Download Lecture 8: Forces & The Laws of Motion

Lecture 17:
Torque &
Rotational Equilibrium
Questions of Yesterday
You are riding on a Ferris wheel moving at constant speed.
1a) At what point is the net force acting on you the greatest?
a) the top
b) the bottom
c) halfway between top and bottom
d) the force is the same over the whole motion
1b) Is the net force doing work on you?
a) YES
b) NO
2) If the mass of the moon were doubled, what would happen to
its centripetal acceleration?
a) it would increase
b) it would decrease
c) it would stay the same
Force & Angular Acceleration
What causes an object to accelerate?
F = ma
Can an object be accelerating if there is no net force acting on
the object?
a = Dw
Dt
aT = ra
What causes ANGULAR acceleration?
Force is needed to change the rate of rotation of an object
Is that all?
Force & Angular Acceleration
F
F
F
r
F
r
F
ANGLE of force with respect to the radial direction affects
angular acceleration
POSITION of force with respect to the axis of rotation affects
angular acceleration
Force & Angular Acceleration
F
F
F
r
F
r
F
The rate of rotation of an object can’t change unless a FORCE
is applied at a certain ANGLE and at a DISTANCE from the
axis of rotation
Torque
The rate of rotation of an object can’t change unless the object
is acted on by a net TORQUE (t)
F
r
F
t=?
FT
q
Fr
Distance from
axis of rotation
Torque
r F sinq
t = r*Fsinq
Force acting
on object
Angle between force
and displacement
from axis of rotation
The rate of rotation of an object can’t change unless a NET
FORCE is applied at a certain ANGLE and at a DISTANCE
from the axis of rotation
Torque Direction
+
F
F
r
r
F
+
-
Direction of Rotation depends on
Direction of Force Vector F
Direction of displacement vector r
pointing FROM axis of rotation TO applied force
Torque Direction
+
F
F
r
r
F
+
-
Torque is a VECTOR!!
Direction tells you direction of rotation
+ = Counterclockwise Motion
- = Clockwise Motion
Right-Hand Rule
+
F
r
r
F
F
+
-
Curl your FINGERS of your
RIGHT hand in direction of MOTION
THUMB points in direction of
TORQUE Vector
CCW (+) Motion -> Torque = OUT of PAGE
CW (-) Motion -> Torque = INTO PAGE
TORQUE Direction is PERPENDICULAR to plane formed by
FORCE and DISPLACEMENT Vectors
Right-Hand Rule
If you have a FORCE acting on an object and a specific
AXIS of rotation…
What is the direction of the TORQUE?
1) POINT the fingers of your right hand
F
r
r
F
F
in the direction of r
2) CURL your fingers in direction of F
3) THUMB points in direction of t
TORQUE Direction is PERPENDICULAR to plane formed by
FORCE and DISPLACEMENT Vectors
Torque
What is the TORQUE on the seesaw plank from each person?
What is the NET torque on the seesaw plank?
FP
mass = m
r
mass = m
r
Fgs
Fg
Fg
Does the seesaw rotate?
What is the net Force on the seesaw plank?
Torque
What is the TORQUE on the seesaw plank from each person?
What is the NET torque on the seesaw plank?
FP
mass = m
r
mass = m
r
Fgs
Fg
Fg
Force of gravity of a HOMOGENEOUS, SYMMETRIC body
acts at the center of the object
Equilibrium Conditions
An object at rest or moving at constant linear velocity and/or
angular velocity (rotation) is in equilibrium
FP
mass = m
r
mass = m
r
Fgs
Fg
Equilibrium
(a = 0, a = 0)
conditions
F = 0
t = 0
Fg
Equilibrium Conditions
Is the system in equilibrium?
FP
mass = m
r/2
mass = m
r
Fgs
Fg
Fg
Equilibrium
(a = 0, a = 0)
conditions
F = 0
t = 0
Equilibrium Conditions
Is the system in equilibrium now?
FP
mass = m
r
mass = m
r
Fgs
Fg
Fg
Equilibrium Conditions
Is the system in equilibrium now?
FP
mass = m
r
mass = m
r
Fgs rS
Fg
t = + Fgr - Fgr + Fgsrs
Fg
Equilibrium Conditions
Is the system in equilibrium now?
FP
mass = m
r1
mass = m
r2
Fgs r3
Fg
Fg
Is there a point where the pivot could be placed to make the
system in equilibrium?
t = + Fgr1 - Fgr2 + Fgsr3 = 0
Choosing your Axis of Rotation
Do the torques exerted on the object depend on the chosen axis
of rotation? What about the net torque?
FP
mass = m
r1
mass = m
r2
Fgs r3
Fg
Fg
If we chose a different axis of rotation to calculate our net
torque…would the system still be in equilibrium?
Choosing your Axis of Rotation
Do the torques exerted on the object depend on the chosen axis
of rotation? What about the net torque?
axis of
rotation
FP
mass = m
r1
r3
Fg
mass = m
r2
Fgs
Fg
If we chose a different axis of rotation to calculate our net
torque…would the system still be in equilibrium?
Choosing your Axis of Rotation
t = +FPr1 - Fg (r1 + r2) - Fgsr3 = 0
axis of
rotation
FP
mass = m
r1
r3
Fg
mass = m
r2
Fgs
Fg
The net torque of system is independent of the chosen axis of
rotation…choose an axis that makes the problem easiest!!
Choosing your Axis of Rotation
If there is an unknown force F, choose your axis to be at the
position of that unknown force (r = 0 -> t = 0)
axis of
rotation
FP
mass = m
r1
r3
Fg
mass = m
r2
Fgs
Fg
The net torque of system is independent of the chosen axis of
rotation…choose an axis that makes the problem easiest!!
Choosing your Axis of Rotation
T2
T1
40.0 cm
60o
30.0 cm
F
*
10.0 N
A uniform 10.0 N picture frame is supported as shown above.
Find the tension in the cords and the magnitude of F
Center of Mass
FP
mass = m
r
mass = m
r
Fgs
Fg
Fg
Force of gravity of a HOMOGENEOUS, SYMMETRIC body
acts at the CENTER (the axis of symmetry) of the object
Center of Mass
FP
mass = m
r
mass = m
r
Fgs
Fg
C.M.
Fg
CENTER OF MASS
The point on an extended object where a single force Fg = mg
can act to represent the force of gravity acting on the entire
extended object
Center of Mass
FP
mass = m
r
mass = m
r
Fgs
Fg
C.M.
Fg
CENTER OF MASS
The rotation induced from Fg positioned at the center of mass is
the same as that induced from all the Fg’s acting on the
extended object
Center of Mass
CENTER OF MASS of people + seesaw system
Position of axis of rotation where system is in equilibrium
FP
mass = m
r1
mass = m
r2
Fgs rS
Fg
C.M.
Fg
In 2-dimensions the center of mass is defined by an x and y
coordinate (a single point in x-y space)
Center of Mass
mass = m
mass = m
r1
r2
Fgs rS
Fg
C.M.
Fg
m1x1 + m2x2 + m3x3….
xCM =
= ∑mixi
m1 + m2 + m3….
∑mi
m1y1 + m2y2 + m3y3….
∑miyi
yCM =
= ∑m
m1 + m2 + m3….
i
Practice Problem
A 20.0-m, 500-N uniform ladder rest against a frictionless wall,
making an angle of 60.0o with the horizontal.
Find the horizontal and vertical forces exerted on the base of
the ladder by the Earth when an 1000-N firefighter is 5.0 m
from the bottom.
If the ladder is just on the verge of slipping when the firefighter
is 10.0 m up, what is the coefficient of static friction between
ladder and ground?
Practice Problem
A hungry 500-N bear walks out on a beam in an attempt to retrieve
some goodies (50 N) hanging at the end of the beam. The beam is
attached to a wall by a hinge and supported from the other end by a
cable. The beam is uniform, weighs 200 N and is 10 m long.
-Draw a free-body diagram of the beam
-When the bear is at x = 1.00 m, find the
tension in the wire and the components of the
reaction force at the hinge
-If the wire can withstand a maximum tension
of 1000 N what is the maximum distance the
bear can walk before the wire breaks?
Questions of the Day
1) If an object is rotating at a constant angular speed which
statement is true?
a) the system is in equilibrium
b) the net force on the object is ZERO
c) the net torque on the object is ZERO
d) all of the above
2) Student 1 (mass = m) sits on the left end of massless seesaw
of length L and Student 2 (mass = 2m) sits at the right end.
Where must the pivot be placed so the system is in
equilibrium?
a) L/2
b) L/3 from the right (from Student 2)
c) L/3 from the left (from Student 1)
d) the system cant be in equilibrium