Transcript Torque

Torque is a twist
or turn that tends
to produce
rotation. * * *
Applications are
found in many
common tools
around the home
or industry where
it is necessary to
turn, tighten or
loosen devices.
Definition of Torque
Torque is defined as the tendency to
produce a change in rotational motion.
Examples:
Torque is Determined by Three Factors:
• The magnitude of the applied force.
• The direction of the applied force.
• The location of the applied force.
Each
The
40-N
of the
force
20-Nthe
The
forces
nearer
forces
produces
different
the
end ofhas
theatwice
wrench
torque
torque
as
due
does
to the
the
have
greater
torques.
direction
20-N force.
of force.
Magnitude
Locationofofof
force
force
Direction
Force
20 N q
2020
N
20NN
q 20
40NN
20 N
20 N
Units for Torque
Torque is proportional to the magnitude of
F and to the distance r from the axis. Thus,
a tentative formula might be:
t = Fr Sin θ
Units: Nm
t = (40 N)(0.60 m)
= 24.0 Nm, cw
t = 24.0 Nm, cw
6 cm
40 N
Direction of Torque
Torque is a vector quantity that has
direction as well as magnitude.
Turning the handle of a
screwdriver clockwise and
then counterclockwise will
advance the screw first
inward and then outward.
Sign Convention for Torque
By convention, counterclockwise torques are
positive and clockwise torques are negative.
Positive torque:
Counter-clockwise,
out of page
cw
ccw
Negative torque:
clockwise, into page
Example 1: An 80-N force acts at the end of
a 12-cm wrench as shown. Find the torque.
• Extend line of action, draw, calculate r.
r = 12 cm sin 600
= 10.4 cm
t = (80 N)(0.104 m)
= 8.31 N m
Calculating Resultant Torque
• Read, draw, and label a rough figure.
• Draw free-body diagram showing all forces,
distances, and axis of rotation.
• Extend lines of action for each force.
• Calculate moment arms if necessary.
• Calculate torques due to EACH individual force
affixing proper sign. CCW (+) and CW (-).
• Resultant torque is sum of individual torques.
Example 2: Find resultant torque about
axis A for the arrangement shown below:
Find t due to
each force.
Consider 20-N
force first:
negative
30 N
r
300
2m
6m
40 N
20 N
300
A
4m
r = (4 m) sin 300
The torque about A is
clockwise and negative.
t = Fr = (20 N)(2 m)
t20 = -40 N m
= 2.00 m
= 40 N m, cw
Example 2 (Cont.): Next we find torque
due to 30-N force about same axis A.
Find t due to
each force.
Consider 30-N
force next.
r
negative
30 N
300
20 N
300
2m
6m
40 N
A
4m
r = (8 m) sin 300
The torque about A is
clockwise and negative.
t = Fr = (30 N)(4 m)
t30 = -120 N m
= 4.00 m
= 120 N m, cw
Example 2 (Cont.): Finally, we consider
the torque due to the 40-N force.
Find t due to
each force.
Consider 40-N
force next:
r = (2 m) sin 900
= 2.00 m
t = Fr = (40 N)(2 m)
= 80 N m, ccw
positive
30 N
r
300
2m
6m
40 N
20 N
300
A
4m
The torque about A is
CCW and positive.
t40 = +80 N m
Example 2 (Conclusion): Find resultant
torque about axis A for the arrangement
shown below:
Resultant torque
is the sum of
individual torques.
20 N
30 N
300
300
2m
6m
40 N
A
4m
tR = t20 + t20 + t20 = -40 N m -120 N m + 80 N m
tR = - 80 N m
Clockwise
Summary: Resultant Torque
• Read, draw, and label a rough figure.
• Draw free-body diagram showing all forces,
distances, and axis of rotation.
• Extend lines of action for each force.
• Calculate moment arms if necessary.
• Calculate torques due to EACH individual force
affixing proper sign. CCW (+) and CW (-).
• Resultant torque is sum of individual torques.
Newtons 2nd law and rotation
• Define and calculate the moment of inertia for
simple systems.
• Define and apply the concepts of Newton’s
second law.
Inertia of Rotation
Consider Newton’s second law for the inertia of
rotation to be patterned after the law for translation.
F = 20 N
a = 4 m/s2
F = 20 N
R = 0.5 m
a=2
Linear Inertia, m
20 N
m = 4 m/s2 = 5 kg
Rotational Inertia, I
t (20 N)(0.5 m)
2
I =a =
=
2.5
kg
m
4 m/s2
rad/s2
Force does for translation what torque does for rotation:
Rotational Inertia
Rotational Inertia is
how difficult it is to
spin an object. It
depends on the mass
of the object and how
far away the object if
from the axis of
rotation (pivot point).
v = wR
m
w
m1
axis
m
4
m3
m2
Object rotating at constant w.
Rotational Inertia Defined:
I = SmR2
Common Rotational Inertias
L
L
I
I=
1
I
2
3
mL
R
R
mR2
½mR2
Hoop
I=
Disk or cylinder
1
12
2
mL
R
I
2
5
mR
2
Solid sphere
Example 1: A circular hoop and a disk
each have a mass of 3 kg and a radius
of 20 cm. Compare their rotational
inertias.
I  mR  (3 kg)(0.2 m)
2
I = ½mR2
Disk
R
I = mR2
I = 0.120 kg m2
R
2
Hoop
I  mR  (3 kg)(0.2 m)
1
2
2
1
2
I = 0.0600 kg m2
2
Newton 2nd Law
For many problems involving rotation, there is an
analogy to be drawn from linear motion.
x
m
f
t
I
R
4 kg
A resultant force F
produces negative
acceleration a for a
mass m.
F  ma
w w  50 rad/s
o
t = 40 N m
A resultant torque t
produces angular
acceleration a of disk
with rotational inertia I.
t  Ia
Newton’s 2nd Law for Rotation
How many revolutions
required to stop?
t = Ia
FR = (½mR2)a
2F
2(40N)
a

mR (4 kg)(0.2 m)
a = 100 rad/s2
F
R
4 kg
w
wo  50 rad/s
R = 0.20 m
F = 40 N
0
2aq  wf2 - wo2
w02 (50 rad/s)2
q

2a
2(100 rad/s 2 )
q = 12.5 rad = 1.99 rev
Summary – Rotational Analogies
Quantity
Linear
Rotational
Displacement
Displacement x
Radians q
Inertia
Mass (kg)
I (kgm2)
Force
Newtons N
Torque N·m
Velocity
v
“ m/s ”
w
Rad/s
Acceleration
a
“ m/s2 ”
a
Rad/s2
CONCLUSION: Chapter 5A
Torque