one
rotation axis
 ti  0
i
Where τ can be positive or negative (clockwise or counterclockwise rotation)
Example 2:
A 16 kg Garfield sits 2.0 m from the fulcrum (support
about which a lever pivots) of a see-saw, with the feet
not touching the ground.
a) Where must his 32 kg friend Oldie sit for the
system to be in equilibrium?
b) In equilibrium, what is the upwards force provided
by the fulcrum?
PES 1110 Fall 2013, Spendier
Lecture 34/Page 4
The Center of gravity (12.4)
In most equilibrium problems, one of the forces acting on the on the body is its weight.
We need to be able to calculate the torque of this force. The weight doesn’t act at a single
point; it is distributed over the entire body.
Real objects consist of many particles (atoms). When experiencing a gravitational torque,
each individual particle experiences a torque.
Center of Gravity (“cog”) - The position at which the sum of the torques on the individual
particles equals the single torque exerted by the total weight of the object.
Each particle has weight wi and is a distance xi from the axis of rotation
ti  xi Fi sin(f)  xi wi sin(90)  xi wi
The total gravitational torque is found by adding up all the torques on each individual
piece. The center of gravity is
 ti  xcogW
i
xcog …. Location of center of gravity
1
xcog   xi wi
W i
The center of gravity is the point at which the entire force of gravity is concentrated.
Difference between position of center of mass (xcom) and center of gravity (xcog)
xcog 
1
W
1
 x w  Mg
i
i
i
xmg
i
at cog
i
i
i
The acceleration due to gravity decreases with altitude, but if we can ignore this variation
over the vertical dimension of the body, then the body’s center of gravity is identical to
the center of mass (“com”)
1
1
xcog 
xi mi g   xi mi  xcom

Mg i
M i
PES 1110 Fall 2013, Spendier
Example 3:
A 300 N uniform bar is leaning against a wall as
shown.
a) Calculate the gravitational torque (magnitude
and direction)
b) What normal force must the wall exert on the
bar at point A to keep the bar from rotating?
Ignore any friction between the bar and the wall.
Lecture 34/Page 5
PES 1110 Fall 2013, Spendier
Second way of doing it:
Lecture 34/Page 6