net,P
=
I
P
"
, the rotational second law form for a rotation of the rod around its
end pivot point P.
The first ingredient is
I
P
=
1
3
mL
2
for the rod around the axis
perpendicular to and through its end (which we already used in Ch. 9++ to
calculate the rotational kinetic energy).
The torque due to the rod’s weight is
+
mg
L
2
cos
!
because 1) we now use the convention “CW is positive,” 2) the CM
is at L/2, and 3)
F
!
=
mgcos
"
(or, if you prefer, you can use the equivalent torque
formula,
FRsin(
!
2
+
"
)
=
mg
L
2
cos
"
,
which gives the same answer).
The
rotational second law,
!
net,P
=
I
P
"
,
now reads
+
mg
L
2
cos
!
= +
1
3
mL
2
"
or, simplifying and rearranging,
!
=
3
2
g
L
cos
"
C
Inserting this into
F
P, tangential
+
mgcos
!
= +
m
"
L / 2
, we have
F
P, tangential
=
!
mgcos
"
+
m
3
2
(
g
L
cos
"
)
L / 2
,
or
F
P, tangential
=
!
1
4
mgcos
"
D
(example on next page)

Mechanics Cycle 3
Chapter 12++
12-20
P
radial is
horizontal now
tangential is
vertical now
Example:
Suppose a thin uniform rod of
mass m and length L has its frictionless
hinge pivoted on a horizontal axis.
It is
lifted up to the angle of 30
o
with respect to
the horizontal plane and is released from
rest, rotating downward due to gravity.
As a result, it swings down through
#
= 0
having some nonzero angular velocity at
that horizontal position.
Notice from our
general formula
C
obtained from the
rotational second law that
!
0
=
3
3
4
g
L
at the initial 30
o
position
and
!
=
3
2
g
L
at the horizontal position.
What are the forces due to the pivot on the rod at the horizontal position (
#
=0)
,
after falling from rest at the initial angle of 30
0
?
1) First, let’s go after
F
P, radial
where we need
!
2
.
Note from
the figure on the right, this is the
horizontal force on the rod due to
the pivot.
With
!
0
=
0
at
#
0
=
30
o
, the energy result
B
,
!
2
=
!
0
2
+
3
g
L
(sin
"
0
#
sin
"
)
, reduces
to:
!
2
=
"
3
g
L
(0
"
sin
30
o
)
= +
3
2
g
L
So
formula
A,
F
P, radial
- mgsin
!
=
"
m
#
2
L / 2
turns into
F
P, radial
!
0
=
!
m
3
2
g
L
L / 2
,
or
F
P, radial
=
!
3
4
mg
at
"
=
0
(this is only true for the fall from rest at 30
0
)
Indeed, the pivot has to pull back to the left on the rod as it swings down (to
make it go in a circle!)
This component, -
!
mg, is
horizontal and points to the
left
, like a good centripetal force should for the rod in a horizontal position.
(continuing next to the tangential force component)
!
0
= 0,
"
0
!
,
"

Mechanics Cycle 3
Chapter 12++
12-21
2) Now for
F
P, tangential
, which, at
#
= 0, is a vertical component (see the previous
figure).
Plugging into formula
D,
F
P, tangential
=
!
1
4
mgcos
"
,
we get
F
P, tangential
=
!
1
4
mg
at
"
=
0 (but this is independent of the initial conditions)
From the minus sign (down was positive), the pivot thus pushes
vertically up
on
the rod in reaction to its weight, at
#
= 0.
But notice it is NOT
!
1
2
mg
as it would
be if the rod were supported on both ends and wasn’t rotating down.
The
interesting fact is that the pivot supports less than half the weight when the
rod is swinging down!!

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