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Rotational Motion
&
Torque
Angular Displacement

Angular displacement is a measure of
the angle a line from the center to a
point on the outside edge sweeps
through as the object rotates
 We use the greek letter “theta”  to
represent angular displacement
 Angular displacement is measured in
“radians”
Arc Length

Arc length is
represented by
the letter s
 it is the linear
distance a point
on the edge of
the object
rotates through
 measured in
meters
r
s

r
s  r
Angular Velocity
 Angular
velocity is a measure of the
rate of change of the angular position
or the “spin rate”
 We use the greek letter lowercase
“omega” () to represent angular
velocity mathematically
 Angular velocity is measured in
“radians per second” (rad/sec)
Angular Velocity


t
r


r
May also be expressed in rpm
(revolutions per minute) but must
be converted to rad/sec for calculations
rev 2 rad 1min
 rad
1



min
1rev
60 sec
30 sec
Angular Acceleration
 Angular
acceleration is the rate of
change of angular velocity
 We use the greek letter “alpha”  to
represent angular acceleration
 Angular acceleration is measured in
radians per second per second
(rad/sec2)
Angular Acceleration
r
  f  i


t
t


r
Rotational Motion
Relationships
    t
    2
   t  t
f
i
2
2
i
f
i
1
2
2
What is Torque?
 Torque
is a measure of how much a
force acting on an object causes that
object to rotate.
 Torque is dependent on force and
lever arm.
 Torque is measured in Newtonmeters (Nm)
Lever Arm
Distance measured
perpendicularly from the line
of force to the pivot point.
Measured in meters
Lever arm
1
F1
Lever arm
2
pivot
F2
Calculating Torque
  Fl
Torque = Force * lever-arm
The symbol for torque is the greek letter “tau”
pivot

Note: Force and lever arm must be
perpendicular to each other
F
Calculating Torque
  Fl  Fr sin 

pivot
r
“r” is the shortest distance from where the
force is applied to the pivot point
“” is the
angle between
r and the line
of F
F
Calculating Torque
By finding components of F
F - the perpendicular component
of the force will cause a the door F
to pivot about its hinges (torque)

pivot
r
F
F// - the parallel component of the force acts
through the pivot point, so it does not cause
the door to pivot (no torque)
F
  Fl  (F sin  )r
Net Torque &
Rotational Equilibrium

The net torque is the sum of the
individual torques.




In rotational equilibrium, the sum of the
net

torques is equal to zero. In other words,
there is no net torque on the object.
    0
net
OR
 
ccw
Note: “ccw” = counterclockwise and “cw” = clockwise
cw
EXAMPLE PROBLEM ON
TORQUE: The Swinging Door
Question
In a hurry to catch a cab, you rush through a
frictionless swinging door and onto the sidewalk. The
force you extered on the door was 50N, applied
perpendicular to the plane of the door. The door is
1.0m wide. Assuming that you pushed the door at its
edge, what was the torque on the swinging door
(taking the hinge as the pivot point)?
Hints
1. Where is the pivot point?
2. What was the force applied?
3. How far from the pivot point was the force applied?
4. What was the angle between the door and the
direction of force?
Solution The pivot point is at the hinges of the door, opposite to
where you were pushing the door. The force you used was 50N,
at a distance 1.0m from the pivot point. You hit the door
perpendicular to its plane, so the angle between the door and the
direction of force was 90 degrees. Since
= r x F = r F sin( )
then the torque on the door was:
= (1.0m) (50N) sin(90°)
= 50 N m
Rotational Kinetic Energy
Rotational energy that an object
has because it is rotating about an
axis
2
1
K rot 
I
2
“I” is rotational inertia measured in kgm2
Angular Momentum
If momentum can be described as “inertia
in motion” then Angular momentum can be
described as “inertia rotating”


L  I
Like momentum, angular momentum must be
conserved if there are no outside torques
acting on the body
I11  I 2 2