Download Chapter 8 Rotational Motion

Chapter 8
Rotational Motion
Rolling Motion
Consider a car moving with
a linear velocity, v.
The tangential speed of a
point on the outer edge of
the tire is equal to the speed
of the car over the ground.
v = vT = rω
Also, the tangential acceleration
of a point on the outer edge of
the tire is equal to the acceleration
of the car over the ground.
a = aT = r α
Rolling Motion
Example: An Accelerating Car
Starting from rest, the car accelerates
for 20.0 s with a constant linear
acceleration of 0.800 m/s2. The
radius of the tires is 0.330 m.
What is the angle through which
each wheel has rotated?
Rolling Motion
a 0.800 m s 2
α= =
= 2.42 rad s 2
r
0.330 m
θ
α
?
-2.42 rad/s2
θ = ωo t + α t
1
2
θ=
1
2
ω
è 
= -2.42 rad/s2
since the wheels go clockwise
ωo
t
0 rad/s
20.0 s
2
(−2.42 rad s )(20.0 s)
2
2
α
= −484 rad
The Action of Forces and Torques on Rigid Objects
In pure translational motion, all points on an
object travel on parallel paths.
The most general motion is a combination of
translation and rotation.
The Action of Forces and Torques on Rigid Objects
According to Newton’s second law, a net force causes an
object to have an acceleration.
What causes an object to have an angular acceleration?
TORQUE
Torque tells you how effective a given force is at
rotating something about some axis.
The Action of Forces and Torques on Rigid Objects
The amount of torque depends on where and in what direction the
force is applied, as well as the location of the axis of rotation.
The Action of Forces and Torques on Rigid Objects
DEFINITION OF TORQUE
Magnitude of Torque = (Magnitude of the force) x (Lever arm)
τ = F
Direction: The torque is positive when the force tends to produce a
counterclockwise rotation about the axis.
SI Unit of Torque: newton x meter (N·m)
The Action of Forces and Torques on Rigid Objects
The lever arm, l, for some force acting about some rotation
axis is defined as the distance between the rotation axis and
a perpendicular to the force.
The Action of Forces and Torques on Rigid Objects
Example: The Achilles Tendon
The tendon exerts a force of magnitude
720 N with a geometry as shown in
the figure.
Determine the torque (magnitude
and direction) of this force about the
ankle joint.
The Action of Forces and Torques on Rigid Objects
τ = F

cos 55 =
3.6 ×10 − 2 m

First, calculate the magnitude of τ
720 N
τ = ( 720 N ) (3.6 ×10 −2 m ) cos55
= 15 N ⋅ m
Since the force rotates the foot about the ankle
joint in a clockwise direction --> τ negative
τ = - 15 N m , if the direction is included
Newton’s Second Law for Rotational Motion About a Fixed Axis
Consider a powered model airplane on a massless string.
FT = maT
τ = FT r
aT = rα
m
τ = ( mr ) α
2
Moment of Inertia, I
Newton’s Second Law for Rotational Motion About a Fixed Axis
τ 1 = ( m1r12 ) α
∑τ = ∑(mr ) α
2
Net external
torque
Moment of
inertia
τ 2 = (m r ) α
2
2 2

τ N = (m r ) α
2
N N
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 "
∑τ = I α
Requirement: Angular acceleration
must be expressed in radians/s2.
I = ∑ ( mr 2 )
Newton’s Second Law for Rotational Motion About a Fixed Axis
Example: 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.
Newton’s Second Law for Rotational Motion About a Fixed Axis
(a)
2
I = ∑ ( mr ) = m r + m r = m ( 0 ) + m ( L )
2
2
11
2
2 2
m1 = m2 = m
2
I = mL
2
r1 = 0 r2 = L
Newton’s Second Law for Rotational Motion About a Fixed Axis
(b) I = ∑ mr 2 = m r 2 + m r 2 = m ( L 2 )2 + m ( L 2 )2
11
2 2
(
)
m1 = m2 = m
2
I = mL
1
2
r1 = L 2 r2 = L 2
Flat plate with axis
I = 13 Ma 2
along one edge
parallel to axis shown:
Newton’s Second Law for Rotational Motion About a Fixed Axis
Example: 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.
Newton’s Second Law for Rotational Motion About a Fixed Axis
equal
"
F
=
T
∑ y 2 − mg = ma y
a y =  2α
∑τ = T 
T2 = mg + ma y
1 1
− T2  2 = Iα
Newton’s Second Law for Rotational Motion About a Fixed Axis
T11 − ( mg + may )  2 = Iα
a y =  2α
T11 − ( mg + m 2α )  2 = Iα
T11 − mg 2
α=
I + m 22
(2150 N) (0.600 m ) − ( 451 kg) (9.80 m s2 ) (0.200 m )
=
= 6.3rad
2
2
46.0 kg ⋅ m + ( 451 kg) ( 0.200 m )
s2
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