Transcript Rotational_Motion powerpoint

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 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
f
i
    2
2
2
i
f
 t 
i
1
2
 i   f 
 
t
2


t
2
Conversions and signs
•
•
For rotating objects, clockwise (cw) is
negative and counter-clockwise (ccw) is
positive. This applies to Ɵ, ɷ, and α.
Conversions:
s  r
vT  r
aT  r
180   rad
1rev  2 r  2 rad
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 and is measured in
Newton-meters (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
By finding the component of
force perpendicular to d
F - the perpendicular component
of the force
F

pivot
d
F
F// - the parallel component of the force – it
does not cause torque (lever arm = 0)
  Fl  ( F sin  )d
F
Or Calculating Torque
by finding the lever arm
  Fl  Fd sin 

pivot
d
“d” is the distance from where the force is
applied to the pivot point
“” is the
angle between
d and the line
of F
F
Net Torque

The net torque is the sum of all the
individual torques.
 net    1   2  ...

Torque that is clockwise (cw) is negative
and torque that is counter-clockwise (ccw)
is positive.
Rotational Equilibrium
•
In rotational equilibrium, the sum of all
the torques is equal to zero. In other
words, there is no net torque on the object.
• There is no angular acceleration. (  0)
• The object is either not rotating or it is
rotating at a constant speed.
• net   ccw  ( cw )  0 or
ccw
cw
 
Linear Equilibrium
•
•
•
•
In linear equilibrium, the sum of all the
forces is equal to zero. In other words,
there is no net force on the object.
There in no linear acceleration. (a = 0)
The object is either not moving linearly
or it is moving at a constant velocity.
( Fnet , x  0 and Fnet , y  0)
Fnet  0
Total Equilibrium
•
•
In total equilibrium, both net force and
net torque are equal to zero. In other
words, there is no net force or net torque
on the object.
and
Fnet  0
 net  0
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
exerted on the door was 50N, applied perpendicular to
the plane of the door. The door is 1.0 m 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
  Fl  Fd sin 
then the torque on the door was:
τ = (1.0m) (50N) sin(90°)
τ = 50 N m