Transcript Rotation and Torque Lecture 09 Thursday: 12 February 2004

Rotation and Torque
Lecture 09
Thursday: 12 February 2004
ROTATION: DEFINITIONS
• Angular position:
• Angular displacement:

q
q 2 – q 1 = Dq

q 2  q1 Dq
Ave.Angular velocity:  ave 

t2  t1
Dt
 Instantaneous Angular velocity:
 dq

dt
What is the direction
of the angular velocity?
•Use your right hand
•Curl your fingers in the
direction of the rotation
•Out-stretched thumb points
in the direction of the angular
velocity
DEFINITIONS (CONTINUED)
Averageangular acceleration :




 2  1 D
 avg 

t2  t1
Dt

 d
Instantaneous angular acceleration :  
dt
Direction of Angular
Acceleration
The easiest way to get the direction of the
angular acceleration is to determine the
direction of the angular velocity and then…
• If the object is speeding up, velocity and
acceleration must be in the same direction.
• If the object is slowing down, velocity and
acceleration must be in opposite directions.
For constant 
   0  t
 x q
v
 a 
q  q 0   0t   t
1
2
 2   02  2 (q  q 0 )
q  q 0  12 ( 0   )t
q  q 0   t  12  t 2
2
Relating Linear and Angular
Variables
s qr
v  r
at   r
v
 r
2
ac 

 r
r
r
2
2 2
Three Accelerations
1. Centripetal Acceleration
2
v
(radial component of the linear acceleration) a 
c
-always non-zero in circular motion.
r
2. Tangential Acceleration
(component of linear acc. along the direction of the velocity)
-non-zero if the object is speeding up or slowing down.
3. Angular Acceleration
(rate of change in angular velocity) aT   r
-non-zero is the object is speeding up or slowing down.
Energy Considerations
Although its linear velocity v is zero, the rapidly
rotating blade of a table saw certainly has kinetic
energy due to that rotation.
How can we express the energy?
We need to treat the table saw (and any other rotating
rigid body) as a collection of particles with different
linear speeds.
KINETIC ENERGY OF
ROTATION
Ki  m v
2
i i
1
2
K   Ki 
1
2
m v
2
i i
vi  ri
K
1
2
m
i
r 
2 2
i
1
2
K  I
1
2
Where
 m r 
2
i i
2
I   mi ri
2
2
Defining Rotational Inertia
•The larger the mass, the smaller the
acceleration produced by a given force.


F  ma
•The rotational inertia I plays the equivalent role
in rotational motion as mass m in translational
motion.
•I is a measure of how hard it is to get an object
rotating. The larger I, the smaller the angular
acceleration produced by a given force.
Determining the Rotational
Inertia of an Object
I is a function of both the mass and shape of the object.
It also depends on the axis of rotation.
1. For common shapes, rotational inertias are listed in tables. A
simple version of which is in chapter 11 of your text book.
2. For collections of point masses, we can use :
i N
I   mi ri 2
i 1
where r is the distance from the axis (or point) of rotation.
3. For more complicated objects made up of objects from #1 or
#2 above, we can use the fact that rotational inertia is a scalar
and so just adds as mass would.
Comparison to Translation
• x q
• v
• a 
• mI
• K=1/2mv21/2I2
Force and Torque


  I
Torque as a Cross Product
 
  r F
  
  r F sin q

(Like F=Ma)
The direction of the Torque is always in the direction of
the angular acceleration.
• For objects in equilibrium, =0 AND F=0
Torque Corresponds to Force
• Just as Force produces translational acceleration
(causes linear motion in an object starting at rest,
for example)
• Torque produces rotational acceleration (cause a
rotational motion in an object starting from rest,
for example)
• The “cross” or “vector” product is another way to multiply
vectors. Cross product results in a vector (e.g. Torque).
Dot product (goes with cos q) results in a scalar (e.g. Work)
An Example
 
  r F Sin q

  xmg

x
W
Forces on “extended”
bodies can be viewed as
acting on a point mass
(with the same total mass)
At the object’s center of
mass (balancing point)
Determining Direction of A
CROSS PRODUCT
  
  r F
Angular Momentum of a Particle
• Angular momentum of a particle about a point of
 rotation:
 
l rp
  
l  r P Sinq
• This is similar to Torques
 
  rF
  
  r F Sinq

Find the direction of the angular
momentum vector-Right hand rule
P
r
r
P
Does an object have to be moving in a
circle to have angular momentum?
• No.
• Once we define a point (or axis) of rotation (that
 is,
a center), any object with a linear momentum p
that does not move directly through that point has
an angular momentum defined relative to the
chosen center as   
Lrp