Transcript Equilibrium is not just translational, is is also rotational. While a set

The center of gravity
of an object is the
point at which its
weight can be
considered to be
located.
xcg = (W1x1 + W2x2 + . . .)/
(W1 + W2 . . .)
For ordinary-sized objects, the
center of mass is the same as
the center of gravity. (If the
object is large enough such
that “g” is different at different
parts of the object, the center of
gravity and the center of mass
will NOT be the same.)
F = ma
FT = maT and Ƭ = FT r, so
Ƭ = maT r
Ƭ = maT r and aT = ra
Thus Ƭ = m ra r or
Ƭ=
2
mr a.
2
mr a
Ƭ=
The net external torque Ƭ is
directly proportional to the
angular acceleration a. The
constant of proportionality
2
is I = mr , the moment of
2
inertia. (Si unit kgm .)
The moment of inertia
depends on the location
and orientation of the
axis relative to the
particles that make up
the object.
There are different formulas
for moment of inertia. The
moment of inertia depends
on the shape of the object,
the distribution of the mass
in the object, and the
location of the pivot point.
Newton’s second law for a rigid
body rotating about a fixed axis.
Torque = moment of inertia X
angular acceleration
Ƭ=Ia
(a must be in rad/s2)
Example 9: The motor of a saw
brings the circular blade up to
an angular speed of 80.0 rev/s
in 240.0 revs. The blade has a
-3
moment of inertia of 1.41 x 10
kg m2. What net torque must
the motor apply to the blade?
Example 11. A crate that weighs 4420 N
is being lifted by the mechanism in
Fig 9.19a. The two cables are wrapped
around their pulleys, which have radii of
0.600 and 0.200 m. The pulleys form a
dual pulley and have a moment of inertia
of I = 50.0 kg m2. The tension in the
motor cable is maintained at 2150 N. Find
the angular acceleration of the dual
pulley and the tension in the cable
attached to the crate.
Rotational work
and energy can
also be
calculated.
Work is equal to force times displacement,
W = Fx. Angular displacement q is equal to
linear displacement/radius, x/r. So x = rq.
Thus W = Fx becomes W = Frq. Since Fr is
equal to torque Ƭ, rotational work is equal
to torque multiplied by angular
displacement.
WR = Ƭq
(q must be in radians and the work unit is the
joule J.)
2
mv .
Kinetic energy is ½
In rotation vT = rw. So
2
rotational KE = ½ m(rw)
2
2
KE = ½ mr w .
2
mr is moment of inertia, so
2
KE = ½ I w
Rotational KE = ½ I
2
w
(w = must be in rad/s and the
unit is the joule J.)
Total kinetic
energy is not just
2
½ mv , but
2
2
½ mv + ½ I w .
Example 12. A thin-walled hollow
cylinder (mass = mh, radius = rh)
and a solid cylinder (ms, rs) start
from rest at the top of an incline.
Both start at the same vertical
height h0. Ignoring friction energy
loss, which cylinder has the
greatest translational speed at the
bottom.
The rotational analog to
displacement is angular
displacement q, the rotational
analog to velocity is angular velocity
w. For acceleration it is angular
acceleration a. For work it is
rotational work Ƭq. The rotational
analog to kinetic energy is rotational
kinetic energy ½ I w2
and
the rotational analog to
momentum is angular
momentum. The formula for
momentum is p = mv. In angular
momentum m is replaced with
moment of inertia I, and velocity is
replaced with angular velocity w.
Angular momentum L
= I w.
L=Iw
w must be in rad/s and
2
I must be in kgm .
2
L is in kgm /s
If the net force on an
object is zero, the
momentum remains
constant.
Law of Conservation
of Momentum.
Similarly,
If the net torque on an
object is zero, the
angular momentum
remains constant.
Law of Conservation of
Angular Momentum.
Example 13. An ice skater is
spinning with both arms and
one leg outstretched. She then
pulls she arm and leg inward.
Using the principle of
conservation of angular
momentum, explain the change
in her spinning motion.
Example 14. A satellite in
an elliptical orbit around the
6
earth is rP = 8.37 x 10 m from
the earth at perigee, and
6
rA = 25.1 x 10 m from the earth
at apogee. The speed at perigee
is vP = 8450 m/s. What is the
speed at apogee vA?
Example 15. A square crate has a
uniformly distributed weight of
580 N. An obstruction in the floor
keeps it from sliding when pushed,
but a great enough force could
cause it to tip over. What minimum
horizontal force is needed to tip it
over and where should that force
be applied?
Example 16. Two spheres are each
rotating at an angular speed of
24 rad/s about axes that pass through
their centers. Each has a radius of
0.20 m and a mass of 1.5 kg. One is
solid and the other is a thin-walled
spherical shell. A frictional torque of
0.12 Nm is applied to the edge of each
sphere by friction. How long does it
take each sphere to stop?