Chapter 9
Rotational Dynamics
9.3 Center of Gravity
W1 x1 + W2 x2 + L
xcg =
W1 + W2 + L
DEFINITION OF CENTER OF GRAVITY
The center of gravity of a rigid
body is the point at which
its weight can be considered
to act when the torque due
to the weight is being calculated.
9.3 Center of Gravity
Finding the center of gravity of an irregular shape.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Consider a powered model airplane on a massless string.
FT = maT
! = FT r
aT = r!
m
" = (mr )!
2
Moment of Inertia, I
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
" 1 = (m1r12 )!
#" = # (mr )!
2
Net external
torque
Moment of
inertia
" 2 = (m r )!
2
2 2
M
" N = (mN rN2 )!
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
ROTATIONAL ANALOG OF NEWTON’S SECOND LAW FOR
A RIGID BODY ROTATING ABOUT A FIXED AXIS
& Moment of
Net external torque = $$
% inertia
# & Angular
#
!! ' $$
!!
" % acceleration "
#" = I !
Requirement: Angular acceleration
must be expressed in radians/s2.
( )
I = ! mr 2
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Example 9 The Moment of Inertial Depends on Where
the Axis Is.
Two particles each have mass and are fixed at the
ends of a thin rigid rod. The length of the rod is L.
Find the moment of inertia when this object
rotates relative to an axis that is
perpendicular to the rod at
(a) one end and (b) the center.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
(a)
( )
2
2
I = ! mr 2 = m1r12 + m2 r22 = m(0 ) + m(L )
m1 = m2 = m
2
I = mL
r1 = 0 r2 = L
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
( )
(b) I = ! mr 2 = m1r12 + m2 r22 = m(L 2 )2 + m(L 2 )2
m1 = m2 = m
2
I = mL
1
2
r1 = L 2 r2 = L 2
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
I = (2/5)MR2
I = MR2
I = (1/2)MR2
I = (1/12)ML2
I = (1/3)ML2
I = (7/5)MR2
I = (2/3)MR2
I = (1/12)ML2
I = (1/3)ML2
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Example 12 Hoisting a Crate
A motor is used to lift a crate with the dual pulley system shown below.
The combined moment of inertia of the dual pulley is 46.0 kg·m2. The
crate has a mass of 451 kg. A tension of 2150 N is maintained in the cable
attached to the motor. Find the angular acceleration of the dual pulley
and the tension in the cable connected to the crate.
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
equal
"
F
=
T
# y 2 ! mg = ma y
a y = l 2!
$" = T l
T2 = mg + ma y
1 1
# T2 l 2 = I!
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
T1l 1 " (mg + ma y )l 2 = I!
a y = l 2!
T1l 1 " (mg + ml 2! )l 2 = I!
T1l 1 " mgl 2
#=
I + ml 22
(
2150 N )(0.600 m )" (451 kg )(9.80 m s 2 )(0.200 m )
=
= 6.3 rad
2
2
46.0 kg ! m + (451 kg )(0.200 m )
s2
9.4 Newton’s Second Law for Rotational Motion About a Fixed Axis
Tension in the cable connected to the crate:
T2 = mg + may = mg + ml2α
= (451 kg)(9.80 m/s2) + (451 kg)(0.200 m)(6.3 rad/s2)
= 5000 N